value of $a,b$ in trigonometric expressionRation of sum and Product of Trigonometric expression.Value of $sec^2 a+2sec^2 b$Finding value of $k$ in trigonometric equation.Finding smallest positive root $sqrtsin(1-x)=sqrtcos x$Trigonometric series sum involving tangentsFind all integers $m$ and $n$ such that $ncosfracpim= sqrt8+sqrt32+sqrt768$If $cos^4 alpha+4sin^4 beta-4sqrt2cos alpha sin beta +2=0$, then find $alpha$, $beta$ in $(0,fracpi2)$value of $x$ in Trigonometric equationTrigonometric series sum with $sin$ functionfinding wired trigononetric ratio
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value of $a,b$ in trigonometric expression
Ration of sum and Product of Trigonometric expression.Value of $sec^2 a+2sec^2 b$Finding value of $k$ in trigonometric equation.Finding smallest positive root $sqrtsin(1-x)=sqrtcos x$Trigonometric series sum involving tangentsFind all integers $m$ and $n$ such that $ncosfracpim= sqrt8+sqrt32+sqrt768$If $cos^4 alpha+4sin^4 beta-4sqrt2cos alpha sin beta +2=0$, then find $alpha$, $beta$ in $(0,fracpi2)$value of $x$ in Trigonometric equationTrigonometric series sum with $sin$ functionfinding wired trigononetric ratio
$begingroup$
If $a,b$ are positive integers such that $displaystyle sqrt8+sqrt32+sqrt768=acosbigg(fracpibbigg)$
what i try: i am trying to convert it into $displaystyle sqrt2+2cos 8x=sqrt2cos 4x$
How do i solve it Help me please
trigonometry
$endgroup$
add a comment |
$begingroup$
If $a,b$ are positive integers such that $displaystyle sqrt8+sqrt32+sqrt768=acosbigg(fracpibbigg)$
what i try: i am trying to convert it into $displaystyle sqrt2+2cos 8x=sqrt2cos 4x$
How do i solve it Help me please
trigonometry
$endgroup$
add a comment |
$begingroup$
If $a,b$ are positive integers such that $displaystyle sqrt8+sqrt32+sqrt768=acosbigg(fracpibbigg)$
what i try: i am trying to convert it into $displaystyle sqrt2+2cos 8x=sqrt2cos 4x$
How do i solve it Help me please
trigonometry
$endgroup$
If $a,b$ are positive integers such that $displaystyle sqrt8+sqrt32+sqrt768=acosbigg(fracpibbigg)$
what i try: i am trying to convert it into $displaystyle sqrt2+2cos 8x=sqrt2cos 4x$
How do i solve it Help me please
trigonometry
trigonometry
asked Mar 12 at 10:09
jackyjacky
1,186715
1,186715
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add a comment |
2 Answers
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$begingroup$
You're on the right track. The surd is $$2sqrt2(1+sqrt2+sqrt3)=2sqrt2(1+sqrt2(1+costfracpi6))=2sqrt2(1+2costfracpi12)=4costfracpi24.$$So take $a=4,,b=24$.
$endgroup$
add a comment |
$begingroup$
Hint: Your term is equal to $$2sqrt2+2sqrt2+sqrt3$$
$endgroup$
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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active
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votes
$begingroup$
You're on the right track. The surd is $$2sqrt2(1+sqrt2+sqrt3)=2sqrt2(1+sqrt2(1+costfracpi6))=2sqrt2(1+2costfracpi12)=4costfracpi24.$$So take $a=4,,b=24$.
$endgroup$
add a comment |
$begingroup$
You're on the right track. The surd is $$2sqrt2(1+sqrt2+sqrt3)=2sqrt2(1+sqrt2(1+costfracpi6))=2sqrt2(1+2costfracpi12)=4costfracpi24.$$So take $a=4,,b=24$.
$endgroup$
add a comment |
$begingroup$
You're on the right track. The surd is $$2sqrt2(1+sqrt2+sqrt3)=2sqrt2(1+sqrt2(1+costfracpi6))=2sqrt2(1+2costfracpi12)=4costfracpi24.$$So take $a=4,,b=24$.
$endgroup$
You're on the right track. The surd is $$2sqrt2(1+sqrt2+sqrt3)=2sqrt2(1+sqrt2(1+costfracpi6))=2sqrt2(1+2costfracpi12)=4costfracpi24.$$So take $a=4,,b=24$.
edited Mar 12 at 11:30
answered Mar 12 at 10:21
J.G.J.G.
30.3k23148
30.3k23148
add a comment |
add a comment |
$begingroup$
Hint: Your term is equal to $$2sqrt2+2sqrt2+sqrt3$$
$endgroup$
add a comment |
$begingroup$
Hint: Your term is equal to $$2sqrt2+2sqrt2+sqrt3$$
$endgroup$
add a comment |
$begingroup$
Hint: Your term is equal to $$2sqrt2+2sqrt2+sqrt3$$
$endgroup$
Hint: Your term is equal to $$2sqrt2+2sqrt2+sqrt3$$
edited Mar 12 at 10:23
J.G.
30.3k23148
30.3k23148
answered Mar 12 at 10:21
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
77.8k42866
add a comment |
add a comment |
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