First slide

Trigonometric Identities

Question

$\cos θ + \cos^{2} θ = 1, a \sin^{12} θ + b \sin^{10} θ + c \sin^{8} θ + d \sin^{6} θ = 1 \Rightarrow \frac{b + c}{a + d} =$

Moderate

A

0
B

1
C

2
D

3

Solution

$\cos θ = 1 - \cos^{2} θ \Rightarrow \cos θ = \sin^{2} θ \Rightarrow \cos^{2} θ = \sin^{4} θ \Rightarrow (\sin^{4} θ + \sin^{2} θ) = 1$
cubing on both sides ${(\sin^{2} x + \sin^{4} x)}^{3} = 1 \Rightarrow \sin^{12} x + 3 \sin^{10} x + 3 \sin^{8} x + \sin^{6} x = 1$
a=1,b=3,c=3,d=1

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