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Question : 9 of 160
Marks:
+1,
-0
Solution:
‌ (a) ‌sin‌5θ=‌sin‌(3θ+2θ)
=‌sin‌3θ‌cos‌2‌θ+cos‌3‌θ⋅sin‌2θ
=‌(3sin‌θ−4sin‌3θ)⋅(1−2sin‌2θ)+(4cos3θ−3‌cos‌θ)⋅(2sin‌θ‌cos‌θ)
=‌3sin‌θ−6sin‌3θ−4sin‌3θ+8sin‌5θ+8sin‌θcos4θ−6sin‌θcos2θ
=3sin‌θ−10sin‌3θ+8sin‌5θ+8sin‌θcos4θ−6sin‌θcos2θ
=3sin‌θ−10sin‌3θ+8sin‌5θ+8sin‌θ(1−sin‌2θ)2−6sin‌θ(1−sin‌2θ)
=3sin‌θ−10sin‌3θ+8sin‌5θ+8sin‌θ(1+sin‌4θ−2sin‌2θ)−6sin‌θ+6sin‌3θ
=−3sin‌θ−4sin‌3θ+8sin‌5θ+8sin‌θ+8sin‌5θ−16sin‌3θ
‌=5sin‌θ−20sin‌3θ+16sin‌5θ
‌=5sin‌θ−20sin‌θ(1−cos2θ)+16(1−cos2θ)2sin‌θ
‌=5sin‌θ−20sin‌θ+20sin‌θcos2θ+16sin‌θ‌(1+cos4θ−2cos2θ)
‌=−15sin‌θ+20sin‌θcos2θ+16sin‌θ+16cos4θsin‌θ−32cos2θsin‌θ
‌=16cos4θsin‌θ−12cos2θsin‌θ+sin‌θ
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