Simplify trigonometric expression using trigonometric identities
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I have the trigonometric expression: $$2sin x +2sin left(frac{pi} {3} -xright) $$
and it should simplified in: $$sin x + sqrt 3 cos x$$ but I do not know what formulas to apply. Could you tell me how to simplify it?
trigonometry
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add a comment |
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I have the trigonometric expression: $$2sin x +2sin left(frac{pi} {3} -xright) $$
and it should simplified in: $$sin x + sqrt 3 cos x$$ but I do not know what formulas to apply. Could you tell me how to simplify it?
trigonometry
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mathworld.wolfram.com/ProsthaphaeresisFormulas.html
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– lab bhattacharjee
Apr 1 at 15:27
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Just between us: IMO the two expressions are of the same complexity.
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– Yves Daoust
Apr 1 at 15:30
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= 2 cos(x-π/6).
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– Anton Sherwood
Apr 1 at 22:26
add a comment |
$begingroup$
I have the trigonometric expression: $$2sin x +2sin left(frac{pi} {3} -xright) $$
and it should simplified in: $$sin x + sqrt 3 cos x$$ but I do not know what formulas to apply. Could you tell me how to simplify it?
trigonometry
$endgroup$
I have the trigonometric expression: $$2sin x +2sin left(frac{pi} {3} -xright) $$
and it should simplified in: $$sin x + sqrt 3 cos x$$ but I do not know what formulas to apply. Could you tell me how to simplify it?
trigonometry
trigonometry
edited Apr 1 at 15:55
MarianD
2,2611618
2,2611618
asked Apr 1 at 15:19
zaz9999zaz9999
162
162
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mathworld.wolfram.com/ProsthaphaeresisFormulas.html
$endgroup$
– lab bhattacharjee
Apr 1 at 15:27
$begingroup$
Just between us: IMO the two expressions are of the same complexity.
$endgroup$
– Yves Daoust
Apr 1 at 15:30
$begingroup$
= 2 cos(x-π/6).
$endgroup$
– Anton Sherwood
Apr 1 at 22:26
add a comment |
$begingroup$
mathworld.wolfram.com/ProsthaphaeresisFormulas.html
$endgroup$
– lab bhattacharjee
Apr 1 at 15:27
$begingroup$
Just between us: IMO the two expressions are of the same complexity.
$endgroup$
– Yves Daoust
Apr 1 at 15:30
$begingroup$
= 2 cos(x-π/6).
$endgroup$
– Anton Sherwood
Apr 1 at 22:26
$begingroup$
mathworld.wolfram.com/ProsthaphaeresisFormulas.html
$endgroup$
– lab bhattacharjee
Apr 1 at 15:27
$begingroup$
mathworld.wolfram.com/ProsthaphaeresisFormulas.html
$endgroup$
– lab bhattacharjee
Apr 1 at 15:27
$begingroup$
Just between us: IMO the two expressions are of the same complexity.
$endgroup$
– Yves Daoust
Apr 1 at 15:30
$begingroup$
Just between us: IMO the two expressions are of the same complexity.
$endgroup$
– Yves Daoust
Apr 1 at 15:30
$begingroup$
= 2 cos(x-π/6).
$endgroup$
– Anton Sherwood
Apr 1 at 22:26
$begingroup$
= 2 cos(x-π/6).
