Question

Verify that sin2x=2cotsin^2x is an identity

Answer 1

Answer:

Vertify is an identity

Sin2x=2cotx(sin^2x)  

starting from the right-hand side

2cotx(sin^2x)

=2(cosx/sinx)(sin^2x)

=2(cosx/sinx)(sin^2x)

=2sinxcosx=sin2x

ans:right-hand side=left-hand side

Step-by-step explanation:

Answer 2

Step-by-step explanation:

sin^2x = 2cotx sin^2x

Rewrite right side as fractions:

sin^2x = $\frac{2}{1}$ * $\frac{cosx}{sinx}$ * $\frac{(sinx)(sinx)}{1}$

Multiply together $\frac{cosx}{sinx}$ and $\frac{(sinx)(sinx)}{1}$ :

sin^2x = $\frac{2}{1}$ * $\frac{(cosx)(sinx)(sinx)}{sinx}$

Cancel out sinx on top and bottom:

sin^2x = $\frac{2}{1}$ * $\frac{(sinx)(cosx)}{1}$

Multiply together 2 and (sinx)(cosx):

sin^2x = 2sinxcosx

Substitute sin^2x in for 2sinxcosx:

sin^2x = sin^2x