作业帮cos(π/12)^4+sin(π/12)^4=
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cos(π/12)^4+sin(π/12)^4=
cos(π/12)^4+sin(π/12)^4=
cos(π/12)^4+sin(π/12)^4=
cos(π/12)^4+sin(π/12)^4
=(cos²(12\π)+sin²(12\π))²-2sin²(12\π)cos²(12\π)
=1-(√2sin(12\π)cos(12\π))²
=1-(2\√2sin6\π)²
∵sin(6\π)=2\1
∴原式=1-(2\√2•2\1)²
=1-8\1
=8\7
=[(cosπ/12)²+sin(π/12)²]² - 2(cosπ/12)sin(π/12)
=1-sinπ/6
＝1/2
这样的题基本思路就是降幂，凑三角函数。
（sinπ/24)(cosπ/24)(cosπ/12)
（sinπ/24)(cosπ/24)(cosπ/12)
sinπ/12*cosπ/12=?
求值：sin(π/12)+cos(π/12)
sin(π/12)-cos(π/12)=
cos*π/12—sin*π/12
cosπ／12乘以sinπ／12
计算cosπ/12*sinπ/12
cos(π/12)^4+sin(π/12)^4=
4 sinπ/12 cosπ/12=
（cosπ/12-sinπ/12)*(COSπ/12+sinπ/12)等于
(cosπ/12 -sinπ/12)(cosπ/12 +sinπ/12)等于
（cosπ/12-sinπ/12)(cosπ/12+sinπ/12)的值为
（cosπ/12 -sinπ/12）（cosπ/12 +sinπ/12）=
(sinπ/12+cosπ/12)(sinπ/12-cosπ/12)=
已知α=7/12π,那么cosα√（1-sinα）/(1+sinα)+sinα√（1-cosα）/(1-cosα)=
cos^2π/12-sin^2π/12等于
sin(π/12)cos(π/12)的值