All the roots of $5cos x - sin x = 4$ in the interval $0^circ leq x leq 360^circ$?Solving $displaystyle cos(x-alpha)cos(x-beta) = cosalphacosbeta+sin^2x$Expressing $costheta - sqrt3sintheta = rsin(theta - alpha)$Solving $2cos2x-4sin xcos x=sqrt6$Why does $acdot cos(kx)+bcdotsin(kx)=sqrta^2+b^2sin(kx+phi)$ hold?$2sin(theta + 17) = dfrac cos (theta +8)cos (theta + 17)$To Substitute or Not … $2sin 2x = 3tan x$, in interval $0le x<360^circ$Given Triangle $a,b,c$ and angles $theta,alpha,beta$.Find $fraccos(alpha)sin(theta)$Solve: $cos(theta + 25^circ) + sin(theta +65^circ) = 1$Solve $2cos x^circ + 3sin x^circ = -1$ where $0 le x le 360$Solve the trigonometric series equation $1 + sin x + sin^2 x + sin^3 x + cdots = 4 + 2sqrt 3$
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All the roots of $5cos x - sin x = 4$ in the interval $0^circ leq x leq 360^circ$?
Solving $displaystyle cos(x-alpha)cos(x-beta) = cosalphacosbeta+sin^2x$Expressing $costheta - sqrt3sintheta = rsin(theta - alpha)$Solving $2cos2x-4sin xcos x=sqrt6$Why does $acdot cos(kx)+bcdotsin(kx)=sqrta^2+b^2sin(kx+phi)$ hold?$2sin(theta + 17) = dfrac cos (theta +8)cos (theta + 17)$To Substitute or Not … $2sin 2x = 3tan x$, in interval $0le x<360^circ$Given Triangle $a,b,c$ and angles $theta,alpha,beta$.Find $fraccos(alpha)sin(theta)$Solve: $cos(theta + 25^circ) + sin(theta +65^circ) = 1$Solve $2cos x^circ + 3sin x^circ = -1$ where $0 le x le 360$Solve the trigonometric series equation $1 + sin x + sin^2 x + sin^3 x + cdots = 4 + 2sqrt 3$
$begingroup$
This is a problem that I stumbled upon in one of my books.
Representing $5cos x - sin x$ in the form $Rcos(x + alpha)$ (as demanded by the question):
$
rightarrow R = sqrt5^2 + (-1)^2 = sqrt26\
rightarrow Rcos x cos alpha - Rsin x sin alpha = 5cos x - sin x \
rightarrow ➊hspace0.25cm5 = sqrt26cos alpha \
rightarrow ➋hspace0.25cm-1 = sqrt26sin alpha \
$
So here are my question(s): $\$
• Why is only ➊ working out?
• When I find one solution in the interval, which is 27°, why is another solution like $(360 - x)$ not working out (in this case 333°)?
• I see 310.4° as a solution to this equation in my book. How do you get to 310.4° from 27° (which is the solution I got)? Which $cos$ identity am I missing? And surprisingly, my book doesn't include 27° as a solution!
trigonometry
asked Mar 15 at 5:23
RamanaRamana
16510
$endgroup$
|
show 2 more comments
$begingroup$
This is a problem that I stumbled upon in one of my books.
Representing $5cos x - sin x$ in the form $Rcos(x + alpha)$ (as demanded by the question):
$
rightarrow R = sqrt5^2 + (-1)^2 = sqrt26\
rightarrow Rcos x cos alpha - Rsin x sin alpha = 5cos x - sin x \
rightarrow ➊hspace0.25cm5 = sqrt26cos alpha \
rightarrow ➋hspace0.25cm-1 = sqrt26sin alpha \
$
So here are my question(s): $\$
• Why is only ➊ working out?
• When I find one solution in the interval, which is 27°, why is another solution like $(360 - x)$ not working out (in this case 333°)?
• I see 310.4° as a solution to this equation in my book. How do you get to 310.4° from 27° (which is the solution I got)? Which $cos$ identity am I missing? And surprisingly, my book doesn't include 27° as a solution!
trigonometry
asked Mar 15 at 5:23
RamanaRamana
16510
$endgroup$
$begingroup$
You are missing an $R$ after the $-$ in the second arrow. The $alpha$ is chosen to satisfy both (1) and (2).
