8sr6 + 2sr2×2×3 +5sr3 - sr2×3×3×3
=8sr6 + 2(2)sr3 + 5sr3 - 3sr2×3
=8sr6 + 4sr3 + 5sr3 - 3sr6
=8sr6 - 3sr6 + 4sr3 +5sr3
=5sr6 + 9sr3 (ans)
To solve this problem, you have to simplify your answer.
You may also look online to solve your problem.
After simplifying your question, The answer to your question is:
x=9/7.
I hope this helps. This answer is varified.
Q(3) = -2(3)+2 = -4
r(q(3)) = (-4)^2-1 = 15
Recall the sum identity for cosine:
cos(a + b) = cos(a) cos(b) - sin(a) sin(b)
so that
cos(a + b) = 12/13 cos(a) - 8/17 sin(b)
Since both a and b terminate in the first quadrant, we know that both cos(a) and sin(b) are positive. Then using the Pythagorean identity,
cos²(a) + sin²(a) = 1 ⇒ cos(a) = √(1 - sin²(a)) = 15/17
cos²(b) + sin²(b) = 1 ⇒ sin(b) = √(1 - cos²(b)) = 5/13
Then
cos(a + b) = 12/13 • 15/17 - 8/17 • 5/13 = 140/221
