Prove that: 2(sin6θ + cos6θ) - 3(sin4θ + cos4θ) + 1 = 0

1.	Prove that: 2(sin6θ + cos6θ) - 3(sin4θ + cos4θ) + 1 = 0
Answer» 2(sin⁶θ + cos⁶θ) - 3(sin⁴θ + cos⁴θ) + 1 = 0 2((sin³)²θ + (cos³)²θ) - 3((sin²)²θ + (cos²)²θ) + 1 = 0 2(1) - 3[1/(cos²)²+ (cos²)²] + 1 = 0 2 - 3 + 1 = 0 3 - 3 = 0 0 = 0

Answer»

2(sin⁶θ + cos⁶θ) - 3(sin⁴θ + cos⁴θ) + 1 = 0

2((sin³)²θ + (cos³)²θ) - 3((sin²)²θ + (cos²)²θ) + 1 = 0

2(1) - 3[1/(cos²)²+ (cos²)²] + 1 = 0

2 - 3 + 1 = 0

3 - 3 = 0

0 = 0

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Prove that: 2(sin6θ + cos6θ) - 3(sin4θ + cos4θ) + 1 = 0

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