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Question : 21 of 160
Marks:
+1,
-0
Solution:
Consider the trigonometric function.
| sin‌9‌θ |
| cos‌27‌θ |
++ =k(tan‌27‌θ−tan‌θ) Consider the left hand side of the function.
LHS
=| sin‌9‌θ |
| cos‌27‌θ |
++ =| sin‌9‌θ‌cos‌9‌θ+sin‌3‌θ‌cos‌27‌θ |
| cos‌9‌θ‌cos‌27‌θ |
+ =| sin‌18‌θ+sin‌30‌θ−sin‌24‌θ |
| 2‌cos‌9‌θ‌cos‌27‌θ |
+ Solve further,
LHS
=[ | sin‌15‌θ+sin‌33‌θ+4‌sin‌θ‌cos‌9‌θ‌cos‌27‌θ |
| 4‌cos‌3‌θ‌cos‌9‌θ‌cos‌27‌θ |
] =[| 2‌sin‌24‌θ‌cos‌9‌θ+4‌sin‌θ‌cos‌9‌θ‌cos‌27‌θ |
| 4‌cos‌3‌θ‌cos‌9‌θ‌cos‌27‌θ |
] =[| sin‌24‌θ+2‌sin‌θ‌cos‌27‌θ |
| 2‌cos‌3‌θ‌cos‌27‌θ |
] =[| sin(27θ−3θ) |
| 2‌cos‌3‌θ‌cos‌27‌θ |
+] Solve further,
LHS
=[| sin‌27‌θ |
| cos‌27‌θ |
−]+ =| sin‌27‌θ |
| cos‌27‌θ |
−(| sin‌3‌θ−2‌sin‌θ |
| cos‌3‌θ |
) =‌tan‌27‌θ−(| 3‌sin‌θ−4sin3θ−2‌sin‌θ |
| cos‌θ(4cos2θ−3) |
) = [tan‌27‌θ−| sin‌θ(1−4sin2θ) |
| cos‌θ(1−4sin2θ) |
] Solve further,
LHS=[tan‌27‌θ−tan‌θ] Compare with RHS.
k=
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