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`        sinA+sin3A+SIN5A+sin7A `
4 years ago

## Answers : (7)

Latika Leekha
165 Points
```							Hello student,Your question seems to be incomplete as it does not state what exactly are we required to prove. Please check and post the complete question again so that we can provide you a meaningful answer.
```
4 years ago
VAISHNAVI BATHINA
130 Points
```							what is the value of angle A
```
4 years ago
2061 Points
```							MANIDEEPyou question literates that this is incomplete u look out the refrence else u will have club the first sinA and last sin 7A terms     APPROVE IF USEFUL
```
4 years ago
rahul
211 Points
```							take fthe first and last term and the middle first and middle last term to get the answer
```
4 years ago
Lab Bhattacharjee
121 Points
```							As the angles are in Arithmetic Series,  multiply the numerator & the denominator by 2sin(3A-A)/2 and use Werner's formula: 2sinA sinB=cos(A-B)-cos(A+B)
```
4 years ago
Athiyaman
57 Points
```							 Hey Manideep,the answer should be 16 sinA.cos22A.cos2Ayou will have to use these 3 formulas in order to get the answer:1)sinA+sinB=2sin(A+B)/2.cos(A-B)/22)cosA+cosB=2cos(A+B)/2.cos(A-B)/2 and lastly,3)sin2A=2sinA.cosA
```
4 years ago
Athiyaman
57 Points
```							 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 AHere 2sin4A is common=2 sin 4A [ cos 3A + cos A ]Use Identity 2 from the above post,=2 sin 4A * 2 cos 2A cos Asin4A can be split into 2sin2A.cos2A from identity 3,= 4 sin 2A cos 2A * 2 cos 2A cos AAgain use sin2A formula= 8 sin A * cos A *cos 2A * 2 cos 2A cos Aand then just arrange the terms= 16 sin A cos² A * cos² (2A) Please approve the answer if you found it helpful
```
4 years ago
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### Course Features

• 31 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions