f(x)=根号3sin2x+cos2x,求f(π╱8)
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咨询记录 · 回答于2024-01-03
f(x)=根号3sin2x+cos2x,求f(π%q8)
亲, f(π/8) = (√6 + √2)/2过程如下:
1. 将 π/8 代入 x,得
f(π/8) = √3sin(2×π/8) + cos(2×π/8)
2. 化简上式,得
f(π/8) = √3sin(π/4) + cos(π/4)
3. 因为 sin(π/4) = cos(π/4) = √2/2,代入上式得
f(π/8) = √3×√2/2 + √2/2
4. 继续化简上式,得
f(π/8) = √6/2 + √2/2
5. 最后化简上式,得
f(π/8) = (√6 + √2)/2
所以 f(π/8) = (√6 + √2)/2。