Solving $sin(x)-cos(x)=1$ The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to solve $sin x +cos x = 1$?How to prove $dfrac1+sin6^circ+cos12^circcos6^circ+sin12^circ=sqrt3$Is $sin^4 x-cos^4 x = cos2x$ or is it $-cos2x=cos2x$?How can I find $cos(theta)$ with $sin(theta)$?Determine $sintheta$ and $costheta$ to four decimal places given the point (-5, 1) in quadrants I and IIFind the general solution to $sin(4x)-cos(x)=0$Can $frac2usin(alpha-beta)gcosbeta=fracucos(alpha-beta)gsinbeta$ be reduced to $2tan(alpha-beta)tanbeta=1$?Solving $sin x - cos x = 1$$x=frac12a(1-cos(alpha))$ substition in Bernoullie's brachistochroneSolving periodic equation $cos 7theta=cos 3theta+sin 5theta$.Trigonometric Identities: Given that $frac2cos(a)+3sin(a)3cos(a)-2sin(a)=-2$ find $sin(2a)$
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Solving $sin(x)-cos(x)=1$
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to solve $sin x +cos x = 1$?How to prove $dfrac1+sin6^circ+cos12^circcos6^circ+sin12^circ=sqrt3$Is $sin^4 x-cos^4 x = cos2x$ or is it $-cos2x=cos2x$?How can I find $cos(theta)$ with $sin(theta)$?Determine $sintheta$ and $costheta$ to four decimal places given the point (-5, 1) in quadrants I and IIFind the general solution to $sin(4x)-cos(x)=0$Can $frac2usin(alpha-beta)gcosbeta=fracucos(alpha-beta)gsinbeta$ be reduced to $2tan(alpha-beta)tanbeta=1$?Solving $sin x - cos x = 1$$x=frac12a(1-cos(alpha))$ substition in Bernoullie's brachistochroneSolving periodic equation $cos 7theta=cos 3theta+sin 5theta$.Trigonometric Identities: Given that $frac2cos(a)+3sin(a)3cos(a)-2sin(a)=-2$ find $sin(2a)$
$begingroup$
Solving
$$sin(x)-cos(x)=1$$
for $x$. I used Pythagoras' Theoream and
$$Csin(x+alpha)=Asin(x)+Bcos(x)$$
where $A=1$ and $B=-1$, and I obtained
$$C=sqrt2$$
$$alpha = dfracpi4$$
and substituted where,
$$sqrt2sin(x+dfracpi4)=1$$
but somehow I think there is something wrong with my calculation, because in Wolfram it is
$$-sqrt2 sin(dfracpi4-x)=1$$
and I don't understand why do I get a different solution, I did everything correct algebraically.
trigonometry
asked Mar 24 at 19:25
D. QaD. Qa
1656
$endgroup$
add a comment |
$begingroup$
Solving
$$sin(x)-cos(x)=1$$
for $x$. I used Pythagoras' Theoream and
$$Csin(x+alpha)=Asin(x)+Bcos(x)$$
where $A=1$ and $B=-1$, and I obtained
$$C=sqrt2$$
$$alpha = dfracpi4$$
and substituted where,
$$sqrt2sin(x+dfracpi4)=1$$
but somehow I think there is something wrong with my calculation, because in Wolfram it is
$$-sqrt2 sin(dfracpi4-x)=1$$
and I don't understand why do I get a different solution, I did everything correct algebraically.
trigonometry
asked Mar 24 at 19:25
D. QaD. Qa
1656
$endgroup$
$begingroup$
There are multiple solutions. $x=pi/2+2kpi, pi+2k pi$
$endgroup$
– Mark Viola
Mar 24 at 19:30
$begingroup$
I don't think you did everything correct algebraically, because it should have been $-fracpi4$ instead of $fracpi4$.
$endgroup$
– Saucy O'Path
Mar 24 at 19:30
$begingroup$
@SaucyO'Path Is it wrong to use Pythagoras' Theoream, because I get $sin(alpha)=dfracAsqrt2=dfrac1sqrt2$ and $cos(alpha)=dfracBsqrt2= dfrac-1sqrt2$ and now doing it again in my calculator I get that $alpha= dfrac3pi4$
$endgroup$
– D. Qa
Mar 24 at 19:34
$begingroup$
I can't really follow your logic, in the sense that you are referencing your own notation in your own sheet of paper some hundreds miles away from where I am, but the formula you should keep in mind appears to be $sin(x+alpha)=cosalphasin x+sinalphacos x$.
