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2(sin6x + cos6x) 3(sin4x + cos4x) +1 = 0

LHS of the given equation is:
2(sin6x+cos6x)-3(sin4x+cos4x)+1=2[(sin2x)3+(cos2x)3]-3(sin4x+cos4x)+1=2[(sin2x+cos2x)(sin4x+cos4x-sin2xcos2x)]-3(sin4x+cos4x)+1=2(sin4x+cos4x-sin2xcos2x)-3(sin4x+cos4x)+1=-sin4x-cos4x-2sin2xcos2x+1=-[(sin2x)2+(cos2x)2+2sin2xcos2x]+1=-(sin2x+cos2x)2+1=-12+1=-1+1=0
= RHS
hope this helps you

  • 96
View Full Answer

2(sin6x + cos6x) - 3(sin4x + cos4x)+1

=2[(sin2x)3+ (cos2x)3] - 3(sin4x + cos4x)+1

= 2(sin2x + cos2x)(sin4x - sin2xcos2x + cos4x)-3(sin4x + cos4x)+1 [a3+b3=(a+b)(a2-ab+b2)]

= 2(sin4x - sin2xcos2x + cos4x) - 3(sin4x + cos4x)+1 [As sin2x + cos2x=1]

= 2(sin4x + cos4x) - 2sin2xcos2x - 3(sin4x + cos4x)+1

= -(sin4x + cos4x) - 2sin2xcos2x+1

= -(sin4x + cos4x + 2sin2xcos2x)+1

= -[(sin2x)2+ (cos2x)2+ 2(sin2x)(cos2x)]+1

= -(sin2x + cos2x)2+1 [As a2+b2+2ab=(a+b)2]

= -(1)2+1 [As sin2x + cos2x=1]

= -1+1 =0

Hence, 2(sin6x + cos6x) - 3(sin4x + cos4x)+1=0

= -

=-(

=

  • 48

maybe you've missed a minus in between.2(sin^6x + cos^6x) - 3(sin^4x + cos^4x) + 1= 2[ (sin^2x)^3 + (cos^2x)^3) - 3[ (sin^2x + cos^2x) ^2 - 2 sin^2xcos^2x] + 1

Now a^3 + b^3 = (a+b)(a^2-ab + b^2) and sin^2x + cos^2x = 1. Using these,( a = sin^2x , b = cos^2x)= 2 [ (sin^2x + cos^2x) ( sin^4x + cos^4x - sin^2xcos^2x)] - 3(1 - 2sin^2x cos^2x) + 1

Now note that sin^4x + cos^4x - sin^2xcos^2x = (sin^2x + cos^2x) ^2 - 3 sin^2x cos^2xand using sin^2x + cos^2x = 1 we get

2( 1. (sin^2x + cos^2x)^2 - 3 sin^2x cos^2x ) - 3( 1 - 2 sin^2x cos^2x) + 1= 2 ( 1 - 3 sin^2xcos^2x) - 3(1- 2 sin^2xcos^2x) + 1= 2 - 6sin^2xcos^2x - 3 + 6sin^2xcos^2x + 1 = 0. which is true, proved.

  • 11
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