trig identity

**math123456** · December 29th 2008, 09:43 PM

Prove that:

cos(2x-3y)+ cos3y = cot a
sin(2x-3y)+ sin3y

**Soroban** · December 29th 2008, 10:23 PM

Hello, math123456!

You have a typo . . .

Prove that: . $\frac{\cos(2x-3y)+ \cos3y}{\sin(2x-3y) + \sin3y} \:=\:\cot{\color{red}x}$

We need two Sum-to-Product identities:

. . $\sin A + \sin B \:=<br /> \:2\sin\left(\tfrac{A+B}{2}\right)\cos\left(\tfrac {A-B}{2}\right)$

. . $\cos A + \cos B \:=\:2\cos\left(\tfrac{A+B}{2}\right)\cos\left(\tf rac{A-B}{2}\right)$

Numerator:
$\cos(2x-3y)+\cos3y \:=\:2\cos\left(\frac{2x-3y+3y}{2}\right)\cos\left(\frac{2x-3y-3y}{2}\right) \:=\:2\cos x\cos(x-3y)$

Denominator:
$\sin(2x-3y)+\sin3y \:=\:2\sin\left(\frac{2x-3y+3y}{2}\right)\cos\left(\frac{2x-3y-3y}{2}\right) \:=\:2\sin x\cos(x - 3y)$

The fraction becomes: . $\frac{2\,\cos x\,\cos(x - 3y)}{2\,\sin x\,\cos(x - 3y)} \;=\;\frac{\cos x}{\sin x} \;=\;\cot x$

**Prove It** · December 29th 2008, 10:26 PM

Originally Posted by math123456

Prove that:

cos(2x-3y)+ cos3y = cot a
sin(2x-3y)+ sin3y

Use the identities

$\sin{a} + \sin{b} = 2\sin{\frac{a+b}{2}}\cos{\frac{a-b}{2}}$

and

$\cos{a} + \cos{b} = 2\cos{\frac{a+b}{2}}\cos{\frac{a-b}{2}}$ .

So $\frac{\cos{2x - 3y} + \cos{3y}}{\sin{2x - 3y} + \sin{3y}}$

$= \frac{2\cos{\frac{2x - 3y + 3y}{2}}\cos{\frac{2x - 3y - 3y}{2}}}{2\sin{\frac{2x - 3y + 3y}{2}}\cos{\frac{2x - 3y - 3y}{2}}}$

$= \frac{\cos{x}\cos{(x - 3y)}}{\sin{x}\cos{(x - 3y)}}$

$= \frac{\cos{x}}{\sin{x}}$

$= \cot{x}$ .

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