????
Asked by sejal2208
| 11th Feb, 2010,
06:30: PM
Dear Student
a3+b3=(a+b)3-3ab(a+b)
sin 6 θ + cos 6 θ = (sin 2 θ + cos 2 θ )3 -3 sin 2 θ cos 2 θ(sin 2 +θ cos 2 θ)=1- 3sin 2 θ cos 2 θ
2 (sin 6 θ + cos 6 θ )= 2- 6sin 2 θ cos 2 θ-------(1)
a2+b2=(a+b)2-2ab
(sin 4 θ + cos 4 θ )= (sin 2 θ + cos 2 θ )2 - 2sin 2 θ cos 2 θ=1-2sin 2 θ cos 2 θ
3 (sin 4 θ + cos 4 θ )=3- 6sin 2 θ cos 2 θ---------(2)
(2)-(1) ===> 3 (sin 4 θ + cos 4 θ ) - 2 ( sin 6 θ + cos 6 θ ) = 1
Regards
Team
Topperlearning
Answered by
| 11th Feb, 2010,
07:01: PM
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Sign Up
Verify mobile number
Enter the OTP sent to your number
Change