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Question : 190 of 200
Marks:
+1,
-0
Solution:
‌‌=‌‌=‌| sin‌3θ‌cos‌2‌θ+cos‌3‌θ‌s‌i‌n‌2θ |
| sin‌θ |
‌=‌| (3sin‌θ−4sin‌3θ)(cos‌2‌θ)+2‌cos‌3‌θ⋅sin‌θ‌cos‌θ |
| sin‌θ |
‌=(3−4sin‌2θ)(2cos2θ−1)‌=[3−4(1−cos2θ)](2cos2θ−1)+8cos4θ−6cos2θ‌=[4cos2θ−1](2cos2θ−1)+8cos4θ−6cos2θ‌=8cos4θ−6cos2θ+1+8cos4θ−6cos2θ+1‌=16cos4θ−12cos2θ+1
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