यदि A, B और C त्रिभुज एब्स के अन्तः कोण हो, तो दिखाइए कि `sin ((B+C)/(2)) = cos (A)/(2).`

Explanation

`triangleABC` में,
`A + B+ C = 180^(@)`
`rArr B + C = 180^(@) -A`
`rArr ((B+C)/(2)) = (180^(@) -A)/(2)`
` = (180^(@))/(2) - (A)/(2) = 90^(@) - (A)/(2)`
`rArr sin ((B+C)/(2)) = sin (90^(@) - (A)/(2))`
`rArr sin ((B+C)/(2)) = cos (A)/(2)`