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Q.Prove that sin⁡78∘+cos⁡132∘=5−14\sin 78^\circ + \cos 132^\circ = \dfrac{\sqrt{5}-1}{4}

Telangana TsbieTelangana Board of Intermediate Education (Intermediate 1st Year) 2024Subjective· 2mImportance★★★★★
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Rewrite cos⁡132∘\cos 132^\circ as −sin⁡42∘-\sin 42^\circ, combine with sin⁡78∘\sin 78^\circ using the sum-to-product formula, and the result collapses to the known value sin⁡18∘\sin 18^\circ.

Given: Show sin⁡78∘+cos⁡132∘=5−14\sin 78^\circ + \cos 132^\circ = \dfrac{\sqrt{5}-1}{4}.

Step 1. cos⁡132∘=cos⁡(180∘−48∘)=−cos⁡48∘\cos 132^\circ = \cos(180^\circ - 48^\circ) = -\cos 48^\circ, and cos⁡48∘=sin⁡42∘\cos 48^\circ = \sin 42^\circ (co-function identity).

So:

sin⁡78∘+cos⁡132∘=sin⁡78∘−sin⁡42∘\sin 78^\circ + \cos 132^\circ = \sin 78^\circ - \sin 42^\circ

Step 2. Use sin⁡C−sin⁡D=2cos⁡(C+D2)sin⁡(C−D2)\sin C - \sin D = 2\cos\left(\dfrac{C+D}{2}\right)\sin\left(\dfrac{C-D}{2}\right) with C=78∘,D=42∘C = 78^\circ, D = 42^\circ:

sin⁡78∘−sin⁡42∘=2cos⁡60∘sin⁡18∘=2⋅12⋅sin⁡18∘=sin⁡18∘\sin 78^\circ - \sin 42^\circ = 2\cos 60^\circ \sin 18^\circ = 2 \cdot \dfrac{1}{2} \cdot \sin 18^\circ = \sin 18^\circ

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