# Can anyone solve this? Prove cos A / sin B - sin A / cos B = 2 cos (A+B) / sin 2B

Apr 30, 2018

Prove $\cos \frac{A}{\sin} \left(B\right) - \sin \frac{A}{\cos} \left(B\right) = \frac{2 \cos \left(A + B\right)}{\sin} \left(2 B\right)$

Multiply the first term by 1 in the form of $\cos \frac{B}{\cos} \left(B\right)$:

$\cos \frac{A}{\sin} \left(B\right) \cos \frac{B}{\cos} \left(B\right) - \sin \frac{A}{\cos} \left(B\right) = \frac{2 \cos \left(A + B\right)}{\sin} \left(2 B\right)$

Multiply the second term by 1 in the form of $\sin \frac{B}{\sin} \left(B\right)$:

$\cos \frac{A}{\sin} \left(B\right) \cos \frac{B}{\cos} \left(B\right) - \sin \frac{A}{\cos} \left(B\right) \sin \frac{B}{\sin} \left(B\right) = \frac{2 \cos \left(A + B\right)}{\sin} \left(2 B\right)$

Combine the two terms over the common denominator:

$\frac{\cos \left(A\right) \cos \left(B\right) - \sin \left(A\right) \sin \left(B\right)}{\sin \left(B\right) \cos \left(B\right)} = \frac{2 \cos \left(A + B\right)}{\sin} \left(2 B\right)$

Use the identity $\cos \left(A + B\right) = \cos \left(A\right) \cos \left(B\right) - \sin \left(A\right) \sin \left(B\right)$:

$\cos \frac{A + B}{\sin \left(B\right) \cos \left(B\right)} = \frac{2 \cos \left(A + B\right)}{\sin} \left(2 B\right)$

Multiply by 1 in the form of $\frac{2}{2}$:

$\frac{2 \cos \left(A + B\right)}{2 \sin \left(B\right) \cos \left(B\right)} = \frac{2 \cos \left(A + B\right)}{\sin} \left(2 B\right)$

Use the identity $\sin \left(2 B\right) = 2 \sin \left(B\right) \cos \left(B\right)$:

$\frac{2 \cos \left(A + B\right)}{\sin} \left(2 B\right) = \frac{2 \cos \left(A + B\right)}{\sin} \left(2 B\right)$ Q.E.D.