`cos2thetacos2phi+sin^2(theta-phi)-sin^2(theta+phi)=`

cos2 theta cos2 phi+sin^(2)(theta-phi)-sin^(2)(theta+phi)=cos(2 theta+2 phi)

2sin(theta-phi)=sin(theta+phi)=1

"cos"2theta."cos"2phi+"sin"^(2)(theta-phi)-"sin"^(2)(theta+phi) is equal to (i) "sin"2(theta+phi) (ii) "cos"2(theta+phi) (iii) "sin"2(theta-phi) (iv) "cos"2(theta-phi)

Prove that: cos2 theta*cos2 phi+cos^(2)(theta+phi)-cos^(2)(theta-phi)=cos(2 theta+2 phi)

cos 2theta cos 2phi + sin^(2) (theta - phi ) - sin^(2) ( theta + phi) is equal to

prove that (cos3 theta+cos3 phi)/(2cos(theta+phi)-1)=(cos theta+cos phi)cos(theta+phi)-(sin theta+sin phi)sin(theta+phi)

Prove that cos 2 theta cos 2 phi + sin ^(2) ( theta - phi) -sin ^(2) (theta + phi) = cos ( 2 theta + 2 phi).

Compute the following: {:[(sin(theta+phi), cos(theta+phi)),(sin(theta - phi),cos(theta-phi))] +[(sin(theta-phi), cos(theta-phi)),(sin(theta+phi), cos(theta+phi))]