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SINA+SIN3A+SIN5A+SIN7A

SINA+SIN3A+SIN5A+SIN7A
 

Grade:11

3 Answers

Athiyaman
57 Points
8 years ago
Hey Manideep,
the answer should be 16 sinA.cos22A.cos2A
you will have to use these 3 formulas in order to get the answer:
1)sinA+sinB=2sin(A+B)/2.cos(A-B)/2
2)cosA+cosB=2cos(A+B)/2.cos(A-B)/2 and lastly,
3)sin2A=2sinA.cosA
Athiyaman
57 Points
8 years ago
And if you want the answer here it is:
 sinA + sin5A + sin3A + sin7A
= (sinA + sin7A) + (sin5A + sin3A)
After rearranging the terms use identity 1 from the above post,
= 2 sin 4A cos 3A + 2 sin 4A cos A
Here 2sin4A is common
=2 sin 4A [ cos 3A + cos A ]
Use Identity 2 from the above post,
=2 sin 4A * 2 cos 2A cos A
sin4A can be split into 2sin2A.cos2A from identity 3,
= 4 sin 2A cos 2A * 2 cos 2A cos A
Again use sin2A formula
= 8 sin A * cos A *cos 2A * 2 cos 2A cos A
and then just arrange the terms
= 16 sin A cos² A * cos² (2A)
Please approve the answer if you found it helpful
Lab Bhattacharjee
121 Points
8 years ago
http://www.askiitians.com/forums/Trigonometry/sina-sin3a-sin5a-sin7a_125330.htm

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