Write an equivalent equation for 2 cos ^4 x that does not contain powers of trigonometric functions greater than 1

Question

Pls help with my pre calc hw, can it be a step by step answer pls

Answer 1

2cos^4(x) = 2(cosx)(cosx)(cosx)(cosx)

No exponent greater than 1

but you probably want something different

so use the double angle formula

cos2x= 2cos^2(x) -1,

2cos^2(x) = cos2x+1

2cos^2(x)cos^2(x)= [cos2x+1][(cos2x+1)/2]

2co^4(x) = (cos^2(2x) + 2cos2x +1)/2

= [(cos4x+1)/2 +2cos2x +1]/2

=cos4x/4 + 1/4 + cos2x + 1/2

= (1/4)cos4x + cos2x + 3/4

Answer 2

So we'll focus for now on just cos⁴(x). We'll mainly be using the trig identity:

cos(2x) = 2cos²(x) - 1

If we solve for cos²(x), we get

cos²(x) = ( cos(2x) + 1 ) / 2

So let's apply this to our cos⁴(x), which we'll split into two cos²(x):

cos⁴(x) = cos²(x) * cos²(x) = ( ( cos(2x) + 1 ) / 2 )²

Multiply it out and you should get

( cos²(2x) + 2cos(2x) + 1 ) / 4

We still have a single cos²(2x) now, so we'll use the same identity again. This time, instead of multiplying x by 2, we'll multiply 2x by 2 to get 4x.

( ( ( cos(4x) + 1 ) / 2 ) + 2cos(2x) + 1 ) / 4

Multiply top and bottom by 2 to get rid of the fraction at the top

( cos(4x) + 1 + 4cos(2x) + 2 ) / 8

Simplify:

( cos(4x) + 4cos(2x) + 3 ) / 8

But wait! The original problem asked for 2cos⁴(x). So we'll multiply this by 2, which gives us:

( cos(4x) + 4cos(2x) + 3 ) / 4

Hope this helps.