$endgroup$
– user647486
Mar 15 at 5:31
$begingroup$
Edited it just now
$endgroup$
– Ramana
Mar 15 at 5:37
$begingroup$
@user647486 I tried substituting the value derived from (2) into the given equation and it gave 4.9, which is wrong.
$endgroup$
– Ramana
Mar 15 at 5:39
$begingroup$
you are probably correct. Your book is wrong
$endgroup$
– rash
Mar 15 at 5:39
$begingroup$
You are correct. $27.019^circ$ is also a correct answer.
$endgroup$
– John Wayland Bales
Mar 15 at 5:47
|
show 2 more comments
$begingroup$
This is a problem that I stumbled upon in one of my books.
Representing $5cos x - sin x$ in the form $Rcos(x + alpha)$ (as demanded by the question):
$
rightarrow R = sqrt5^2 + (-1)^2 = sqrt26\
rightarrow Rcos x cos alpha - Rsin x sin alpha = 5cos x - sin x \
rightarrow ➊hspace0.25cm5 = sqrt26cos alpha \
rightarrow ➋hspace0.25cm-1 = sqrt26sin alpha \
$
So here are my question(s): $\$
• Why is only ➊ working out?
• When I find one solution in the interval, which is 27°, why is another solution like $(360 - x)$ not working out (in this case 333°)?
• I see 310.4° as a solution to this equation in my book. How do you get to 310.4° from 27° (which is the solution I got)? Which $cos$ identity am I missing? And surprisingly, my book doesn't include 27° as a solution!
trigonometry
asked Mar 15 at 5:23
RamanaRamana
16510
$endgroup$
This is a problem that I stumbled upon in one of my books.
Representing $5cos x - sin x$ in the form $Rcos(x + alpha)$ (as demanded by the question):
$
rightarrow R = sqrt5^2 + (-1)^2 = sqrt26\
rightarrow Rcos x cos alpha - Rsin x sin alpha = 5cos x - sin x \
rightarrow ➊hspace0.25cm5 = sqrt26cos alpha \
rightarrow ➋hspace0.25cm-1 = sqrt26sin alpha \
$
So here are my question(s): $\$
• Why is only ➊ working out?
• When I find one solution in the interval, which is 27°, why is another solution like $(360 - x)$ not working out (in this case 333°)?
• I see 310.4° as a solution to this equation in my book. How do you get to 310.4° from 27° (which is the solution I got)? Which $cos$ identity am I missing? And surprisingly, my book doesn't include 27° as a solution!
trigonometry
trigonometry
asked Mar 15 at 5:23
RamanaRamana
16510
asked Mar 15 at 5:23
RamanaRamana
16510
edited Mar 15 at 5:37
Ramana
asked Mar 15 at 5:23
RamanaRamana
16510
asked Mar 15 at 5:23
RamanaRamana
16510
asked Mar 15 at 5:23
RamanaRamana
16510
16510
$begingroup$
You are missing an $R$ after the $-$ in the second arrow. The $alpha$ is chosen to satisfy both (1) and (2).
$endgroup$
– user647486
Mar 15 at 5:31
$begingroup$
Edited it just now
$endgroup$
– Ramana
Mar 15 at 5:37
$begingroup$
@user647486 I tried substituting the value derived from (2) into the given equation and it gave 4.9, which is wrong.
$endgroup$
– Ramana
Mar 15 at 5:39
$begingroup$
you are probably correct. Your book is wrong
$endgroup$
– rash
Mar 15 at 5:39
$begingroup$
You are correct. $27.019^circ$ is also a correct answer.
$endgroup$
– John Wayland Bales
Mar 15 at 5:47
|
show 2 more comments
$begingroup$
You are missing an $R$ after the $-$ in the second arrow. The $alpha$ is chosen to satisfy both (1) and (2).
$endgroup$
– user647486
Mar 15 at 5:31
$begingroup$
Edited it just now
$endgroup$
– Ramana
Mar 15 at 5:37
$begingroup$
@user647486 I tried substituting the value derived from (2) into the given equation and it gave 4.9, which is wrong.