$endgroup$
– Anton Sherwood
Apr 1 at 22:26
add a comment |
3 Answers
3
active
oldest
votes
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Using the formula for $sin (alpha - beta)$ you obtain
begin{align}
2&sin x +2sin left(frac{pi} {3} -xright)\
= 2&sin x +2left[sin left(frac{pi} {3}right) cos x - cosleft(frac{pi} {3}right) sin xright]\
= 2&sin x + 2left[{sqrt 3 over 2} cos x - frac 1 2 sin xright]\[1ex]
= 2&sin x + sqrt 3 cos x - sin x\[1em] = , &color{red}{sin x + sqrt 3 cos x}
end{align}
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add a comment |
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Use that $$sin(x-y)=sin(x)cos(y)-sin(y)cos(x)$$
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add a comment |
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Hint: $sin(frac{pi}{3}-x)=sin frac{pi}{3}cos x-cos frac{pi}{3}sin x$
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add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Using the formula for $sin (alpha - beta)$ you obtain
begin{align}
2&sin x +2sin left(frac{pi} {3} -xright)\
= 2&sin x +2left[sin left(frac{pi} {3}right) cos x - cosleft(frac{pi} {3}right) sin xright]\
= 2&sin x + 2left[{sqrt 3 over 2} cos x - frac 1 2 sin xright]\[1ex]
= 2&sin x + sqrt 3 cos x - sin x\[1em] = , &color{red}{sin x + sqrt 3 cos x}
end{align}
$endgroup$
add a comment |
$begingroup$
Using the formula for $sin (alpha - beta)$ you obtain
begin{align}
2&sin x +2sin left(frac{pi} {3} -xright)\
= 2&sin x +2left[sin left(frac{pi} {3}right) cos x - cosleft(frac{pi} {3}right) sin xright]\
= 2&sin x + 2left[{sqrt 3 over 2} cos x - frac 1 2 sin xright]\[1ex]
= 2&sin x + sqrt 3 cos x - sin x\[1em] = , &color{red}{sin x + sqrt 3 cos x}
end{align}
$endgroup$
add a comment |
$begingroup$
Using the formula for $sin (alpha - beta)$ you obtain
begin{align}
2&sin x +2sin left(frac{pi} {3} -xright)\
= 2&sin x +2left[sin left(frac{pi} {3}right) cos x - cosleft(frac{pi} {3}right) sin xright]\
= 2&sin x + 2left[{sqrt 3 over 2} cos x - frac 1 2 sin xright]\[1ex]
= 2&sin x + sqrt 3 cos x - sin x\[1em] = , &color{red}{sin x + sqrt 3 cos x}
end{align}
$endgroup$
Using the formula for $sin (alpha - beta)$ you obtain
begin{align}
2&sin x +2sin left(frac{pi} {3} -xright)\
= 2&sin x +2left[sin left(frac{pi} {3}right) cos x - cosleft(frac{pi} {3}right) sin xright]\
= 2&sin x + 2left[{sqrt 3 over 2} cos x - frac 1 2 sin xright]\[1ex]
= 2&sin x + sqrt 3 cos x - sin x\[1em] = , &color{red}{sin x + sqrt 3 cos x}
end{align}
edited Apr 1 at 15:59
answered Apr 1 at 15:44
MarianDMarianD
2,2611618
2,2611618
add a comment |
add a comment |
$begingroup$
Use that $$sin(x-y)=sin(x)cos(y)-sin(y)cos(x)$$
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add a comment |
$begingroup$
Use that $$sin(x-y)=sin(x)cos(y)-sin(y)cos(x)$$
$endgroup$
add a comment |
$begingroup$
Use that $$sin(x-y)=sin(x)cos(y)-sin(y)cos(x)$$
$endgroup$
Use that $$sin(x-y)=sin(x)cos(y)-sin(y)cos(x)$$
answered Apr 1 at 15:26
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
79k42867
add a comment |
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Hint: $sin(frac{pi}{3}-x)=sin frac{pi}{3}cos x-cos frac{pi}{3}sin x$
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add a comment |
$begingroup$
Hint: $sin(frac{pi}{3}-x)=sin frac{pi}{3}cos x-cos frac{pi}{3}sin x$
$endgroup$
add a comment |
$begingroup$
Hint: $sin(frac{pi}{3}-x)=sin frac{pi}{3}cos x-cos frac{pi}{3}sin x$
$endgroup$
Hint: $sin(frac{pi}{3}-x)=sin frac{pi}{3}cos x-cos frac{pi}{3}sin x$
answered Apr 1 at 15:25
VasyaVasya
4,4671618
4,4671618
add a comment |
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$begingroup$
mathworld.wolfram.com/ProsthaphaeresisFormulas.html
$endgroup$
– lab bhattacharjee
Apr 1 at 15:27
$begingroup$
Just between us: IMO the two expressions are of the same complexity.
$endgroup$
– Yves Daoust
Apr 1 at 15:30
$begingroup$
= 2 cos(x-π/6).
$endgroup$
– Anton Sherwood
Apr 1 at 22:26