(pi)sin=-(sqrt(3))/(2)" का हल जात कीजिए। "

" (a) "int_(0)^( pi/6)sqrt(1-sin2x)dx" का मान ज्ञात कीजिए "

Simplified value of sin [ (pi)/(2) - sin^(-1) (- (sqrt3)/(2)) ] is :

Prove that (i) "sin " (7pi)/(12) " cos " (pi)/(2) - "cos " *(7pi)/(12) " sin " (pi)/(4) = (sqrt(3))/(2) (ii) " sin " (pi)/(4) " cos " (pi)/(2) + "cos"(pi)/(4) " sin " (pi)/(12) = (sqrt(3))/(2) (iii) " cos " (2pi)/(3) " cos " (pi)/(4) - " sin " (2pi)/(3) " sin " (pi)/(4) =(-(sqrt(3) +1))/(2sqrt(2))

The value of sin [ (pi)/(3) - sin^(-1) (- (sqrt(3))/(2))]

Prove that (i) " sin " .(pi)/(6). " cos " .(pi)/(6) = (sqrt(3))/(4) " " (iii) cos^(2) .(pi)/(2) - sin^(2) .(pi)/(2) =(sqrt(3))/(2)

Find the values of sin{(pi)/(6)-sin^(-1)(-(sqrt(3))/(2))}

Prove that sin^-1 (-(sqrt3)/(2)) = - (pi/3)

sin {(pi) / (3) -sin ^ (- 1) (- (sqrt (3)) / (2))}

The value of sin[(pi)/2-sin^(-1)(-(sqrt(3))/2)]=