$endgroup$
– Ramana
Mar 15 at 5:39
$begingroup$
you are probably correct. Your book is wrong
$endgroup$
– rash
Mar 15 at 5:39
$begingroup$
You are correct. $27.019^circ$ is also a correct answer.
$endgroup$
– John Wayland Bales
Mar 15 at 5:47
$begingroup$
You are missing an $R$ after the $-$ in the second arrow. The $alpha$ is chosen to satisfy both (1) and (2).
$endgroup$
– user647486
Mar 15 at 5:31
$begingroup$
You are missing an $R$ after the $-$ in the second arrow. The $alpha$ is chosen to satisfy both (1) and (2).
$endgroup$
– user647486
Mar 15 at 5:31
$begingroup$
Edited it just now
$endgroup$
– Ramana
Mar 15 at 5:37
$begingroup$
Edited it just now
$endgroup$
– Ramana
Mar 15 at 5:37
$begingroup$
@user647486 I tried substituting the value derived from (2) into the given equation and it gave 4.9, which is wrong.
$endgroup$
– Ramana
Mar 15 at 5:39
$begingroup$
@user647486 I tried substituting the value derived from (2) into the given equation and it gave 4.9, which is wrong.
$endgroup$
– Ramana
Mar 15 at 5:39
$begingroup$
you are probably correct. Your book is wrong
$endgroup$
– rash
Mar 15 at 5:39
$begingroup$
you are probably correct. Your book is wrong
$endgroup$
– rash
Mar 15 at 5:39
$begingroup$
You are correct. $27.019^circ$ is also a correct answer.
$endgroup$
– John Wayland Bales
Mar 15 at 5:47
$begingroup$
You are correct. $27.019^circ$ is also a correct answer.
$endgroup$
– John Wayland Bales
Mar 15 at 5:47
|
show 2 more comments
3 Answers
3
active
oldest
votes
$begingroup$
begineqnarray
5cos x - sin x &=& 4\
frac5sqrt26cos x-frac1sqrt26sin x&=&frac4sqrt26\
cosalphacos x-sinalphasin x&=&frac4sqrt26\
endeqnarray
Giving
$$cos(x+alpha)=frac4sqrt26$$
So either
$$ x+alpha=arccosleft(frac4sqrt26right)=38.33^circ $$
or
$$ x+alpha=360^circ-arccosleft(frac4sqrt26right)=321.67^circ $$
with
$$alpha=arccosleft(frac5sqrt26right)=11.31$$
So you get that either $x=27.02^circ$ or $x=310.36^circ$.
answered Mar 15 at 6:06
John Wayland BalesJohn Wayland Bales
14.8k21238
$endgroup$
$begingroup$
i think you got it wrong
$endgroup$
– rash
Mar 15 at 6:10
$begingroup$
No, it is correct.
$endgroup$
– John Wayland Bales
Mar 15 at 6:11
$begingroup$
It is supposed to $arccos(frac5sqrt26)$
$endgroup$
– rash
Mar 15 at 6:11
$begingroup$
if it is what u took then it should be $x - alpha$ according to the sum formula
$endgroup$
– rash
Mar 15 at 6:12
$begingroup$
@rash Yes, that's a typo. I will fix it. Thanks.
$endgroup$
– John Wayland Bales
Mar 15 at 6:14
|
show 1 more comment
$begingroup$
This is just my way of saying what everyone else has already said.
beginalign
R &=sqrt1^2+5^2=sqrt26 \
cos alpha &= dfrac5sqrt26 \
sin alpha &= dfrac1sqrt26 \
alpha &approx 11.31^circ \
hline
5cos x - 1 sin x &= 4 \
cos x cos alpha - sin x sin alpha &= dfrac4sqrt26 \
cos(x + alpha) &= dfrac4sqrt26 \
x + 11.31^circ &approx pm 38.33^circ + n 360^circ \
hline
x &approx 38.33^circ - 11.31^circ = 27.02^circ \
x &approx -38.33^circ - 11.31^circ + 360^circ = 310.36^circ
endalign
answered Mar 15 at 6:31
steven gregorysteven gregory
18.3k32358
$endgroup$
add a comment |
$begingroup$
It seems that you know why $R=sqrt26$,
Hence $5cos x - sin x$ can be expressed as $sqrt26(frac5sqrt26cos x - frac1sqrt26sin x)$
$$textLet, cosalpha = frac5sqrt26 text and sinalpha = frac1sqrt26$$
Hence, $5cos x - sin x$ can now be expressed as $sqrt26cos (x + alpha)$.
