要求用倍角公式解2sin67°30′cos67°30′cos²π/8-sin²π/8还有好多类似的还有已知asin(x+y)=bcos(x+z),求tanx=bsinz-asiny/acosy-bcosz

Question

要求用倍角公式解
2sin67°30′cos67°30′ cos²π/8 - sin²π/8
还有好多类似的
还有已知asin(x+y)=bcos(x+z),求tanx=bsinz-asiny/acosy-bcosz

Answer 1

67°30′=3π/8
所以2sin67°30′cos67°30′ cos²π/8 - sin²π/8
=2sin(π/2-π/8)cos(π/2-π/8)cos²π/8 - sin²π/8
=2cos(π/8)sin(π/8)cos²π/8 - sin²π/8
=sin(π/4)cos²π/8 - sin²π/8
=cos(π/4)cos²π/8 - sin²π/8
=(2cos²π/8-1) cos²π/8 - sin²π/8
=2cos²π/8*cos²π/8 -1
=(cosπ/4+1)(cosπ/4+1)/2 -1
=√2/2-0.25
=0.457
二、
asin(x+y)=bcos(x+z),求tanx=bsinz-asiny/acosy-bcosz
asinxcosy+acosxsiny=bcosxcosz-bsinxsinz
atanxcosy+asiny=bcosz-btanxsinz
tanx=(bcosz-asiny)/(acosy+bsinz)
题目有误!