$endgroup$
– Saucy O'Path
Mar 24 at 19:58
$begingroup$
See math.stackexchange.com/questions/618192/…
$endgroup$
– lab bhattacharjee
Mar 25 at 3:36
add a comment |
$begingroup$
Solving
$$sin(x)-cos(x)=1$$
for $x$. I used Pythagoras' Theoream and
$$Csin(x+alpha)=Asin(x)+Bcos(x)$$
where $A=1$ and $B=-1$, and I obtained
$$C=sqrt2$$
$$alpha = dfracpi4$$
and substituted where,
$$sqrt2sin(x+dfracpi4)=1$$
but somehow I think there is something wrong with my calculation, because in Wolfram it is
$$-sqrt2 sin(dfracpi4-x)=1$$
and I don't understand why do I get a different solution, I did everything correct algebraically.
trigonometry
asked Mar 24 at 19:25
D. QaD. Qa
1656
$endgroup$
Solving
$$sin(x)-cos(x)=1$$
for $x$. I used Pythagoras' Theoream and
$$Csin(x+alpha)=Asin(x)+Bcos(x)$$
where $A=1$ and $B=-1$, and I obtained
$$C=sqrt2$$
$$alpha = dfracpi4$$
and substituted where,
$$sqrt2sin(x+dfracpi4)=1$$
but somehow I think there is something wrong with my calculation, because in Wolfram it is
$$-sqrt2 sin(dfracpi4-x)=1$$
and I don't understand why do I get a different solution, I did everything correct algebraically.
trigonometry
trigonometry
asked Mar 24 at 19:25
D. QaD. Qa
1656
asked Mar 24 at 19:25
D. QaD. Qa
1656
asked Mar 24 at 19:25
D. QaD. Qa
1656
asked Mar 24 at 19:25
D. QaD. Qa
1656
asked Mar 24 at 19:25
D. QaD. Qa
1656
1656
$begingroup$
There are multiple solutions. $x=pi/2+2kpi, pi+2k pi$
$endgroup$
– Mark Viola
Mar 24 at 19:30
$begingroup$
I don't think you did everything correct algebraically, because it should have been $-fracpi4$ instead of $fracpi4$.
$endgroup$
– Saucy O'Path
Mar 24 at 19:30
$begingroup$
@SaucyO'Path Is it wrong to use Pythagoras' Theoream, because I get $sin(alpha)=dfracAsqrt2=dfrac1sqrt2$ and $cos(alpha)=dfracBsqrt2= dfrac-1sqrt2$ and now doing it again in my calculator I get that $alpha= dfrac3pi4$
$endgroup$
– D. Qa
Mar 24 at 19:34
$begingroup$
I can't really follow your logic, in the sense that you are referencing your own notation in your own sheet of paper some hundreds miles away from where I am, but the formula you should keep in mind appears to be $sin(x+alpha)=cosalphasin x+sinalphacos x$.
$endgroup$
– Saucy O'Path
Mar 24 at 19:58
$begingroup$
See math.stackexchange.com/questions/618192/…
$endgroup$
– lab bhattacharjee
Mar 25 at 3:36
add a comment |
$begingroup$
There are multiple solutions. $x=pi/2+2kpi, pi+2k pi$
$endgroup$
– Mark Viola
Mar 24 at 19:30
$begingroup$
I don't think you did everything correct algebraically, because it should have been $-fracpi4$ instead of $fracpi4$.
$endgroup$
– Saucy O'Path
Mar 24 at 19:30
$begingroup$
@SaucyO'Path Is it wrong to use Pythagoras' Theoream, because I get $sin(alpha)=dfracAsqrt2=dfrac1sqrt2$ and $cos(alpha)=dfracBsqrt2= dfrac-1sqrt2$ and now doing it again in my calculator I get that $alpha= dfrac3pi4$
$endgroup$
– D. Qa
Mar 24 at 19:34
$begingroup$
I can't really follow your logic, in the sense that you are referencing your own notation in your own sheet of paper some hundreds miles away from where I am, but the formula you should keep in mind appears to be $sin(x+alpha)=cosalphasin x+sinalphacos x$.
$endgroup$
– Saucy O'Path
Mar 24 at 19:58
$begingroup$
See math.stackexchange.com/questions/618192/…
$endgroup$
– lab bhattacharjee
Mar 25 at 3:36
$begingroup$
There are multiple solutions. $x=pi/2+2kpi, pi+2k pi$
$endgroup$
– Mark Viola
Mar 24 at 19:30
$begingroup$
There are multiple solutions. $x=pi/2+2kpi, pi+2k pi$
$endgroup$
– Mark Viola
Mar 24 at 19:30
$begingroup$
I don't think you did everything correct algebraically, because it should have been $-fracpi4$ instead of $fracpi4$.
$endgroup$
– Saucy O'Path
Mar 24 at 19:30
$begingroup$
I don't think you did everything correct algebraically, because it should have been $-fracpi4$ instead of $fracpi4$.
