Question

a minus b whole square cos square C by 2 + a + b whole square sin square C by 2 is equals to C square to prove

Answer 1

Answer:

we need to prove that:

$(a - b)^{2} cos^{2}\frac{c}{2}+(a+b)^{2}sin^{2}\frac{c}{2} = c^{2}$

Left hand side

$(a - b)^{2} cos^{2}\frac{c}{2}+(a+b)^{2}sin^{2}\frac{c}{2}$

since,$(a+b)^{2} =a^{2}+b^{2}+2ab\;\text{and}\;(a-b)^{2} =a^{2}+b^{2}-2ab$

$(a^{2} +b^{2} -2ab)cos^{2}\frac{c}{2}+(a^{2} +b^{2} -2ab)sin^{2}\frac{c}{2}$

$(a^{2} +b^{2})(cos^{2}\frac{c}{2}+sin^{2}\frac{c}{2})-2ab(cos^{2}\frac{c}{2}-sin^{2}\frac{c}{2})$

Since, $sin^{2}\theta+cos^{2}\theta=1$ and $cos^{2}\theta+sin^{2}\theta=cos2\theta$

$(a^{2} +b^{2})(1)-2ab(cos c)$

$a^{2} +b^{2}-2ab(cos c)$

by the law of cosines $(a^{2} +b^{2}-2ab(cos c)=c^{2})$

$c^{2}$ =Right hand side

Hence, proved

Answer 2

Answer:

i hope this may help you..