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if sinA +sin^2 A = 1 then prove that cos ^8 A + 2cos^6 A + cos ^4 A=1

if sinA +sin^2 A = 1 then prove that cos ^8 A + 2cos^6 A + cos ^4 A=1

Grade:11

3 Answers

Arun
25750 Points
6 years ago
We can also prove that
cos^8A + 2 cos^6A + cos^4A - 1 = 0
Hence
sinA+sin^2A=1
=>sinA=1-sin^2A=cos^2A
Now
cos^12A+3cos^10A+3cos^8A+cos^6A-1
Using cos^2A=sinA in this expression we get
sin^6A+cos^6A+3cos^10A+3cos^8A-1
=1-3sin^2Acos^2A+3cos^10A+3cos^8A-1
=-3sin^2Acos^2A+3cos^10A+3cos^8A
=-3sin^2Acos^2A+3sin^4A+3cos^10A
=3sin^2A(sin^2A-cos^2A)+3cos^10A
=-3sin^2Acos2A+3cos^10A
=-3sin^2Acos2A+3cos^4Acos^6A
=-3sin^2Acos2A+3sin^2Acos^6A
=3sin^2A(cos^6A-cos2A)
=3sin^2A(sin^2Acos^2A-cos2A)
=3sin^2A(cos^2Asin^2A-cos^2A+sin^2A)
=3sin^2A{cos^2A(sin^2A-1)+sin^2A}
=3sin^2A{-sinAcos^2A+sin^2A}
=3sin^2A{-sinAsinA+sin^2A}
=3sin^2A{=sin^2A+sin^2A}
=0
shaunak
19 Points
6 years ago
sinA+sin^2A=1
=>sinA=1-sin^2A=cos^2A
Now
cos^12A + 3cos^10A + 3cos^8A + cos^6A-1
Using cos^2A = sinA in this expression we get
sin^6A + cos^6A + 3cos^10A + 3cos^8A-1
=1 - 3sin^2 Acos^2A + 3cos^10A + 3cos^8A-1
=-3sin^2A cos^2A + 3cos^10A + 3cos^8A
=-3sin^2A cos^2A + 3sin^4A + 3cos^10A
=3sin^2A (sin^2A-cos^2A) + 3cos^10A
=-3sin^2A cos2A + 3cos^10A
=-3sin^2A cos2A + 3cos^4A cos^6A
=-3sin^2A cos2A + 3sin^2A cos^6A
=3sin^2A (cos^6A-cos2A)
=3sin^2A (sin^2A cos^2A - cos2A)
=3sin^2A (cos^2A sin^2A - cos^2A + sin^2A)
=3sin^2A {cos^2A (sin^2A-1) + sin^2A}
=3sin^2A {-sinA cos^2A + sin^2A}
=3sin^2A {-sinA sinA + sin^2A}
=3sin^2A {-sin^2A + sin^2A}
=0
Nikhileshsing Kailassing Rajput
11 Points
6 years ago
As, sina+sin^2a=1So, sina=cos^2aAs, cos^8a+2cos^6a+cos^4=sin^4a+2sin^2.cos^2a+cos^4a=(sin^2a+cos^2a)^2

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