Question

verify that the following equation is an identity.

\\( \sin 64 \alpha = 4 \sin 16 \alpha \cos 16 \alpha \cos 32 \alpha \\)

to verify the identity, start with the more complicated side and transform it to look like the other side. choose the correct transformation and transform the expression at each step

\\( 4 \sin 16 \alpha \cos 16 \alpha \cos 32 \alpha = \square \cos 32 \alpha \\)

Explanation:

Step1: Use double - angle identity

The double - angle identity for sine is \(\sin(2\theta)=2\sin\theta\cos\theta\). Let \(\theta = 16\alpha\), then \(2\sin16\alpha\cos16\alpha=\sin(2\times16\alpha)=\sin32\alpha\). So \(4\sin16\alpha\cos16\alpha\cos32\alpha = 2\times(2\sin16\alpha\cos16\alpha)\cos32\alpha\). Substituting \(2\sin16\alpha\cos16\alpha=\sin32\alpha\) into the expression, we get \(2\sin32\alpha\cos32\alpha\).

Step2: Use double - angle identity again

Let \(\theta = 32\alpha\) in the double - angle identity \(\sin(2\theta)=2\sin\theta\cos\theta\). Then \(2\sin32\alpha\cos32\alpha=\sin(2\times32\alpha)=\sin64\alpha\).

Answer:

The left - hand side \(4\sin16\alpha\cos16\alpha\cos32\alpha\) is transformed to \(\sin64\alpha\) using the double - angle identity (\(\sin(2\theta) = 2\sin\theta\cos\theta\)) twice. So \(\sin64\alpha=4\sin16\alpha\cos16\alpha\cos32\alpha\) is an identity.