若sin(π/6-a)=1/3 则cos(2/3π+2a)的值为
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若sin(π/6-a)=1/3 则cos(2/3π+2a)的值为
若sin(π/6-a)=1/3 则cos(2/3π+2a)的值为
若sin(π/6-a)=1/3 则cos(2/3π+2a)的值为
cos(2π/3+2a) =cos[2(π/3+a)] =2cos (π/3+a)-1 =2sin [π/2-(π/3+a)]-1 =2sin (π/6-a)-1 =-7/9 -7/9 构建三角
2/3π+2a = π- (π/3 -2a)
cos(2/3π+2a)=cos[ π- (π/3 -2a)]
= - cos (π/3 -2a)
= - cos2(π/6 -a)
= -(1- 2(sin(π/6 - a ))^2)
=-(1 - 2 * 1/9)= - 7 / 9
由cos(a+b)=cos a cos b-sin a sin b cos(a-b)=cos a cos b+sin a sin b解题设a为锐角,证:1、2分之根3乘cos a + 2分之1乘sin a=cos(6分之π-a)2、cos a-sin a=根号2cos(4分之π+a)
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