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```
Sin theta + sin^2theta + sin ^3theta = 1 prove that cos^6theta - 4cos^4theta + 8cos^2theta=4

```
one year ago

## Answers : (2)

2069 Points
```							hello anurag,Given relation  sin x + sin 2 x + sin 3 x = 1  ⇒ sin x + sin 3 x = 1 − sin 2 x , now squaring ⇒ ( sin x + sin 3 x ) 2 = ( 1 − sin 2 x ) 2  ⇒ sin 2 x + sin 6 x + 2 sin 4 x = cos 4 x  ⇒ 1 − cos 2 x + ( 1 − cos 2 x ) 3 + 2 ( 1 − cos 2 x ) 2 = cos 4 x  ⇒ 1 − cos 2 x + 1 − 3 cos 2 x + 3 cos 4 x − cos 6 x + 2 − 4 cos 2 x + 2 cos 4 x = cos 4 x  ⇒ cos 6 x − 4 cos 4 x + 8 cos 2 x = 4kindly approve :)
```
one year ago
Vikas TU
13783 Points
```							Dear student Here is the correct explanation sinA + sin^2A + sin^3A = 1=> sinA + sin^3A = 1 - sin^2A=> sinA ( 1+sin^2A ) = cos​^2A=> sinA ( 1 + 1 - cos^2A ) = cos^2A=> sinA ( 2 - cos^2A ) = cos^2A=> sinA = cos^2A / ( 2 - cos^2A)=> sqrt ( 1 - cos^2A ) = cos^2A / 2- cos^2A=> 1 - cos^2A = ( cos^2A )2 /  ( 2 - cos^2A )2=> 1 - cos^2A = cos^4A / ( 4 + cos^4A - 4 cos^2A )=>  ( 1 - cos^2A ) ( 4 + cos^4A - 4 cos^2A ) = cos^4A=>  ( ​4 + cos^4A - 4 cos^2A -4cos​^2A - cos^6A + 4cos^4A = cos^4A=>  ( 4 -8cos^2A - cos^6A + 4cos^4A ) = cos^4A - cos^4A=>  ( 4 -8cos^2A - cos^6A + 4 cos^4A ) = 0=>  4 = 8cos^2A + cos^6A - 4cos^4A
```
one year ago
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• 31 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions