© examsiri.com
Question : 9 of 160
Marks:
+1,
-0
Solution:
We have,
‌cos‌5‌θ=cos(4θ+θ)‌=cos‌4‌θ‌cos‌θ−sin‌4θsin‌θ‌=(2cos22θ−1)‌cos‌θ−2sin‌2θ‌cos‌2‌θ‌s‌i‌n‌θ
‌=2‌cos‌θ[2cos2θ−1]2−cos‌θ−2(2sin‌θ‌cos‌θ)
‌‌‌(2cos2θ−1)sin‌θ ‌=2‌cos‌θ[4cos4θ+1−4cos2θ]‌‌‌−cos‌θ−4sin‌2θ‌cos‌θ(2cos2θ−1)‌=8cos5θ+2‌cos‌θ−8cos3θ−cos‌θ‌‌‌−(8cos3θ−4‌cos‌θ)sin‌2θ‌=8cos5θ+cos‌θ−8cos3θ‌‌‌−(8cos3θ−4‌cos‌θ)(1−cos2θ)‌=8cos5θ+cos‌θ−8cos3θ‌‌‌−8cos3θ+4‌cos‌θ+8cos5θ−4cos3θ‌=16cos5θ−20cos3θ+5‌cos‌θ
© examsiri.com
Go to Question: