sin²π/8-cos²π/8=

sin²π/8 -cos²π/8 设f(α)=2sinαcosα+cosα/1+sin²α+cos(3π/2+α)-sin²(π/2+ 求SINπ/64 COSπ/64 COSπ/32 COSπ/16 COSπ/8 已知tan（3π+α）=2,求：1、（sinα+cosα）²；2、sinα-cosα/2sinα+cosα α属于（0,π/2）且2sin²α-sinαcosα-3cos²α=0求sin(α+π/4)/sin 计算：cos（π/8）*sin（π/8）=? 求证:1-2sinαcosα/cos²α-sin²α=tan（π/4-α） 求证(1-2sinαcosα)/(cos²α-sin²α)=tan(π/4-α) 求下列各式的值.第一题sin 15°cos 15° 第二题cos²π/8-sin²π/8 f(x)=cos²(π/6)+sinπ/6cosπ/6 (cosπ/8+sinπ/8)（cos^3π/8-sin^3π/8） (cos π/8+ sin π/8 ) (cos^3 π/8+sin^3 π/8 )的值.