$endgroup$
– Saucy O'Path
Mar 24 at 19:30
$begingroup$
@SaucyO'Path Is it wrong to use Pythagoras' Theoream, because I get $sin(alpha)=dfracAsqrt2=dfrac1sqrt2$ and $cos(alpha)=dfracBsqrt2= dfrac-1sqrt2$ and now doing it again in my calculator I get that $alpha= dfrac3pi4$
$endgroup$
– D. Qa
Mar 24 at 19:34
$begingroup$
@SaucyO'Path Is it wrong to use Pythagoras' Theoream, because I get $sin(alpha)=dfracAsqrt2=dfrac1sqrt2$ and $cos(alpha)=dfracBsqrt2= dfrac-1sqrt2$ and now doing it again in my calculator I get that $alpha= dfrac3pi4$
$endgroup$
– D. Qa
Mar 24 at 19:34
$begingroup$
I can't really follow your logic, in the sense that you are referencing your own notation in your own sheet of paper some hundreds miles away from where I am, but the formula you should keep in mind appears to be $sin(x+alpha)=cosalphasin x+sinalphacos x$.
$endgroup$
– Saucy O'Path
Mar 24 at 19:58
$begingroup$
I can't really follow your logic, in the sense that you are referencing your own notation in your own sheet of paper some hundreds miles away from where I am, but the formula you should keep in mind appears to be $sin(x+alpha)=cosalphasin x+sinalphacos x$.
$endgroup$
– Saucy O'Path
Mar 24 at 19:58
$begingroup$
See math.stackexchange.com/questions/618192/…
$endgroup$
– lab bhattacharjee
Mar 25 at 3:36
$begingroup$
See math.stackexchange.com/questions/618192/…
$endgroup$
– lab bhattacharjee
Mar 25 at 3:36
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
I would rather use the substitution $$cos(x)=(pm)sqrt1-sin^2(x)$$ to obtain $$sin(x)-cos(x)=1iffldotsiff sin(x)-1=(pm)sqrt1-sin^2(x)$$
Squaring $$sin^2(x)-2sin(x)+1=1-sin^2(x)iff 2sin^2(x)-2sin(x)=0iff colorbluesin^2(x)-sin(x)=0$$
Can you end it now?
answered Mar 24 at 19:33
Dr. MathvaDr. Mathva
3,493630
$endgroup$
1
$begingroup$
Well, I set $v=sin(x)$ and then wrote $2v^2-2v=0$ and solved for $v$ where I would get $v=0,1$ and so, $sin(x)=1$ and $sin(x)=0$, I will get $x=dfracpi2+2pi n$ and $x=pi + 2pi n$
$endgroup$
– D. Qa
Mar 24 at 20:10
1
$begingroup$
Exactly @D. Qa ;)
$endgroup$
– Dr. Mathva
Mar 24 at 20:14
add a comment |
$begingroup$
You made a mistake:
$$
sin x - cos x=sqrt2sinleft(xcolorred-fracpi4right).
$$
The correctness of the last expression can be easily verified by trigonometric summation formula:
$$
sin(x+y)=sin x cos y+cos x sin y.
$$
answered Mar 24 at 19:37
useruser
6,48511031
$endgroup$
$begingroup$
I am now confused, because when using Pythagoras' Theoream, you get $sin(alpha)=dfrac1sqrt2$ and $cos(alpha)=-dfrac1sqrt2$, and so $alpha=-dfracpi4$ would be correct for sine but not for cosine because the value in cosine is negative, since $B=-1$
$endgroup$
– D. Qa
Mar 24 at 19:39
$begingroup$
Well, checking again, $cos(-dfracpi4) neq dfrac-1sqrt2$ and $sin(-dfracpi4) neq dfrac1sqrt2$
$endgroup$
– D. Qa
Mar 24 at 19:40
1
$begingroup$
@D.Qa I have added the formula for the sine of a sum. Just substitute $y=pmfracpi4$ and compare the results.
$endgroup$
– user
Mar 24 at 19:42
2
$begingroup$
@D.Qa $$beginalign sin(xcolorred+fracpi4)&=sin x cosleft(+fracpi4right)+cos x sin left(+fracpi4right)=fracsin xcolorred+cos xsqrt2\ sin(xcolorred-fracpi4)&=sin x cosleft(-fracpi4right)+cos x sin left(-fracpi4right)=fracsin xcolorred-cos xsqrt2endalign $$ Hopefully this helps.
$endgroup$
– user
Mar 24 at 20:08
$begingroup$
Yes, I was trying it on my own, and it did actually, thank you, I think now I understood what is going on. :)
$endgroup$
– D. Qa
Mar 24 at 20:11
add a comment |
$begingroup$
No, you're good.
Remember that $sin(-x)=-sin(x)$.
