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Exercise 3.4 · Q68

Q.Prove the following: sin⁡18°cos⁡39°+sin⁡6°cos⁡15°=sin⁡24°cos⁡33°\sin18°\cos39°+\sin6°\cos15°=\sin24°\cos33°

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Step 1: Convert each product to a sum using 2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B)2\sin A\cos B=\sin(A+B)+\sin(A-B), halved: sin⁡18°cos⁡39°=12[sin⁡57°+sin⁡(−21°)]=12(sin⁡57°−sin⁡21°)\sin18°\cos39°=\dfrac12[\sin57°+\sin(-21°)]=\dfrac12(\sin57°-\sin21°).

Step 2: Similarly sin⁡6°cos⁡15°=12[sin⁡21°+sin⁡(−9°)]=12(sin⁡21°−sin⁡9°)\sin6°\cos15°=\dfrac12[\sin21°+\sin(-9°)]=\dfrac12(\sin21°-\sin9°).

Step 3: Sum: 12(sin⁡57°−sin⁡21°+sin⁡21°−sin⁡9°)=12(sin⁡57°−sin⁡9°)\dfrac12(\sin57°-\sin21°+\sin21°-\sin9°)=\dfrac12(\sin57°-\sin9°). …

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