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Naily [24]
3 years ago
9

If sin θ = cot θ, then, what is the value of cos θ + 2 cos^2 θ + 2 cos^3 θ + cos^ 4 θ ?

Mathematics
1 answer:
VARVARA [1.3K]3 years ago
7 0

Notice that

cos(θ) + 2 cos²(θ) + 2 cos³(θ) + cos⁴(θ)

= cos(θ) (1 + 2 cos(θ) + 2 cos²(θ) + cos³(θ))

= cos(θ) ([1 + cos(θ)] + [cos(θ) + cos²(θ)] + [cos²(θ) + cos³(θ)])

= cos(θ) ([1 + cos(θ)] + [cos(θ) (1 + cos(θ))] + [cos²(θ) (1 + cos(θ))])

= cos(θ) (1 + cos(θ)) (1 + cos(θ) + cos²(θ))

Given that sin(θ) = cot(θ), by definition of cotangent this tells us that

sin(θ) = cos(θ)/sin(θ)   ⇒   cos(θ) = sin²(θ)

and by the Pythagorean identity

cos²(θ) + sin²(θ) = 1

it follows that

cos(θ) = sin²(θ) = 1 - cos²(θ)

Substituting these results into the factorization above gives

cos(θ) (1 + cos(θ)) (1 + cos(θ) + cos²(θ))

= cos(θ) (1 + cos(θ)) (1 + [1 - cos²(θ)] + cos²(θ))

= 2 cos(θ) (1 + cos(θ))

= 2 sin²(θ) (1 + cos(θ))

= 2 (1 - cos²(θ)) (1 + cos(θ))

= 2 (1 + cos(θ) - cos²(θ) - cos³(θ))

= 2 (cos(θ) + cos(θ) - cos³(θ))

= 2 (2 cos(θ) - cos³(θ))

= 2 cos(θ) (2 - cos²(θ))

= 2 cos(θ) (1 + cos(θ))

= 2 (cos(θ) + cos²(θ))

= 2 (1 - cos²(θ) + cos²(θ))

= 2

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