cos{theta+(n-(3)/(2))phi}-cos{theta+(n+(3)/(2))phi}

If 0

If sin2 theta+sin2 phi=(1)/(2) and cos2 theta+cos2 phi=(3)/(2), then cos^(2)(theta-phi)=

2cos theta=x+(1)/(x) and 2cos phi=y+(1)/(y) then (A)x^(n)+(1)/(x^(n))=2cos(n theta)n in Z(B)(x)/(y)+(y)/(x)=2cos(theta-phi)(C)xy+(1)/(Xy)=2cos(theta+phi)(D)x^(m)y^(n)+(1)/(x^(m)y^(n))=2(cos m theta+n phi),m,n in Z

The value of [cos 2 (theta+ phi)+ 4 cos (theta + phi) sin theta sin phi+2 sin ^(2) phi] is-

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

If 0

If 2 tan theta = cot phi, "show that" cos(theta - phi) = 3 cos ( theta + phi).

If cos(theta-phi) , cos(theta) , cos(theta+phi) are in HP, the the value of cos(theta)sec(phi/2) is

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))]

(sin (theta+phi) -2 sintheta+ sin (theta-phi))/( cos(theta+phi) -2 costheta+ cos (theta-phi))= tan theta