If 5θ = 5cosθ - 20 ^ 3 θ + a ^ 5 θ + b , then the value of a + b is equal to

Correct answer: B. 16 5θ = (4θ + θ) = 4θ cos θ - 4θ sin θ = (2 [2co s ^ 2 θ - 1] ^ 2 - 1)cos θ - 2.2sin θ cos θ[2co s ^ 2 θ - 1]sin θ = cos θ[2(4 ^ 4 θ + 1 - 4 ^ 2 θ) - 1] - 4 ^ 2 θ cos θ[2 ^ 2 θ - 1] = 8 ^ 5 θ - 8 ^ 3 θ + 2cos θ - cos θ - 4(1 - ^ 2 θ)cos θ(2 ^ 2 θ - 1) = 8 ^ 5 θ - 8 ^ 3 θ + cos θ - 4cos θ [2 ^ 2 θ - 1 - 2 ^ 4 θ + ^ 2 θ] = 16 ^ 5 θ - 20 ^ 3 θ + 5cos θ Clearly, a = 16, b = 0

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

NTA Abhyas 2020 · Mathematics · Trigonometric Ratios & Identities

If 5θ = 5cosθ - 20 ^ 3 θ + a ^ 5 θ + b , then the value of a + b is...

\cos 5\theta = 5cos\theta - 20\cos^{3} \theta + a\cos^{5} \theta + b

, then the value of

a + b

is equal to

Options

A $20$
B $16$
C $- 16$
D $15$

Correct answer

B. $16$

Step-by-step solution

\cos 5\theta = \cos \left(4\theta + \theta\right)

= \cos 4\theta cos \theta - \sin 4\theta sin \theta

= \left(2{\left[2co{s}^{2}\theta - 1\right]}^{2} - 1\right)cos \theta - 2.2sin \theta cos \theta\left[2co{s}^{2}\theta - 1\right]sin \theta

= cos \theta\left[2\left(4\cos^{4} \theta + 1 - 4\cos^{2} \theta\right) - 1\right]

- 4\sin^{2} \theta cos \theta\left[2\cos^{2} \theta - 1\right]

= 8\cos^{5} \theta - 8\cos^{3} \theta + 2cos \theta - cos \theta

- 4\left(1 - \cos^{2}\theta\right)cos \theta\left(2\cos^{2} \theta - 1\right)

= 8\cos^{5} \theta - 8\cos^{3} \theta + cos \theta - 4cos \theta

\left[2\cos^{2} \theta - 1 - 2\cos^{4} \theta + \cos^{2} \theta\right]

= 16\cos^{5} \theta - 20\cos^{3} \theta + 5cos \theta

Clearly,

a = 16, b = 0

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