Prove that the following identities :2(sin6 θ – cos6 θ) – 3(si

1.	Prove that the following identities :2(sin6 θ – cos6 θ) – 3(sin4 θ + cos4 θ ) + (sin2 θ + cos2 θ)
Answer» Taking LHS = 2(sin⁶ θ – cos⁶ θ) – 3(sin⁴ θ+ cos⁴ θ ) + (sin² θ + cos² θ) = 2[(sin² θ)³ – (cos² θ)³ ] – 3[(sin² θ)² + (cos² θ)² ]+1 [∵ cos² θ + sin² θ = 1] Now, we use these identities, (a³ – b³)= (a + b)³ – 3ab(a + b) and (a² + b²) = (a +b)² – 2ab] = 2[(sin² θ + cos² θ)³ – 3sin²θ cos²θ (sin² θ+ cos² θ)] –3[(sin² θ + cos² θ) – 2 sin² θ cos² θ]+ 1 =2[(1) – 3sin²θ cos²θ (1)] – 3[(1) – 2 sin² θ cos² θ] + 1 [∵ cos² θ + sin² θ = 1] =2(1 – 3 sin² θ cos² θ )– 3 + 6sin² θ cos² θ+ 1 = 2– 6sin²θ cos²θ – 2 + 6sin² θ cos² θ = 0 = RHS Hence Proved

Prove that the following identities :2(sin6 θ – cos6 θ) – 3(sin4 θ + cos4 θ ) + (sin2 θ + cos2 θ)

Answer»

Taking LHS

= 2(sin⁶ θ – cos⁶ θ) – 3(sin⁴ θ+ cos⁴ θ ) + (sin² θ + cos² θ)

= 2[(sin² θ)³ – (cos² θ)³ ] – 3[(sin² θ)² + (cos² θ)² ]+1 [∵ cos² θ + sin² θ = 1]

Now, we use these identities, (a³ – b³)= (a + b)³ – 3ab(a + b) and (a² + b²) = (a +b)² – 2ab]

= 2[(sin² θ + cos² θ)³ – 3sin²θ cos²θ (sin² θ+ cos² θ)] –3[(sin² θ + cos² θ) – 2 sin² θ cos² θ]+ 1

=2[(1) – 3sin²θ cos²θ (1)] – 3[(1) – 2 sin² θ cos² θ] + 1 [∵ cos² θ + sin² θ = 1]

=2(1 – 3 sin² θ cos² θ )– 3 + 6sin² θ cos² θ+ 1

= 2– 6sin²θ cos²θ – 2 + 6sin² θ cos² θ

= 0

= RHS

Hence Proved