已知a,b都是锐角,cos a=1/7 cos(a+b)=-(11/14),求 cos b的值

题目

已知a,b都是锐角,cos a=1/7 cos(a+b)=-(11/14),求 cos b的值
化简：(1) 3√15sin x+3√5cos x
(2) √2/4sin(兀/4 - x)+√6/4cos(兀/4 - x)

答案

a,b都是锐角,cosa=1/7,sina=√(1－cos^2a)=4√3/7
sin(a+b)=√[1－cos^2(a+b)]=5√3/14
cosb=cos[(a+b)-a]=cos(a+b)cosa+sin(a+b)sina
=(-11/14)*(1/7)+(5√3/14)*(4√3/7)=1/2
3√15sin x+3√5cos x=6√5(cos30*sinx+sin30*cosx)=6√5sin(x+30)
√2/4sin(兀/4 - x)+√6/4cos(兀/4 - x)
= √2/2[cos(兀/3)*sin(兀/4 - x)+sin(兀/3)*cos(兀/4 - x)]
=√2/2sin(7兀/12-x)=√2/2cos(x-兀/12)