Answer
$1 - 8 ~sin^2~\frac{x}{2}~cos^2~\frac{x}{2} = cos~2x$
![](https://gradesaver.s3.amazonaws.com/uploads/solution/2d0f8c94-1f2e-4e6a-9f88-879bbf9de9fb/result_image/1555316028.jpg?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIAJVAXHCSURVZEX5QQ%2F20240727%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240727T011533Z&X-Amz-Expires=900&X-Amz-SignedHeaders=host&X-Amz-Signature=11a40e3caf09e7d60e439abb2e7bcd9cd57cfa1d93e024a7a9162834880bad56)
Work Step by Step
$1 - 8 ~sin^2~\frac{x}{2}~cos^2~\frac{x}{2}$
When we graph this function, we can see that it looks like the graph of $~~cos~2x$
Note that: $~sin~2a = 2~sin~a~cos~a$
Also: $~~cos~2b = 1-2~sin^2~b$
We can verify this algebraically:
$1 - 8 ~sin^2~\frac{x}{2}~cos^2~\frac{x}{2} = 1 - 2~(4~sin^2~\frac{x}{2}~cos^2~\frac{x}{2})$
$1 - 8 ~sin^2~\frac{x}{2}~cos^2~\frac{x}{2} = 1 - 2~(2~sin~\frac{x}{2}~cos~\frac{x}{2})^2$
$1 - 8 ~sin^2~\frac{x}{2}~cos^2~\frac{x}{2} = 1 - 2~(sin~x)^2$
$1 - 8 ~sin^2~\frac{x}{2}~cos^2~\frac{x}{2} = cos~2x$
![](https://gradesaver.s3.amazonaws.com/uploads/solution/2d0f8c94-1f2e-4e6a-9f88-879bbf9de9fb/steps_image/small_1555316028.jpg?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIAJVAXHCSURVZEX5QQ%2F20240727%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240727T011533Z&X-Amz-Expires=900&X-Amz-SignedHeaders=host&X-Amz-Signature=233a5b9bb351c6a75373b5c5873d27acd30d52be0560b826ae31359a79f46411)