cos4θ-cos2θ/sin4θ+sin2θ

Answers:
A. cotθtanθ
B. -cotθ
C. -tanθ
D. -1

Recall the double angle identities:

cos(2θ) = 1 - 2 sin²(θ)

sin(2θ) = 2 sin(θ) cos(θ)

With some rewriting, we can get

cos(4θ) - cos(2θ) = 2 cos²(2θ) - 1 - cos(2θ) = (2 cos(2θ) + 1) (cos(2θ) - 1)

and

sin(4θ) + sin(2θ) = 2 sin(2θ) cos(2θ) + sin(2θ) = sin(2θ) (2 cos(2θ) + 1)

The numerator and denominator both share a factor of 2 cos(2θ) + 1; canceling them leaves us with

(cos(2θ) - 1) / sin(2θ)

Replace cos(2θ) = 1 - 2 sin²(θ) and sin(2θ) = 2 sin(θ) cos(θ) :

((1 - 2 sin²(θ)) - 1) / (2 sin(θ) cos(θ))

-2 sin²(θ) / (2 sin(θ) cos(θ))

- sin(θ) / cos(θ)

- tan(θ)

so the answer is C.

(cos4θ-cos2θ)/(sin4θ+sin2θ)

C. -tanθ is your answer.

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !

answer:

your answer is: c. 196 7/8 in^3.