Question

Prove that :

$2(Sin^6\theta + Cos^6\theta)- 3 ( Sin^4\theta + Cos^4\theta) + 1 = 0$

Answer 1

⭕️Sin6Φ=(sin²Φ)³

Similarly....

⭕️ cos6Φ=(cos²Φ)³

⭕️Now multiply expand (sin²Φ)² and (cos²Φ)²
using the identity (A+B)²=A²+2AB+B²

and simplify...

⭕️you will get answer as 2-3+1 where terms will get cancelled out !

Finally we get 0 as the answer!
hence proved !!!

$\huge\bf{\boxed {\red{\mathfrak{thank \: you :)}}}}$

Answer 2

Answer:

0

Step-by-step explanation:

Given Equation is 2(sin⁶θ + cos⁶θ) - 3(sin⁴θ + cos⁴θ) + 1

⇒ 2[(sin²θ)³ + (cos²θ)³] - 3(sin⁴θ + cos⁴θ) + 1

⇒ 2[(sin²θ + cos²θ)(sin⁴θ - sin²θcos²θ + cos⁴θ) - 3(sin⁴θ + cos⁴θ) + 1

⇒ 2[(1)(sin⁴θ - sin²θcos²θ + cos⁴θ)] - 3(sin⁴θ + cos⁴θ) + 1

⇒ 2(sin⁴θ - sin²θcos²θ + cos⁴θ) - 3(sin⁴θ + cos⁴θ) + 1

⇒ 2sin⁴θ - 2sin²θcos²θ + 2cos⁴θ - 3sin⁴θ - 3cos⁴θ + 1

⇒ -sin⁴θ - 2sin²θcos²θ - cos⁴θ + 1

⇒ -(sin⁴θ + 2sin²θcos²θ + cos⁴θ) + 1

⇒ -(sin²θ + cos²θ)² + 1

⇒ -(1)² + 1

⇒ -1 + 1

⇒ 0.

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