Question

Prove that (sin3A sinA)sinA+ (cos3A -cosA)cosA= 0

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Given:

An equation that is $(\sin3A \sin A)\sin A+ (\cos3A -\cos A)\cos A$

To prove:

The above equation must be equal to 0

Solution:

The equation is $(\sin3A \sin A)\sin A+ (\cos3A -\cos A)\cos A$

Solving for LHS:

$\sin3A\sin A+\sin^2 A+\cos3A\cos A-\cos^2 A$

Rearranging the equation:

$\cos3A\cos A+\sin3A\sin A-(\cos^2 A-\sin^2 A)$

Using the identities:

• $\cos (x-y)=\cos x \cos y+\sin x \sin y$
• $\cos 2A=\cos^2 A-\sin ^2A$

The equation forms:

$=\cos (3A-A)-\cos 2A\\\\=\cos 2A-\cos 2A\\\\=0$

Hence proved