That means that $-sqrt2 sin(dfracpi4-x)$ is equivalent to $sqrt2 sin(x-dfracpi4)$
answered Mar 24 at 19:34
Saketh MalyalaSaketh Malyala
7,4561534
$endgroup$
$begingroup$
The OP came up with $sqrt2sinleft(x+frac]pi4right),$ not $sqrt2sinleft(x-frac]pi4right).$
$endgroup$
– Cameron Buie
Mar 24 at 20:29
add a comment |
$begingroup$
Hint: Use the so-called Weierstrass substitution:
$$sin(x)=frac2t1+t^2$$
$$cos(x)=frac1-t^21+t^2$$
answered Mar 24 at 19:39
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
$endgroup$
add a comment |
$begingroup$
One thing you could do is square both sides. This yields $$sin^2x-2sin xcos x+cos^2x=1\1-2sin xcos x=1\sin xcos x=0\sin x=0text or cos x=0.$$ This provides us with all integer multiples of $fracpi2$ as possible solutions. However, we must be a bit wary, since some of these may actually be solutions to $$sin x-cos x=-1.$$ Fortunately, periodicity means that we only have to check $0,fracpi2,pi,frac3pi2$ to see which of those works, then conclude the rest accordingly.
answered Mar 24 at 20:34
Cameron BuieCameron Buie
86.8k773161
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I would rather use the substitution $$cos(x)=(pm)sqrt1-sin^2(x)$$ to obtain $$sin(x)-cos(x)=1iffldotsiff sin(x)-1=(pm)sqrt1-sin^2(x)$$
Squaring $$sin^2(x)-2sin(x)+1=1-sin^2(x)iff 2sin^2(x)-2sin(x)=0iff colorbluesin^2(x)-sin(x)=0$$
Can you end it now?
answered Mar 24 at 19:33
Dr. MathvaDr. Mathva
3,493630
$endgroup$
1
$begingroup$
Well, I set $v=sin(x)$ and then wrote $2v^2-2v=0$ and solved for $v$ where I would get $v=0,1$ and so, $sin(x)=1$ and $sin(x)=0$, I will get $x=dfracpi2+2pi n$ and $x=pi + 2pi n$
$endgroup$
– D. Qa
Mar 24 at 20:10
1
$begingroup$
Exactly @D. Qa ;)
$endgroup$
– Dr. Mathva
Mar 24 at 20:14
add a comment |
$begingroup$
I would rather use the substitution $$cos(x)=(pm)sqrt1-sin^2(x)$$ to obtain $$sin(x)-cos(x)=1iffldotsiff sin(x)-1=(pm)sqrt1-sin^2(x)$$
Squaring $$sin^2(x)-2sin(x)+1=1-sin^2(x)iff 2sin^2(x)-2sin(x)=0iff colorbluesin^2(x)-sin(x)=0$$
Can you end it now?
answered Mar 24 at 19:33
Dr. MathvaDr. Mathva
3,493630
$endgroup$
1
$begingroup$
Well, I set $v=sin(x)$ and then wrote $2v^2-2v=0$ and solved for $v$ where I would get $v=0,1$ and so, $sin(x)=1$ and $sin(x)=0$, I will get $x=dfracpi2+2pi n$ and $x=pi + 2pi n$
$endgroup$
– D. Qa
Mar 24 at 20:10
1
$begingroup$
Exactly @D. Qa ;)
$endgroup$
– Dr. Mathva
Mar 24 at 20:14
add a comment |
$begingroup$
I would rather use the substitution $$cos(x)=(pm)sqrt1-sin^2(x)$$ to obtain $$sin(x)-cos(x)=1iffldotsiff sin(x)-1=(pm)sqrt1-sin^2(x)$$
Squaring $$sin^2(x)-2sin(x)+1=1-sin^2(x)iff 2sin^2(x)-2sin(x)=0iff colorbluesin^2(x)-sin(x)=0$$
Can you end it now?
answered Mar 24 at 19:33
Dr. MathvaDr. Mathva
3,493630
$endgroup$
I would rather use the substitution $$cos(x)=(pm)sqrt1-sin^2(x)$$ to obtain $$sin(x)-cos(x)=1iffldotsiff sin(x)-1=(pm)sqrt1-sin^2(x)$$
Squaring $$sin^2(x)-2sin(x)+1=1-sin^2(x)iff 2sin^2(x)-2sin(x)=0iff colorbluesin^2(x)-sin(x)=0$$
Can you end it now?
