How to prove the identity \(\displaystyle{3}{{\sin}^{{4}}{x}}-{2}{{\sin}^{{6}}{x}}={1}-{3}{{\cos}^{{4}}{x}}+{2}{{\cos}^{{6}}{x}}\)

Vanessa Mccarty

Vanessa Mccarty

Answered question

2022-04-06

How to prove the identity
3sin4x2sin6x=13cos4x+2cos6x

Answer & Explanation

wyjadaczeqa8

wyjadaczeqa8

Beginner2022-04-07Added 14 answers

3sin4x2sin6x=3(sin2x)22(sin2x)3
=(sin2x)2(32sin2x)
Now we can use the Pythagorean Identity:
sin2x+cos2x=1sin2x=1cos2x
3sin4x2sin6x=3(sin2x)22(sin2x)3
=(sin2x)2(32sin2x)
=(1cos2x)2(32(1cos2x))
=(12cos2x+cos4x)(1+2cos2x)
=13cos4x+2cos6x
tutaonana223a

tutaonana223a

Beginner2022-04-08Added 15 answers

L.H.S=3sin4x2sin6x
=3(sin2x)22sin2x)3
=3(1cos2x)22(1cos2x)3
=3(1+cos4x2cos2x)2(1cos6x3cos2x+3cos4x)
=3+3cos4x6cos2x2+2cos6x+6cos2x6cos4x)
=13cos4x+2cos6x
=R.H.S

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