To solve your initial equation, we can now write it as,
$$sqrt26cos (x + alpha) =4$$
$$cos (x + alpha) = frac4sqrt26$$
$$x +alpha = cos^-1 frac4sqrt26$$
If $cosalpha = frac5sqrt26$, then $alpha = cos^-1 frac5sqrt26$,
$$x = cos^-1 frac4sqrt26 - alpha rightarrow x = cos^-1 frac4sqrt26 - cos^-1 frac5sqrt26$$
$$therefore x = 27^o text or 321.40^o$$
answered Mar 15 at 6:07
rashrash
524115
$endgroup$
add a comment |
Your Answer
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3 Answers
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3 Answers
3
active
oldest
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active
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active
oldest
votes
$begingroup$
begineqnarray
5cos x - sin x &=& 4\
frac5sqrt26cos x-frac1sqrt26sin x&=&frac4sqrt26\
cosalphacos x-sinalphasin x&=&frac4sqrt26\
endeqnarray
Giving
$$cos(x+alpha)=frac4sqrt26$$
So either
$$ x+alpha=arccosleft(frac4sqrt26right)=38.33^circ $$
or
$$ x+alpha=360^circ-arccosleft(frac4sqrt26right)=321.67^circ $$
with
$$alpha=arccosleft(frac5sqrt26right)=11.31$$
So you get that either $x=27.02^circ$ or $x=310.36^circ$.
answered Mar 15 at 6:06
John Wayland BalesJohn Wayland Bales
14.8k21238
$endgroup$
$begingroup$
i think you got it wrong
$endgroup$
– rash
Mar 15 at 6:10
$begingroup$
No, it is correct.
$endgroup$
– John Wayland Bales
Mar 15 at 6:11
$begingroup$
It is supposed to $arccos(frac5sqrt26)$
$endgroup$
– rash
Mar 15 at 6:11
$begingroup$
if it is what u took then it should be $x - alpha$ according to the sum formula
$endgroup$
– rash
Mar 15 at 6:12
$begingroup$
@rash Yes, that's a typo. I will fix it. Thanks.
$endgroup$
– John Wayland Bales
Mar 15 at 6:14
|
show 1 more comment
$begingroup$
begineqnarray
5cos x - sin x &=& 4\
frac5sqrt26cos x-frac1sqrt26sin x&=&frac4sqrt26\
cosalphacos x-sinalphasin x&=&frac4sqrt26\
endeqnarray
Giving
$$cos(x+alpha)=frac4sqrt26$$
So either
$$ x+alpha=arccosleft(frac4sqrt26right)=38.33^circ $$
or
$$ x+alpha=360^circ-arccosleft(frac4sqrt26right)=321.67^circ $$
with
$$alpha=arccosleft(frac5sqrt26right)=11.31$$
So you get that either $x=27.02^circ$ or $x=310.36^circ$.
answered Mar 15 at 6:06
John Wayland BalesJohn Wayland Bales
14.8k21238
$endgroup$
$begingroup$
i think you got it wrong
$endgroup$
– rash
Mar 15 at 6:10
$begingroup$
No, it is correct.
$endgroup$
– John Wayland Bales
Mar 15 at 6:11
$begingroup$
It is supposed to $arccos(frac5sqrt26)$
$endgroup$
– rash
Mar 15 at 6:11
$begingroup$
if it is what u took then it should be $x - alpha$ according to the sum formula
$endgroup$
– rash
Mar 15 at 6:12
$begingroup$
@rash Yes, that's a typo. I will fix it. Thanks.
