Verify the following trig identity please?

Question

1-sin^2(2x) = (1-2sin^2(x))^2

please show steps..thank you!

Answer 1

1-sin^2(2x) = cos^2(2x)                                    [ Since sin^2(u) + cos^2(u) = 1 ]

                    = ( cos (2x) )^2                               [ Write cos^2(2u) as ( cos (2u) )^2 ]

                    = (1-2sin^2(x))^2                           [ Since cos(2u) = 1-2sin^2(u) ]

Therefore it is proved that 1-sin^2(2x) = (1-2sin^2(x))^2.

Answer 2

The identity is ,

Check whether the identity is true or not.

Let us take RHS.

                            

                                            

.                                        

It is LHS.

 So, the identity is true.

.

Comments

The trigonometric identity is [1 - sin²(2x)] = [1 - 2sin²(x)]².

To prove the identity, either left or right hand side expression can be considered for simplification.

Left hand side expression is [1 - sin²(2x)].

= 1 - [sin(2x)]²

= 1 - [2 sin(x) cos(x)]²

= 1 - 4 sin²(x) cos²(x)

= 1 - 4 sin²(x) [1 - sin²(x)]

= 1 - 4 sin²(x) - 4 sin⁴(x)

= [1 - 2 sin²(x)]²                     [ (a - b)² = a² - 2ab + b² ]

= Right hand side expression.

Therefore [1 - sin²(2x)] = [1 - 2sin²(x)]²