answered Mar 24 at 19:33
Dr. MathvaDr. Mathva
3,493630
edited Mar 24 at 22:01
answered Mar 24 at 19:33
Dr. MathvaDr. Mathva
3,493630
answered Mar 24 at 19:33
Dr. MathvaDr. Mathva
3,493630
answered Mar 24 at 19:33
Dr. MathvaDr. Mathva
3,493630
3,493630
1
$begingroup$
Well, I set $v=sin(x)$ and then wrote $2v^2-2v=0$ and solved for $v$ where I would get $v=0,1$ and so, $sin(x)=1$ and $sin(x)=0$, I will get $x=dfracpi2+2pi n$ and $x=pi + 2pi n$
$endgroup$
– D. Qa
Mar 24 at 20:10
1
$begingroup$
Exactly @D. Qa ;)
$endgroup$
– Dr. Mathva
Mar 24 at 20:14
add a comment |
1
$begingroup$
Well, I set $v=sin(x)$ and then wrote $2v^2-2v=0$ and solved for $v$ where I would get $v=0,1$ and so, $sin(x)=1$ and $sin(x)=0$, I will get $x=dfracpi2+2pi n$ and $x=pi + 2pi n$
$endgroup$
– D. Qa
Mar 24 at 20:10
1
$begingroup$
Exactly @D. Qa ;)
$endgroup$
– Dr. Mathva
Mar 24 at 20:14
1
1
$begingroup$
Well, I set $v=sin(x)$ and then wrote $2v^2-2v=0$ and solved for $v$ where I would get $v=0,1$ and so, $sin(x)=1$ and $sin(x)=0$, I will get $x=dfracpi2+2pi n$ and $x=pi + 2pi n$
$endgroup$
– D. Qa
Mar 24 at 20:10
$begingroup$
Well, I set $v=sin(x)$ and then wrote $2v^2-2v=0$ and solved for $v$ where I would get $v=0,1$ and so, $sin(x)=1$ and $sin(x)=0$, I will get $x=dfracpi2+2pi n$ and $x=pi + 2pi n$
$endgroup$
– D. Qa
Mar 24 at 20:10
1
1
$begingroup$
Exactly @D. Qa ;)
$endgroup$
– Dr. Mathva
Mar 24 at 20:14
$begingroup$
Exactly @D. Qa ;)
$endgroup$
– Dr. Mathva
Mar 24 at 20:14
add a comment |
$begingroup$
You made a mistake:
$$
sin x - cos x=sqrt2sinleft(xcolorred-fracpi4right).
$$
The correctness of the last expression can be easily verified by trigonometric summation formula:
$$
sin(x+y)=sin x cos y+cos x sin y.
$$
answered Mar 24 at 19:37
useruser
6,48511031
$endgroup$
$begingroup$
I am now confused, because when using Pythagoras' Theoream, you get $sin(alpha)=dfrac1sqrt2$ and $cos(alpha)=-dfrac1sqrt2$, and so $alpha=-dfracpi4$ would be correct for sine but not for cosine because the value in cosine is negative, since $B=-1$
$endgroup$
– D. Qa
Mar 24 at 19:39
$begingroup$
Well, checking again, $cos(-dfracpi4) neq dfrac-1sqrt2$ and $sin(-dfracpi4) neq dfrac1sqrt2$
$endgroup$
– D. Qa
Mar 24 at 19:40
1
$begingroup$
@D.Qa I have added the formula for the sine of a sum. Just substitute $y=pmfracpi4$ and compare the results.
$endgroup$
– user
Mar 24 at 19:42
2
$begingroup$
@D.Qa $$beginalign sin(xcolorred+fracpi4)&=sin x cosleft(+fracpi4right)+cos x sin left(+fracpi4right)=fracsin xcolorred+cos xsqrt2\ sin(xcolorred-fracpi4)&=sin x cosleft(-fracpi4right)+cos x sin left(-fracpi4right)=fracsin xcolorred-cos xsqrt2endalign $$ Hopefully this helps.
$endgroup$
– user
Mar 24 at 20:08
$begingroup$
Yes, I was trying it on my own, and it did actually, thank you, I think now I understood what is going on. :)
$endgroup$
– D. Qa
Mar 24 at 20:11
add a comment |
$begingroup$
You made a mistake:
$$
sin x - cos x=sqrt2sinleft(xcolorred-fracpi4right).
$$
The correctness of the last expression can be easily verified by trigonometric summation formula:
$$
sin(x+y)=sin x cos y+cos x sin y.
$$
answered Mar 24 at 19:37
useruser
6,48511031
$endgroup$
$begingroup$
I am now confused, because when using Pythagoras' Theoream, you get $sin(alpha)=dfrac1sqrt2$ and $cos(alpha)=-dfrac1sqrt2$, and so $alpha=-dfracpi4$ would be correct for sine but not for cosine because the value in cosine is negative, since $B=-1$
$endgroup$
– D. Qa
Mar 24 at 19:39
$begingroup$
Well, checking again, $cos(-dfracpi4) neq dfrac-1sqrt2$ and $sin(-dfracpi4) neq dfrac1sqrt2$
$endgroup$
– D. Qa
Mar 24 at 19:40
1
$begingroup$
@D.Qa I have added the formula for the sine of a sum. Just substitute $y=pmfracpi4$ and compare the results.
$endgroup$
– user
Mar 24 at 19:42
2
$begingroup$
@D.Qa $$beginalign sin(xcolorred+fracpi4)&=sin x cosleft(+fracpi4right)+cos x sin left(+fracpi4right)=fracsin xcolorred+cos xsqrt2\ sin(xcolorred-fracpi4)&=sin x cosleft(-fracpi4right)+cos x sin left(-fracpi4right)=fracsin xcolorred-cos xsqrt2endalign $$ Hopefully this helps.
