Trigo Time
Geometry
Level
2
Given:
(sin^2) x + 2(sin x)(cos x) + (cos^2) x = 9
Then, what is:
(sin^4) x + 2(sin x)(cos x) + (cos^4) x ?
15
-5
23
-23
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1 solution
(sin^4) x + 2(sin x)(cos x) + (cos^4) x -------> [(sin^2)x + (cos^2)x]^2 - 2(sin^2)x (cos^2)x + 2(sin x)(cos x)
By:
(sin^2) x + 2(sin x)(cos x) + (cos^2) x = 9 ,
1 + 2(sin x)(cos x) = 9
(sin x)(cos x) = 4 -------> 2(sin^2)x (cos^2)x = (4^2)*2 = 32
Then,
(sin^4) x + 2(sin x)(cos x) + (cos^4) x -------> [(sin^2)x + (cos^2)x]^2 - 2(sin^2)x (cos^2)x + 2(sin x)(cos x) ------> (1)^2 - 32 + 8 = 1 - 32 + 8 = -23
Final Answer: -23