1-2sin²π/12
1-2sin²π/12
sin²α+√3sinαcosα-2cos²α=0,α∈(π/6,5π/12),求:(1)sin(2
计算:sin²1°+sin²2°+sin²3°...+sin²45°+sin
数列求和 sin²1°+sin²2°+sin²3°+.+sin²88°+sin&
化简sin²α-2sinαcosα+1
sin²1°+sin²2°+sin²89°+sin²88°=
sin²1°+sin²2°+……sin²88°+sin²89°
泰勒公式的为什么㏑( 1 + sin X ) = sin X - ( sin X )²/2 +(sin X )
sinπ/24*cosπ/24*sinπ/12=—— 答案是这么写的:1/2(2sinπ/24*cosπ/24)cosπ
已知tan(θ-π)=-1/2,求下列各式的值.(1)sin²θ-2sin(π-θ)×sin(θ+2/π)-c
设f(α)=2sinαcosα+cosα/1+sin²α+cos(3π/2+α)-sin²(π/2+
1、已知tan(3π+α)=2,求:(1)(sinα+cosα)²;(2)sinα-cosα/2sinα+co