RRB NTPC · Question 63
2(sin^6θ + cos^6θ) - 3(sin4θ + cos4θ) is equal to:
Show answer & explanation
Answer: (B) -1
Explanation
sin⁶θ + cos⁶θ = 1 − 3sin²θcos²θ and sin⁴θ + cos⁴θ = 1 − 2sin²θcos²θ. So the expression = 2 − 6s²c² − 3 + 6s²c² = −1.
RRB NTPC · Question 63
Answer: (B) -1
sin⁶θ + cos⁶θ = 1 − 3sin²θcos²θ and sin⁴θ + cos⁴θ = 1 − 2sin²θcos²θ. So the expression = 2 − 6s²c² − 3 + 6s²c² = −1.