Question

How do I verify this identity? I know you should write sin2a as sin(a+a).

Answer 1

Given the question in the image, the following are the solution steps to verify the identity

STEP 1: Write the given identity

`\sin 2\alpha=2\sin \alpha\cos \alpha`

STEP 2: Verify the identity

`\begin{gathered} \sin 2\alpha=2\sin \alpha\cos \alpha \\ \text{Consider the left hand side of the above trigonometry identity.} \\ \text{That is, }\sin 2\alpha\text{.} \\ \text{ Rewrite }\sin 2\alpha\text{ as }\sin (\alpha+a) \\ \text{ It is known that }\sin (a+b)=\sin (a)\cos b+\cos (a)\sin (b) \\ U\sin g\text{ this statement above, we have;} \\ \sin (\alpha+a)=\sin a\cos \alpha+\cos a\sin \alpha \\ \text{It is known that }xy+yx=xy+xy=2* xy=2xy \\ U\sin g\text{ this statement above, we have;} \\ \sin a\cos \alpha+\cos a\sin \alpha=\sin a\cos \alpha+\sin \alpha\cos \alpha=2*\sin \alpha\cos \alpha=2\sin \alpha\cos \alpha \\ \text{Hence, }\sin 2\alpha=2\sin \alpha\cos \alpha \end{gathered}`

The verification of the identity is as seen above.