Is Henri right about this?

Geometry Level 3

Henri makes the following statement:

∃ \exists no θ ∈ R \theta \in \mathbb{R} such that ∣ b ⋅ sin ⁡ ( π 2 − θ ) ∣ ≠ b 2 − a 2 \left|b\cdot\sin{\left(\dfrac{\pi}{2}-\theta\right)}\right| \neq \sqrt{b^2-a^2} when sin ⁡ θ = a b . \sin{\theta} = \dfrac{a}{b}.

Is Henri correct?

No Yes Cannot be determined

Akeel Howell by Akeel Howell , 22, USA

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1 solution

Akeel Howell
Feb 2, 2017

Assume that sin ⁡ ( π 2 − θ ) ≠ b 2 − a 2 b . \sin{\left(\dfrac{\pi}{2}-\theta\right)} \neq \dfrac{\sqrt{b^2-a^2}}{b}.

sin ⁡ ( π 2 − θ ) = cos ⁡ θ . \sin{\left(\dfrac{\pi}{2}-\theta\right)} = \cos{\theta}.

sin ⁡ 2 θ + cos ⁡ 2 θ = 1 ⟹ cos ⁡ 2 θ = 1 − sin ⁡ 2 θ = 1 − a 2 b 2 . \sin^2{\theta}+\cos^2{\theta} = 1 \implies \cos^2{\theta} = 1 - \sin^2{\theta} = 1- \dfrac{a^2}{b^2}.

So ∣ cos ⁡ θ ∣ = b 2 b 2 − a 2 b 2 = b 2 − a 2 b 2 {|\cos{\theta}}| = \sqrt{\dfrac{b^2}{b^2}-\dfrac{a^2}{b^2}} = \sqrt{\dfrac{b^2-a^2}{b^2}}

∴ ∣ cos ⁡ θ ∣ = b 2 − a 2 b ⟹ ∣ b ⋅ cos ⁡ θ ∣ = b 2 − a 2 . \therefore |\cos{\theta}| = \dfrac{\sqrt{b^2-a^2}}{b}\\ \implies \left|b\cdot\cos{\theta}\right| = \sqrt{b^2-a^2}.

So Henri was correct.

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Note that cos ⁡ 2 θ = ∣ cos ⁡ θ ∣ \sqrt{\cos^2\theta} = |\cos\theta| , not necessarily simply cos ⁡ θ \cos\theta .

Brian Moehring - 4 years, 4 months ago

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updated the solution.

Akeel Howell - 4 years, 4 months ago

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