Question

1338

高一數學，求解謝謝

B₁Q
2. 設f(n) = sin"0+cos"0,試求3f(4)-2f(6)的值為
(A) 2 (B) 1 (C) O
D) -1 E-2
14

Answer 1

✨ 最佳解答 ✨

qn

( (a+b)²=a²+2ab+b² )
(sin²θ+cos²θ)²=sin⁴θ+2sin²θcos²θ+cos⁴θ
f(4)=sin⁴θ+cos⁴θ=1-2sin²θcos²θ

( a³+b³=(a+b)(a²-ab+b²) )
sin⁶θ+cos⁶θ=(sin²θ+cos²θ)(sin⁴θ-sin²θcos²θ+cos⁴θ)
f(6)=sin⁶θ+cos⁶θ=1-3sin²θcos²θ

3f(4)-2f(6)=(3-6sin²θcos²θ)-(2-6sin²θcos²θ)
=3-2
=1