Question

If a,b,c are interior angles of triangle and show that cos^2a/2+cos^2(b+c)/2=1

Answer 1

Solution

as,a,b,c are interior angles of triangle so,

a+b+c=180°

a=180°-(b+c)

=>a/2=90°-[(b+c)/2]

now...

$l.h.s. = \cos {}^{2} ( \frac{a}{2} ) + \cos {}^{2} ( \frac{b + c}{2} ) \\ \: \: \: \: \: \: \: \: \: \: \: = \cos {}^{2} ( \frac{a}{2} ) + \sin {}^{2} (90 - ( \frac{b + c}{2} ) ) \\ \: \: \: \: \: \: \: \: \: \: \: = \cos {}^{2} ( \frac{a}{2} ) + \sin {}^{2} ( \frac{a}{2} ) = 1 = r.h.s.(proved)$