`triangleABC` में सिद्ध करे कि
`b^(2) sin2C + c^(2)sin2B = 2bc sinA = 4triangle`

Explanation

`b^(2)sin2C + c^(2)sin2B`
` = b^(2).2sinCcosC + c^(2).2sinBcosB`
` = 2b^(2).kc cosC + 2c^(2). kbcosB = 2bck (bcosC + ccosB)`
`[(a)/(sinA) = (b)/(sinB) = (c)/(sinC) = (1)/(k)]`
` = 2bck.a = 2bc sinA`
पुनः `2bcsinA = 4.(1)/(2) bc sinA = 4triangle =` दायाँ पक्ष
अतः (1) और (2) से, `b^(2)sinC + c^(2)sin2B = 2bc sinA = 4triangle`