(sqrt(2+cos x)-1)/((pi-x)^(2))

If f (x) is continuous at x=pi, where f(x) =(sqrt(2+cos x)-1)/((pi-2)^(2)), "for" x ne pi, find f(pi).

(i) lim_(x to (pi)/(2)) (1+cos 2x)/((pi-2x)^(2)) (ii) lim_( x to 0) (1-cos x.sqrt(cos 2x))/(x^(2)) (iii) lim_(thetararr(pi)/(6)) (sin(theta-(pi)/(6)))/(sqrt(3)-2 cos theta)

int_((pi)/(3))^((pi)/(2))(sqrt(1+cos x))/((1-cos x)^((5)/(2)))dx

Evaluate lim_(x->pi)(sqrt(1-cos6x))/(sqrt2(pi/3-x))

cos^(-1)x sqrt(3)+cos^(-1)x=(pi)/(2)

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

If pi

Prove that (i) "cos " ((pi)/(3) +x) =(1)/(2) ( " cos " x - sqrt(3) sin x) (ii) " sin " ((pi)/(4) + x) + " sin " ((pi)/(4)-x) =sqrt(2) " cos " x (iii) (1)/(sqrt(2)) " cos ((pi)/(4) + x) = (1)/(2) " (cos x - sin x) " (iv) " cos x + cos " ((2pi)/(3) +x) + " cos " ((2pi)/(3)-x) =0

Domain of the function f(x)=sqrt(cos(sin x))+sin(x^(2)-1) is [-1,1][-2,2]c*[-pi,sqrt(2)]uu[sqrt(2),pi][-sqrt(2),-sqrt(2)]