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Algebra Difficulty 2.5 Junior Find the answer

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}, what is the value of θ\theta?

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Solution

Since cos⁡60∘=12\cos 60^{\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: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.