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CEMC Fermat, 2015

Algebra Difficulty 2.4 Junior Find the answer Canada

If cos⁡60∘=cos⁡45∘cos⁡θ\cos 60^\circ = \cos 45^\circ \cos\theta with 0∘≤θ≤90∘0^\circ \leq \theta \leq 90^\circ, then θ\theta equals

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Solution

Since cos⁡60∘=12\cos60^\circ = \frac{1}{2} and cos⁡45∘=12\cos 45^\circ = \frac{1}{\sqrt{2}}, then the given equation cos⁡60∘=cos⁡45∘cos⁡θ\cos 60^\circ = \cos 45^\circ \cos \theta becomes 12=12cos⁡θ\frac{1}{2} = \frac{1}{\sqrt{2}}\cos \theta.

Therefore, cos⁡θ=22=12\cos \theta = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}.

Since 0∘≤θ≤90∘0^\circ \leq \theta \leq 90^\circ, then θ=45∘\theta = 45^\circ.

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.