cos(a-b)=-(4/5)cos(a+b)=12/13a-b=(π/2、π) (a+b)=(3π/2 2π) 求：cos2a cos2b

Cos(a+b)*cos(a-b)=1/5 求cos ^2-sin^2 已知cos(a+b)=4/5,cos(a-b)=-4/5,3π/2 .已知Cos(a+b)=4/5,cos(a-b)= -4/5,3/2π 设cos(a-b)=-4/5,cos(a+b)=12/13,45° 5cos^2a+4cos^2β=4cosα则cos^2a+cos^2β 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) 已知5cos²a+4cos²b=4cosa,则cos²a+cos²b的取值范围是____________ sin(a-b)*cos a-cos(a-b)*sin a=1/5,则cos 2b的值是 在三角形ABC中,已知cos A =4/5,cos B =12/13,求cos C. 证明：cos(a+b)*cos(a-b)=cos2b-cos2a a,b为锐角,cos(a+b)=12/13 cos(2a+b)=4/5 求cosa 已知锐角a、b满足cos a=4/5,tan(a-b)=-1/3,求cos b |cos(a/2)|=b,5pai cos^2(a+b)=? 已知a,b为锐角,cos a=4/5,cos(a+b)=-16/65,求cos b的值? 求证cos(a+b)cos(a－b)=cos^2b－sin^2a 求证cos(a+b)cos(a-b)=cos^2a-sin^2b 求证:cos(a+b)cos(a-b)=cos平方b-sin平方a