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Trigonometric identities

Please help! I always seem to get sinx + cosx instead of (sinx + cos)/sin2x
Screen Shot 2016-09-26 at 21.34.30.png17.0KB
Reply 1
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metrize
18
What have you done so far?
Reply 2
Avatar for Stanspam
Stanspam
OP
13
Original post by metrize
What have you done so far?


it's reallyyy lengthy, but i can get it simplified down to this
(edited 7 years ago)
Reply 3
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metrize
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Original post by Stanspam
it's reallyyy lengthy...

Just take a picture, otherwise we can't tell you where you went wrong
Reply 4
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Stanspam
OP
13
this
Screen Shot 2016-09-26 at 21.48.59.png13.7KB
Reply 5
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Stanspam
OP
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oh, it's supposed to be 4 on the denominator instead of 2
Original post by Stanspam
oh, it's supposed to be 4 on the denominator instead of 2


Well just use double angle on it.

Do you know how to add fractions? ab+cd=ad+bcbd\displaystyle \frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd} ??? Use it.

The denominator would go to 1−cos⁡2(2x)1-\cos^2(2x) which is equivalent to sin⁡2(2x)\sin^2(2x)
(edited 7 years ago)
Reply 7
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B_9710
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Original post by Stanspam
this


This can still be simplified to the expression the question asks for.
Reply 8
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Stanspam
OP
13
Original post by B_9710
This can still be simplified to the expression the question asks for.


nope. sorry, still not getting it
Original post by Stanspam
nope. sorry, still not getting it


2sin⁡(x)cos⁡(x)≡sin⁡(2x)2\sin(x)\cos(x) \equiv \sin(2x)

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