$endgroup$
– user
Mar 24 at 20:08
$begingroup$
Yes, I was trying it on my own, and it did actually, thank you, I think now I understood what is going on. :)
$endgroup$
– D. Qa
Mar 24 at 20:11
add a comment |
$begingroup$
You made a mistake:
$$
sin x - cos x=sqrt2sinleft(xcolorred-fracpi4right).
$$
The correctness of the last expression can be easily verified by trigonometric summation formula:
$$
sin(x+y)=sin x cos y+cos x sin y.
$$
answered Mar 24 at 19:37
useruser
6,48511031
$endgroup$
You made a mistake:
$$
sin x - cos x=sqrt2sinleft(xcolorred-fracpi4right).
$$
The correctness of the last expression can be easily verified by trigonometric summation formula:
$$
sin(x+y)=sin x cos y+cos x sin y.
$$
answered Mar 24 at 19:37
useruser
6,48511031
edited Mar 24 at 19:40
answered Mar 24 at 19:37
useruser
6,48511031
answered Mar 24 at 19:37
useruser
6,48511031
answered Mar 24 at 19:37
useruser
6,48511031
6,48511031
$begingroup$
I am now confused, because when using Pythagoras' Theoream, you get $sin(alpha)=dfrac1sqrt2$ and $cos(alpha)=-dfrac1sqrt2$, and so $alpha=-dfracpi4$ would be correct for sine but not for cosine because the value in cosine is negative, since $B=-1$
$endgroup$
– D. Qa
Mar 24 at 19:39
$begingroup$
Well, checking again, $cos(-dfracpi4) neq dfrac-1sqrt2$ and $sin(-dfracpi4) neq dfrac1sqrt2$
$endgroup$
– D. Qa
Mar 24 at 19:40
1
$begingroup$
@D.Qa I have added the formula for the sine of a sum. Just substitute $y=pmfracpi4$ and compare the results.
$endgroup$
– user
Mar 24 at 19:42
2
$begingroup$
@D.Qa $$beginalign sin(xcolorred+fracpi4)&=sin x cosleft(+fracpi4right)+cos x sin left(+fracpi4right)=fracsin xcolorred+cos xsqrt2\ sin(xcolorred-fracpi4)&=sin x cosleft(-fracpi4right)+cos x sin left(-fracpi4right)=fracsin xcolorred-cos xsqrt2endalign $$ Hopefully this helps.
$endgroup$
– user
Mar 24 at 20:08
$begingroup$
Yes, I was trying it on my own, and it did actually, thank you, I think now I understood what is going on. :)
$endgroup$
– D. Qa
Mar 24 at 20:11
add a comment |
$begingroup$
I am now confused, because when using Pythagoras' Theoream, you get $sin(alpha)=dfrac1sqrt2$ and $cos(alpha)=-dfrac1sqrt2$, and so $alpha=-dfracpi4$ would be correct for sine but not for cosine because the value in cosine is negative, since $B=-1$
$endgroup$
– D. Qa
Mar 24 at 19:39
$begingroup$
Well, checking again, $cos(-dfracpi4) neq dfrac-1sqrt2$ and $sin(-dfracpi4) neq dfrac1sqrt2$
$endgroup$
– D. Qa
Mar 24 at 19:40
1
$begingroup$
@D.Qa I have added the formula for the sine of a sum. Just substitute $y=pmfracpi4$ and compare the results.
$endgroup$
– user
Mar 24 at 19:42
2
$begingroup$
@D.Qa $$beginalign sin(xcolorred+fracpi4)&=sin x cosleft(+fracpi4right)+cos x sin left(+fracpi4right)=fracsin xcolorred+cos xsqrt2\ sin(xcolorred-fracpi4)&=sin x cosleft(-fracpi4right)+cos x sin left(-fracpi4right)=fracsin xcolorred-cos xsqrt2endalign $$ Hopefully this helps.
$endgroup$
– user
Mar 24 at 20:08
$begingroup$
Yes, I was trying it on my own, and it did actually, thank you, I think now I understood what is going on. :)
$endgroup$
– D. Qa
Mar 24 at 20:11
$begingroup$
I am now confused, because when using Pythagoras' Theoream, you get $sin(alpha)=dfrac1sqrt2$ and $cos(alpha)=-dfrac1sqrt2$, and so $alpha=-dfracpi4$ would be correct for sine but not for cosine because the value in cosine is negative, since $B=-1$
$endgroup$
– D. Qa
Mar 24 at 19:39
$begingroup$
I am now confused, because when using Pythagoras' Theoream, you get $sin(alpha)=dfrac1sqrt2$ and $cos(alpha)=-dfrac1sqrt2$, and so $alpha=-dfracpi4$ would be correct for sine but not for cosine because the value in cosine is negative, since $B=-1$
$endgroup$
– D. Qa
Mar 24 at 19:39
$begingroup$
Well, checking again, $cos(-dfracpi4) neq dfrac-1sqrt2$ and $sin(-dfracpi4) neq dfrac1sqrt2$
$endgroup$
– D. Qa
Mar 24 at 19:40
$begingroup$
Well, checking again, $cos(-dfracpi4) neq dfrac-1sqrt2$ and $sin(-dfracpi4) neq dfrac1sqrt2$
$endgroup$
– D. Qa
Mar 24 at 19:40
1
1
$begingroup$
@D.Qa I have added the formula for the sine of a sum. Just substitute $y=pmfracpi4$ and compare the results.