$endgroup$
– John Wayland Bales
Mar 15 at 6:14
|
show 1 more comment
$begingroup$
begineqnarray
5cos x - sin x &=& 4\
frac5sqrt26cos x-frac1sqrt26sin x&=&frac4sqrt26\
cosalphacos x-sinalphasin x&=&frac4sqrt26\
endeqnarray
Giving
$$cos(x+alpha)=frac4sqrt26$$
So either
$$ x+alpha=arccosleft(frac4sqrt26right)=38.33^circ $$
or
$$ x+alpha=360^circ-arccosleft(frac4sqrt26right)=321.67^circ $$
with
$$alpha=arccosleft(frac5sqrt26right)=11.31$$
So you get that either $x=27.02^circ$ or $x=310.36^circ$.
answered Mar 15 at 6:06
John Wayland BalesJohn Wayland Bales
14.8k21238
$endgroup$
begineqnarray
5cos x - sin x &=& 4\
frac5sqrt26cos x-frac1sqrt26sin x&=&frac4sqrt26\
cosalphacos x-sinalphasin x&=&frac4sqrt26\
endeqnarray
Giving
$$cos(x+alpha)=frac4sqrt26$$
So either
$$ x+alpha=arccosleft(frac4sqrt26right)=38.33^circ $$
or
$$ x+alpha=360^circ-arccosleft(frac4sqrt26right)=321.67^circ $$
with
$$alpha=arccosleft(frac5sqrt26right)=11.31$$
So you get that either $x=27.02^circ$ or $x=310.36^circ$.
answered Mar 15 at 6:06
John Wayland BalesJohn Wayland Bales
14.8k21238
edited Mar 15 at 6:21
answered Mar 15 at 6:06
John Wayland BalesJohn Wayland Bales
14.8k21238
answered Mar 15 at 6:06
John Wayland BalesJohn Wayland Bales
14.8k21238
answered Mar 15 at 6:06
John Wayland BalesJohn Wayland Bales
14.8k21238
14.8k21238
$begingroup$
i think you got it wrong
$endgroup$
– rash
Mar 15 at 6:10
$begingroup$
No, it is correct.
$endgroup$
– John Wayland Bales
Mar 15 at 6:11
$begingroup$
It is supposed to $arccos(frac5sqrt26)$
$endgroup$
– rash
Mar 15 at 6:11
$begingroup$
if it is what u took then it should be $x - alpha$ according to the sum formula
$endgroup$
– rash
Mar 15 at 6:12
$begingroup$
@rash Yes, that's a typo. I will fix it. Thanks.
$endgroup$
– John Wayland Bales
Mar 15 at 6:14
|
show 1 more comment
$begingroup$
i think you got it wrong
$endgroup$
– rash
Mar 15 at 6:10
$begingroup$
No, it is correct.
$endgroup$
– John Wayland Bales
Mar 15 at 6:11
$begingroup$
It is supposed to $arccos(frac5sqrt26)$
$endgroup$
– rash
Mar 15 at 6:11
$begingroup$
if it is what u took then it should be $x - alpha$ according to the sum formula
$endgroup$
– rash
Mar 15 at 6:12
$begingroup$
@rash Yes, that's a typo. I will fix it. Thanks.
$endgroup$
– John Wayland Bales
Mar 15 at 6:14
$begingroup$
i think you got it wrong
$endgroup$
– rash
Mar 15 at 6:10
$begingroup$
i think you got it wrong
$endgroup$
– rash
Mar 15 at 6:10
$begingroup$
No, it is correct.
$endgroup$
– John Wayland Bales
Mar 15 at 6:11
$begingroup$
No, it is correct.
$endgroup$
– John Wayland Bales
Mar 15 at 6:11
$begingroup$
It is supposed to $arccos(frac5sqrt26)$
$endgroup$
– rash
Mar 15 at 6:11
$begingroup$
It is supposed to $arccos(frac5sqrt26)$
$endgroup$
– rash
Mar 15 at 6:11
$begingroup$
if it is what u took then it should be $x - alpha$ according to the sum formula
$endgroup$
– rash
Mar 15 at 6:12
$begingroup$
if it is what u took then it should be $x - alpha$ according to the sum formula
$endgroup$
– rash
Mar 15 at 6:12
$begingroup$
@rash Yes, that's a typo. I will fix it. Thanks.
$endgroup$
– John Wayland Bales
Mar 15 at 6:14
$begingroup$
@rash Yes, that's a typo. I will fix it. Thanks.
