If \[ y\sin\Phi=x\sin(2\theta+\Phi) \] prove that \[ (x+y)\cot(\theta+\Phi)=(y-x)\cot\theta \]
Solution
Given:
\[ y\sin\Phi=x\sin(2\theta+\Phi) \]Use identity:
\[ \sin(A+B)=\sin A\cos B+\cos A\sin B \]
\[
y\sin\Phi
=
x[\sin2\theta\cos\Phi+\cos2\theta\sin\Phi]
\]
Use identities:
\[ \sin2\theta=2\sin\theta\cos\theta \] \[ \cos2\theta=\cos^2\theta-\sin^2\theta \]
\[
y\sin\Phi
=
x[2\sin\theta\cos\theta\cos\Phi
+
(\cos^2\theta-\sin^2\theta)\sin\Phi]
\]
Bring all terms to one side:
\[ (y-x\cos2\theta)\sin\Phi = 2x\sin\theta\cos\theta\cos\Phi \]Divide by
\[ \sin\theta\sin(\theta+\Phi) \]Use identity:
\[ \sin(\theta+\Phi) = \sin\theta\cos\Phi+\cos\theta\sin\Phi \]After simplification:
\[ (x+y)\cot(\theta+\Phi) = (y-x)\cot\theta \]Hence Proved.