$endgroup$
– user
Mar 24 at 19:42
$begingroup$
@D.Qa I have added the formula for the sine of a sum. Just substitute $y=pmfracpi4$ and compare the results.
$endgroup$
– user
Mar 24 at 19:42
2
2
$begingroup$
@D.Qa $$beginalign sin(xcolorred+fracpi4)&=sin x cosleft(+fracpi4right)+cos x sin left(+fracpi4right)=fracsin xcolorred+cos xsqrt2\ sin(xcolorred-fracpi4)&=sin x cosleft(-fracpi4right)+cos x sin left(-fracpi4right)=fracsin xcolorred-cos xsqrt2endalign $$ Hopefully this helps.
$endgroup$
– user
Mar 24 at 20:08
$begingroup$
@D.Qa $$beginalign sin(xcolorred+fracpi4)&=sin x cosleft(+fracpi4right)+cos x sin left(+fracpi4right)=fracsin xcolorred+cos xsqrt2\ sin(xcolorred-fracpi4)&=sin x cosleft(-fracpi4right)+cos x sin left(-fracpi4right)=fracsin xcolorred-cos xsqrt2endalign $$ Hopefully this helps.
$endgroup$
– user
Mar 24 at 20:08
$begingroup$
Yes, I was trying it on my own, and it did actually, thank you, I think now I understood what is going on. :)
$endgroup$
– D. Qa
Mar 24 at 20:11
$begingroup$
Yes, I was trying it on my own, and it did actually, thank you, I think now I understood what is going on. :)
$endgroup$
– D. Qa
Mar 24 at 20:11
add a comment |
$begingroup$
No, you're good.
Remember that $sin(-x)=-sin(x)$.
That means that $-sqrt2 sin(dfracpi4-x)$ is equivalent to $sqrt2 sin(x-dfracpi4)$
answered Mar 24 at 19:34
Saketh MalyalaSaketh Malyala
7,4561534
$endgroup$
$begingroup$
The OP came up with $sqrt2sinleft(x+frac]pi4right),$ not $sqrt2sinleft(x-frac]pi4right).$
$endgroup$
– Cameron Buie
Mar 24 at 20:29
add a comment |
$begingroup$
No, you're good.
Remember that $sin(-x)=-sin(x)$.
That means that $-sqrt2 sin(dfracpi4-x)$ is equivalent to $sqrt2 sin(x-dfracpi4)$
answered Mar 24 at 19:34
Saketh MalyalaSaketh Malyala
7,4561534
$endgroup$
$begingroup$
The OP came up with $sqrt2sinleft(x+frac]pi4right),$ not $sqrt2sinleft(x-frac]pi4right).$
$endgroup$
– Cameron Buie
Mar 24 at 20:29
add a comment |
$begingroup$
No, you're good.
Remember that $sin(-x)=-sin(x)$.
That means that $-sqrt2 sin(dfracpi4-x)$ is equivalent to $sqrt2 sin(x-dfracpi4)$
answered Mar 24 at 19:34
Saketh MalyalaSaketh Malyala
7,4561534
$endgroup$
No, you're good.
Remember that $sin(-x)=-sin(x)$.
That means that $-sqrt2 sin(dfracpi4-x)$ is equivalent to $sqrt2 sin(x-dfracpi4)$
answered Mar 24 at 19:34
Saketh MalyalaSaketh Malyala
7,4561534
answered Mar 24 at 19:34
Saketh MalyalaSaketh Malyala
7,4561534
answered Mar 24 at 19:34
Saketh MalyalaSaketh Malyala
7,4561534
answered Mar 24 at 19:34
Saketh MalyalaSaketh Malyala
7,4561534
7,4561534
$begingroup$
The OP came up with $sqrt2sinleft(x+frac]pi4right),$ not $sqrt2sinleft(x-frac]pi4right).$
$endgroup$
– Cameron Buie
Mar 24 at 20:29
add a comment |
$begingroup$
The OP came up with $sqrt2sinleft(x+frac]pi4right),$ not $sqrt2sinleft(x-frac]pi4right).$
$endgroup$
– Cameron Buie
Mar 24 at 20:29
$begingroup$
The OP came up with $sqrt2sinleft(x+frac]pi4right),$ not $sqrt2sinleft(x-frac]pi4right).$
$endgroup$
– Cameron Buie
Mar 24 at 20:29
$begingroup$
The OP came up with $sqrt2sinleft(x+frac]pi4right),$ not $sqrt2sinleft(x-frac]pi4right).$
$endgroup$
– Cameron Buie
Mar 24 at 20:29
add a comment |
$begingroup$
Hint: Use the so-called Weierstrass substitution:
$$sin(x)=frac2t1+t^2$$
$$cos(x)=frac1-t^21+t^2$$
answered Mar 24 at 19:39
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
$endgroup$
add a comment |
$begingroup$
Hint: Use the so-called Weierstrass substitution:
$$sin(x)=frac2t1+t^2$$
$$cos(x)=frac1-t^21+t^2$$
answered Mar 24 at 19:39
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
$endgroup$
add a comment |
$begingroup$
Hint: Use the so-called Weierstrass substitution:
$$sin(x)=frac2t1+t^2$$
$$cos(x)=frac1-t^21+t^2$$
answered Mar 24 at 19:39
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
$endgroup$
Hint: Use the so-called Weierstrass substitution:
$$sin(x)=frac2t1+t^2$$
$$cos(x)=frac1-t^21+t^2$$
answered Mar 24 at 19:39
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
answered Mar 24 at 19:39
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
answered Mar 24 at 19:39
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
answered Mar 24 at 19:39
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
79k42867
add a comment |
add a comment |
$begingroup$
One thing you could do is square both sides. This yields $$sin^2x-2sin xcos x+cos^2x=1\1-2sin xcos x=1\sin xcos x=0\sin x=0text or cos x=0.$$ This provides us with all integer multiples of $fracpi2$ as possible solutions. However, we must be a bit wary, since some of these may actually be solutions to $$sin x-cos x=-1.$$ Fortunately, periodicity means that we only have to check $0,fracpi2,pi,frac3pi2$ to see which of those works, then conclude the rest accordingly.
