`cos((3pi)/4)-isin((3pi)/4)=?`

cos((3pi)/4+x)-sin(pi/4-x)=?

The polar form of the complex number (i^(25))^3 is (a) cos(pi/2)+isin(pi/2) (b) cospi+isinpi (c) cospi-isinpi (d) cos((3pi)/2)+isin((3pi)/2)

cos((3pi)/(4)+x)-cos((3pi)/(4)-x) = -sqrt(2)sinx

Prove that: cos((3pi)/(4)+A)-cos((3pi)/(4)-A)=-sqrt(2)sinA

Prove the following: (i) cos((3pi)/(4)+x)-cos((3pi)/(4)-x)=-sqrt2 sin x (ii) cos"((pi)/(4)+x)+cos((pi)/(4)-x)=sqrt2 cos x .

cos(pi/4)+isin(-pi/4)=?

Find the rang of f(x)=log_sqrt5(3+cos((3pi)/4+x)+cos((pi)/4+x)+cos((pi)/4-x)-cos((3pi)/4-x))

If a=cos((4pi)/3)+isin((4pi)/3) then |(1,1,1),(1,a,a^2),(1,a^2,a)|

Prove that: i) sin(5pi)/(18) - cos(4pi)/(9) = sqrt(3)sinpi/9 ii) cos(3pi)/4+A-cos((3pi)/(4)-A)=-sqrt(2)sinA