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```
Prove that cos^6 A - sin^6 A = cos 2A(1-1sin^2 2A)........

```
one year ago

Saurabh Koranglekar
10233 Points
```							Dear studentLHSa^ 6 - b^ 6 = ( a^ 2 - b^ 2)(a^ 4 + b^4 + (ab)^2)here LHS = cos2A ( sin^4A + cos^4A + ( sinA cos A)^2)Adjusting second bracket a^ 2 + b^2 + 2ab -2ab + ab = (a+b)^2 -abWe getCos 2A*( 1- sin^2(2A)) = RHSRegards
```
one year ago
Arun
25256 Points
```							2a)2a/4 )= cos2a(1-1/4sin2a*sin2a )= (cos2a)(1 – 4cos2a*sin2a)= (cos2a)(1 – cos2a*sin2a + cos2a*sin2a)= (cos2a)(1 – 2 cos2a*sin2a + cos4a + sin4a )(cos2a – sin2= (cos3a)2 – (sin3a)2cos
```
one year 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