answered Mar 24 at 20:34
Cameron BuieCameron Buie
86.8k773161
$endgroup$
add a comment |
$begingroup$
One thing you could do is square both sides. This yields $$sin^2x-2sin xcos x+cos^2x=1\1-2sin xcos x=1\sin xcos x=0\sin x=0text or cos x=0.$$ This provides us with all integer multiples of $fracpi2$ as possible solutions. However, we must be a bit wary, since some of these may actually be solutions to $$sin x-cos x=-1.$$ Fortunately, periodicity means that we only have to check $0,fracpi2,pi,frac3pi2$ to see which of those works, then conclude the rest accordingly.
answered Mar 24 at 20:34
Cameron BuieCameron Buie
86.8k773161
$endgroup$
add a comment |
$begingroup$
One thing you could do is square both sides. This yields $$sin^2x-2sin xcos x+cos^2x=1\1-2sin xcos x=1\sin xcos x=0\sin x=0text or cos x=0.$$ This provides us with all integer multiples of $fracpi2$ as possible solutions. However, we must be a bit wary, since some of these may actually be solutions to $$sin x-cos x=-1.$$ Fortunately, periodicity means that we only have to check $0,fracpi2,pi,frac3pi2$ to see which of those works, then conclude the rest accordingly.
answered Mar 24 at 20:34
Cameron BuieCameron Buie
86.8k773161
$endgroup$
One thing you could do is square both sides. This yields $$sin^2x-2sin xcos x+cos^2x=1\1-2sin xcos x=1\sin xcos x=0\sin x=0text or cos x=0.$$ This provides us with all integer multiples of $fracpi2$ as possible solutions. However, we must be a bit wary, since some of these may actually be solutions to $$sin x-cos x=-1.$$ Fortunately, periodicity means that we only have to check $0,fracpi2,pi,frac3pi2$ to see which of those works, then conclude the rest accordingly.
answered Mar 24 at 20:34
Cameron BuieCameron Buie
86.8k773161
answered Mar 24 at 20:34
Cameron BuieCameron Buie
86.8k773161
answered Mar 24 at 20:34
Cameron BuieCameron Buie
86.8k773161
answered Mar 24 at 20:34
Cameron BuieCameron Buie
86.8k773161
86.8k773161
add a comment |
add a comment |
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$begingroup$
There are multiple solutions. $x=pi/2+2kpi, pi+2k pi$
$endgroup$
– Mark Viola
Mar 24 at 19:30
$begingroup$
I don't think you did everything correct algebraically, because it should have been $-fracpi4$ instead of $fracpi4$.
$endgroup$
– Saucy O'Path
Mar 24 at 19:30
$begingroup$
@SaucyO'Path Is it wrong to use Pythagoras' Theoream, because I get $sin(alpha)=dfracAsqrt2=dfrac1sqrt2$ and $cos(alpha)=dfracBsqrt2= dfrac-1sqrt2$ and now doing it again in my calculator I get that $alpha= dfrac3pi4$
$endgroup$
– D. Qa
Mar 24 at 19:34
$begingroup$
I can't really follow your logic, in the sense that you are referencing your own notation in your own sheet of paper some hundreds miles away from where I am, but the formula you should keep in mind appears to be $sin(x+alpha)=cosalphasin x+sinalphacos x$.
$endgroup$
– Saucy O'Path
Mar 24 at 19:58
$begingroup$
See math.stackexchange.com/questions/618192/…
$endgroup$
– lab bhattacharjee
Mar 25 at 3:36