sin `pi/3` = 2sin `pi/6` cos `pi/6`

sin^(-1)((-1)/2) (a) pi/3 (b) -pi/3 (c) pi/6 (d) -pi/6

lim_ (x rarr (pi) / (6)) (2 (sin x cos ((pi) / (6)) - cos x sin ((pi) / (6)))) / (x- (pi) / (6))

Prove that: 2(sin pi)/(6)sec(pi)/(3)-4(sin(5 pi))/(6)(cos pi)/(4)=1

The value of 2 sin (pi/8) sin((2pi)/8) sin((3pi)/8) sin ((5pi)/8) sin ((6pi)/8) sin((7pi)/8) is :

sin(pi+pi/6)=

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))

Prove that (i) " tan"^(2) .(pi)/(3) + 2cos^(2) .(pi)/(4)+ 3 sec^(2).(pi)/(6)+ 4 cos^(2).(pi)/(2)=8 (ii) " sin ".(pi)/(6) " cos 0 + sin ".(pi)/(4) " cos " .(pi)(4) + " sin " .(pi)/(3) "cos " .(pi)/(6) =(7)/(4) (iii) " 4sin " (pi)/(6) " sin"^(2) (pi)/(3) + 3 " cos " .(pi)/(3) " tan ".(pi)/(4) = " cosec"^(2).(pi)/(2)=4