Question

( begin{array}{l}qquad begin{aligned} L H S &=sin ^{2}left(frac{pi}{8}+frac{A}{2}right)-sin ^{2}left(frac{pi}{8}-frac{A}{2}right) therefore & cos 2 x=1-2 sin ^{2} x Rightarrow & sin ^{2} x=frac{1}{2}[1-cos 2 x] L H S &=frac{1}{2}[1-cos (pi / 4+A)] &=frac{-1}{2}[1-cos (pi / 4-A)] &=frac{1}{R}-frac{1}{2} cos (pi y+A) frac{-1}{2}+frac{1}{2} cos (pi / 4+) &=frac{1}{2}[cos (pi / 4-A)-cos (pi / 4+A)] &=frac{1}{2}left[-2 sin left(frac{pi}{4}right) sin (-A)right] therefore & cos x-cos y=-2 sin left(frac{x+y}{2}right) sin left(frac{x-y}{2}right) end{aligned} therefore H S= & +frac{1}{sqrt{2}} sin A & =frac{1}{R+1} Send{array} )

# C.-4. Show that : sina (6+) - sine (6 A) = 2 )sin A C-5.a Show that : cosa + cos2 (a + b) - 2cos a cos B cos (a + b) = sin2B.

Solution