$endgroup$
– John Wayland Bales
Mar 15 at 6:14
|
show 1 more comment
$begingroup$
This is just my way of saying what everyone else has already said.
beginalign
R &=sqrt1^2+5^2=sqrt26 \
cos alpha &= dfrac5sqrt26 \
sin alpha &= dfrac1sqrt26 \
alpha &approx 11.31^circ \
hline
5cos x - 1 sin x &= 4 \
cos x cos alpha - sin x sin alpha &= dfrac4sqrt26 \
cos(x + alpha) &= dfrac4sqrt26 \
x + 11.31^circ &approx pm 38.33^circ + n 360^circ \
hline
x &approx 38.33^circ - 11.31^circ = 27.02^circ \
x &approx -38.33^circ - 11.31^circ + 360^circ = 310.36^circ
endalign
answered Mar 15 at 6:31
steven gregorysteven gregory
18.3k32358
$endgroup$
add a comment |
$begingroup$
This is just my way of saying what everyone else has already said.
beginalign
R &=sqrt1^2+5^2=sqrt26 \
cos alpha &= dfrac5sqrt26 \
sin alpha &= dfrac1sqrt26 \
alpha &approx 11.31^circ \
hline
5cos x - 1 sin x &= 4 \
cos x cos alpha - sin x sin alpha &= dfrac4sqrt26 \
cos(x + alpha) &= dfrac4sqrt26 \
x + 11.31^circ &approx pm 38.33^circ + n 360^circ \
hline
x &approx 38.33^circ - 11.31^circ = 27.02^circ \
x &approx -38.33^circ - 11.31^circ + 360^circ = 310.36^circ
endalign
answered Mar 15 at 6:31
steven gregorysteven gregory
18.3k32358
$endgroup$
add a comment |
$begingroup$
This is just my way of saying what everyone else has already said.
beginalign
R &=sqrt1^2+5^2=sqrt26 \
cos alpha &= dfrac5sqrt26 \
sin alpha &= dfrac1sqrt26 \
alpha &approx 11.31^circ \
hline
5cos x - 1 sin x &= 4 \
cos x cos alpha - sin x sin alpha &= dfrac4sqrt26 \
cos(x + alpha) &= dfrac4sqrt26 \
x + 11.31^circ &approx pm 38.33^circ + n 360^circ \
hline
x &approx 38.33^circ - 11.31^circ = 27.02^circ \
x &approx -38.33^circ - 11.31^circ + 360^circ = 310.36^circ
endalign
answered Mar 15 at 6:31
steven gregorysteven gregory
18.3k32358
$endgroup$
This is just my way of saying what everyone else has already said.
beginalign
R &=sqrt1^2+5^2=sqrt26 \
cos alpha &= dfrac5sqrt26 \
sin alpha &= dfrac1sqrt26 \
alpha &approx 11.31^circ \
hline
5cos x - 1 sin x &= 4 \
cos x cos alpha - sin x sin alpha &= dfrac4sqrt26 \
cos(x + alpha) &= dfrac4sqrt26 \
x + 11.31^circ &approx pm 38.33^circ + n 360^circ \
hline
x &approx 38.33^circ - 11.31^circ = 27.02^circ \
x &approx -38.33^circ - 11.31^circ + 360^circ = 310.36^circ
endalign
answered Mar 15 at 6:31
steven gregorysteven gregory
18.3k32358
answered Mar 15 at 6:31
steven gregorysteven gregory
18.3k32358
answered Mar 15 at 6:31
steven gregorysteven gregory
18.3k32358
answered Mar 15 at 6:31
steven gregorysteven gregory
18.3k32358
18.3k32358
add a comment |
add a comment |
$begingroup$
It seems that you know why $R=sqrt26$,
Hence $5cos x - sin x$ can be expressed as $sqrt26(frac5sqrt26cos x - frac1sqrt26sin x)$
$$textLet, cosalpha = frac5sqrt26 text and sinalpha = frac1sqrt26$$
Hence, $5cos x - sin x$ can now be expressed as $sqrt26cos (x + alpha)$.
