`" sin " (x-(pi)/(6)) + " cos " (x-(pi)/(3)) =sqrt(3) " sin " x`

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

cos ((3pi)/( 4) + x) - cos((3pi)/( 4) -x) =- sqrt2 sin x

cos ((3pi)/( 4) + x) - cos((3pi)/( 4) -x) =- sqrt2 sin x

cos ((3pi)/( 4) + x) - cos((3pi)/( 4) -x) =- sqrt2 sin x

cos((3pi)/(4)+x)-cos ((3pi)/(4)-x)=-sqrt(2) sin x

lim_ (x rarr (pi) / (6)) (sin (x- (pi) / (6))) / (sqrt (3) -cos x)

sin(x-pi/6)+cos(x-pi/3)=sqrt3 sinx

Prove that cos ((3pi)/(4)+x)-cos ((3pi)/(4)-x)=-sqrt2 sin x

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

Find the extreme values of the following function over R. (i) sin^(2) (60^(0) + x) + sin^(2) (60^(0) - x) (ii) cos (x+ (Pi)/(3) ) + 2sqrt(2) sin (x+ (pi)/(3))-3 (iii) 3 sin^(2) x+ 5 cos^(2) x (iv) cos 2 x+ cos^(2) x (v ) cos x cos ((2pi)/(3) + x) cos ((2pi)/(3)-x)