Mixed Trig.

Geometry Level 3

If sin ⁡ θ = x cos ⁡ ϕ \sin\theta = x\cos\phi and cos ⁡ θ = y sin ⁡ ϕ \cos\theta = y\sin\phi .

What is the value of cos ⁡ 2 ϕ = ? \cos^2\phi= ?

\quad A. x − y 2 x 2 − y 2 \quad \dfrac{x- y^2}{x^2-y^2}

\quad B. 1 − y 2 x 2 − y 2 \quad \dfrac{1-y^2}{x^2-y^2}

\quad C. 1 − y 2 x 2 + y 2 \quad \dfrac{1-y^2}{x^2+y^2}

\quad D. 1 + y 2 x 2 + y 2 \quad \dfrac{1+y^2}{x^2+y^2}

\quad E. ( 1 + y 2 ) y 2 x 2 − y 2 \quad \dfrac{(1+y^2)y^2}{x^2-y^2}

E B C D A

Hana Wehbi by Hana Wehbi , 51, USA

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1 solution

{ sin ⁡ θ = x cos ⁡ ϕ ⟹ sin ⁡ 2 θ = x 2 cos ⁡ 2 ϕ . . . ( 1 ) cos ⁡ θ = y sin ⁡ ϕ ⟹ cos ⁡ 2 θ = y 2 sin ⁡ 2 ϕ . . . ( 2 ) \begin{cases} \sin \theta = x \cos \phi & \implies \sin^2 \theta = x^2 \cos^2 \phi & ...(1) \\ \cos \theta = y \sin \phi & \implies \cos^2 \theta = y^2 \sin^2 \phi & ...(2) \end{cases}

From ( 1 ) + ( 2 ) (1)+(2) :

sin ⁡ 2 θ + cos ⁡ 2 θ = x 2 cos ⁡ 2 ϕ + y 2 sin ⁡ 2 ϕ Note that sin ⁡ 2 θ + cos ⁡ 2 θ = 1 ⟹ x 2 cos ⁡ 2 ϕ + y 2 sin ⁡ 2 ϕ = 1 x 2 cos ⁡ 2 ϕ + y 2 ( 1 − cos ⁡ 2 ϕ ) = 1 x 2 cos ⁡ 2 ϕ − y 2 cos ⁡ 2 ϕ = 1 − y 2 ⟹ cos ⁡ 2 ϕ = 1 − y 2 x 2 − y 2 \begin{aligned} \color{#3D99F6} \sin^2 \theta + \cos^2 \theta & = x^2 \cos^2 \phi + y^2 \sin^2 \phi & \small \color{#3D99F6} \text{Note that } \sin^2 \theta + \cos^2 \theta = 1 \\ \implies x^2 \cos^2 \phi + y^2 \sin^2 \phi & = \color{#3D99F6} 1 \\ x^2 \cos^2 \phi + y^2 \left(1-\cos^2 \phi \right) & = 1 \\ x^2 \cos^2 \phi - y^2\cos^2 \phi & = 1-y^2 \\ \implies \cos^2 \phi & = \boxed{\dfrac {1-y^2}{x^2-y^2}} \end{aligned}

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Thank you for the nice solution.

Hana Wehbi - 3 years, 11 months ago

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