Write cos(3x) - cos x as a product

-2sin(x) * sin(2x)

or any equivalent form, such as 4cos(x)[cos²(x)-1]

Step-by-step explanation:

simplify cos(3x) first:

cos(3x) = cos(2x+x) = cos(2x)cos(x) -sin(2x)sin(x)

using trig identities

= [2cos²(x)-1]cos(x) - [2sin(x)cos(x)]sin(x)

= 2cos³(x) - cos(x) - 2sin²(x)cos(x)

substituting using trig identity sin²(x) + cos²(x) = 1

2cos³(x) - cos(x) - 2[1-cos²(x)]cos(x)

2cos³(x) - cos(x) - 2cos(x)+2cos³(x)

4cos³(x) - 3cos(x)

remember this cos(3x), we still have to subtract cos(x)

4cos³(x) - 3cos(x) - cos(x) = 4cos³(x) - 4cos(x)

we can factor 4cos(x) to write this as a product of:

4cos(x)[cos²(x)-1]

further simplification if you want

trig identity sin²(x) + cos²(x) = 1

simplifying: sin²(x) = 1-cos²(x)

simplifying: -sin²(x) = cos²(x)-1

4cos(x)[cos²(x)-1]

4cos(x)[-sin²(x)]

-4cos(x)sin²(x)

trig identity: sin(2a) = 2cos(a)sin(a)

-2sin(x) * 2cos(x)sin(x)

-2sin(x)*sin(2x)

the answer is a

step-by-step explanation:

because if you put 1 in to the square root of x it equals 300.