Answer:
[tex]6-8cos2x+2cos4x[/tex]
Step-by-step explanation:
We are given that
[tex]16sin^4 x[/tex]
We can write the given expression as
[tex]16(sin^2x \times sin^2 x)[/tex]
[tex]16(\frac{1-cos2x}{2})(\frac{1-cos2x}{2})[/tex]
By using the formula
[tex]sin^2\theta=\frac{1-cos2\theta}{2}[/tex]
[tex]4(1-cos2x)^2[/tex]
[tex]4(1-2cos2x+cos^2(2x)[/tex]
Using the identity
[tex](a-b)^2=a^2+b^2-2ab[/tex]
[tex]4(1-2cos2x+\frac{1+cos4x}{2})[/tex]
[tex]4-8cos2x+2+2cos4x[/tex]
[tex]6-8cos2x+2cos4x[/tex]
This is required expression.
