Question

prove that 2(sin*6Φ+cos*6Φ)-3(sin*4Φ+cos4Φ)+1=0​

Answer 1

Answer:

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Answer 2

Answer:

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Step-by-step explanation:

2(sin6θ + cos6θ) - 3 (sin4θ + cos4θ) + 1  

=2[(sin2θ)3 + (cos2θ)3] -3[(sin2θ)2 + (cos2θ)2]+1

=2[(sin2θ + cos2θ)3 -3sin2θcos2θ (sin2θ + cos2θ)] -3[(sin2θ + cos2θ)2

-2sin2θcos2θ]+1

The algebraic identity

a3 + b3 = (a+b)3 - 3ab(a+b) and  

a2 + b2 = (a+b)2 - 2ab

are used in the above step where

a = sin2θ and b = cos2θ.

writing sin2θ + cos2θ = 1, we have  

= 2[1-3 sin2θ cos2θ] -3[-2 sin2θcos2θ] + 1

= 2-6 sin2θcos2θ -3 + 6 sin2θcos2θ + 1

= -3+3=0