To solve your initial equation, we can now write it as,
$$sqrt26cos (x + alpha) =4$$
$$cos (x + alpha) = frac4sqrt26$$
$$x +alpha = cos^-1 frac4sqrt26$$
If $cosalpha = frac5sqrt26$, then $alpha = cos^-1 frac5sqrt26$,
$$x = cos^-1 frac4sqrt26 - alpha rightarrow x = cos^-1 frac4sqrt26 - cos^-1 frac5sqrt26$$
$$therefore x = 27^o text or 321.40^o$$
answered Mar 15 at 6:07
rashrash
524115
$endgroup$
add a comment |
$begingroup$
It seems that you know why $R=sqrt26$,
Hence $5cos x - sin x$ can be expressed as $sqrt26(frac5sqrt26cos x - frac1sqrt26sin x)$
$$textLet, cosalpha = frac5sqrt26 text and sinalpha = frac1sqrt26$$
Hence, $5cos x - sin x$ can now be expressed as $sqrt26cos (x + alpha)$.
To solve your initial equation, we can now write it as,
$$sqrt26cos (x + alpha) =4$$
$$cos (x + alpha) = frac4sqrt26$$
$$x +alpha = cos^-1 frac4sqrt26$$
If $cosalpha = frac5sqrt26$, then $alpha = cos^-1 frac5sqrt26$,
$$x = cos^-1 frac4sqrt26 - alpha rightarrow x = cos^-1 frac4sqrt26 - cos^-1 frac5sqrt26$$
$$therefore x = 27^o text or 321.40^o$$
answered Mar 15 at 6:07
rashrash
524115
$endgroup$
add a comment |
$begingroup$
It seems that you know why $R=sqrt26$,
Hence $5cos x - sin x$ can be expressed as $sqrt26(frac5sqrt26cos x - frac1sqrt26sin x)$
$$textLet, cosalpha = frac5sqrt26 text and sinalpha = frac1sqrt26$$
Hence, $5cos x - sin x$ can now be expressed as $sqrt26cos (x + alpha)$.
To solve your initial equation, we can now write it as,
$$sqrt26cos (x + alpha) =4$$
$$cos (x + alpha) = frac4sqrt26$$
$$x +alpha = cos^-1 frac4sqrt26$$
If $cosalpha = frac5sqrt26$, then $alpha = cos^-1 frac5sqrt26$,
$$x = cos^-1 frac4sqrt26 - alpha rightarrow x = cos^-1 frac4sqrt26 - cos^-1 frac5sqrt26$$
$$therefore x = 27^o text or 321.40^o$$
answered Mar 15 at 6:07
rashrash
524115
$endgroup$
It seems that you know why $R=sqrt26$,
Hence $5cos x - sin x$ can be expressed as $sqrt26(frac5sqrt26cos x - frac1sqrt26sin x)$
$$textLet, cosalpha = frac5sqrt26 text and sinalpha = frac1sqrt26$$
Hence, $5cos x - sin x$ can now be expressed as $sqrt26cos (x + alpha)$.
To solve your initial equation, we can now write it as,
$$sqrt26cos (x + alpha) =4$$
$$cos (x + alpha) = frac4sqrt26$$
$$x +alpha = cos^-1 frac4sqrt26$$
If $cosalpha = frac5sqrt26$, then $alpha = cos^-1 frac5sqrt26$,
$$x = cos^-1 frac4sqrt26 - alpha rightarrow x = cos^-1 frac4sqrt26 - cos^-1 frac5sqrt26$$
$$therefore x = 27^o text or 321.40^o$$
answered Mar 15 at 6:07
rashrash
524115
edited Mar 15 at 6:30
answered Mar 15 at 6:07
rashrash
524115
answered Mar 15 at 6:07
rashrash
524115
answered Mar 15 at 6:07
rashrash
524115
524115
add a comment |
add a comment |
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$begingroup$
You are missing an $R$ after the $-$ in the second arrow. The $alpha$ is chosen to satisfy both (1) and (2).
$endgroup$
– user647486
Mar 15 at 5:31
$begingroup$
Edited it just now
$endgroup$
– Ramana
Mar 15 at 5:37
$begingroup$
@user647486 I tried substituting the value derived from (2) into the given equation and it gave 4.9, which is wrong.
$endgroup$
– Ramana
Mar 15 at 5:39
$begingroup$
you are probably correct. Your book is wrong
$endgroup$
– rash
Mar 15 at 5:39
$begingroup$
You are correct. $27.019^circ$ is also a correct answer.
$endgroup$
– John Wayland Bales
Mar 15 at 5:47