数学题求解答手写拍照 求解第九题，求手写拍照解答
\u4e24\u9053\u9ad8\u4e2d\u6570\u5b66\u9898\uff0c\u6c42\u89e3\u7b54\uff0c\u65b9\u4fbf\u7684\u8bdd\u624b\u5199\u8fc7\u7a0b\u62cd\u4e00\u4e0b\u8c22\u8c22\u5566\u7b2c\u4e8c\u9898\u7684\u7b2c\u4e00\u6b65\u6211\u4e5f\u8bf4\u4e0d\u660e\u767d\uff0c\u53ea\u80fd\u8bf4\u662f\u611f\u89c9😂
f(x)=2-8(sinx)^2*(cosx)^2
=2-2(2sinx*cosx)^2
=2-2(sin2x)^2
=2-(1-cos4x)
=1+cos4x
f(x)=1+cos4x
T=2\u03c0/4=\u03c0/2
2)g(x)=1+cos4(x+\u03c0/8)=1+cos(4x+\u03c0/2)=1-sin4x
h(x)=g(x)+f(x)=1-sin4x+1+cos4x=2+cos4x-sin4x
=2+\u221a2cos(4x+\u03c0/4)
x\u5728[0\uff0c\u03c0/4]
4x+\u03c0/4\u5728[\u03c0/4\uff0c5\u03c0/4]
cos(4x+\u03c0/4)\u7684\u503c\u57df\u4e3a[\uff0d1\uff0c\u221a2/2]
2+\u221a2cos(4x+\u03c0/4)\u7684\u503c\u57df\u4e3a[2-\u221a2\uff0c1+\u221a2]
h(x)=g(x)+f(x)\uff0cx\u5728[0\uff0c\u03c0/4]\u6700\u5c0f\u503c\u4e3a\uff1a2-\u221a2
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