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Miscellaneous Exercise 3 (II) · Q117

Q.Prove the following: sin⁡47°+sin⁡61°−sin⁡11°−sin⁡25°=cos⁡7°\sin47°+\sin61°-\sin11°-\sin25°=\cos7°

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Step 1: sin⁡61°−sin⁡11°=2cos⁡36°sin⁡25°\sin61°-\sin11°=2\cos36°\sin25° and sin⁡47°−sin⁡25°=2cos⁡36°sin⁡11°\sin47°-\sin25°=2\cos36°\sin11° (both sum-to-product with mean angle 36°36°).

Step 2: Sum =2cos⁡36°(sin⁡25°+sin⁡11°)=2\cos36°(\sin25°+\sin11°).

Step 3: sin⁡25°+sin⁡11°=2sin⁡18°cos⁡7°\sin25°+\sin11°=2\sin18°\cos7°, so the total is 2cos⁡36°⋅2sin⁡18°cos⁡7°=4sin⁡18°cos⁡36°cos⁡7°2\cos36°\cdot2\sin18°\cos7°=4\sin18°\cos36°\cos7°. …

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