show that (b-c)square cos square A/2+(b+c)square sin square A/2=asquar
Answers
Answer:
we need to prove that:
(a - b)^{2} cos^{2}\frac{c}{2}+(a+b)^{2}sin^{2}\frac{c}{2} = c^{2}(a−b)
2
cos
2
2
c
+(a+b)
2
sin
2
2
c
=c
2
Left hand side
(a - b)^{2} cos^{2}\frac{c}{2}+(a+b)^{2}sin^{2}\frac{c}{2}(a−b)
2
cos
2
2
c
+(a+b)
2
sin
2
2
c
since,(a+b)^{2} =a^{2}+b^{2}+2ab\;\text{and}\;(a-b)^{2} =a^{2}+b^{2}-2ab(a+b)
2
=a
2
+b
2
+2aband(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
−2ab)cos
2
2
c
+(a
2
+b
2
−2ab)sin
2
2
c
(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})(a
2
+b
2
)(cos
2
2
c
+sin
2
2
c
)−2ab(cos
2
2
c
−sin
2
2
c
)
Since, sin^{2}\theta+cos^{2}\theta=1sin
2
θ+cos
2
θ=1 and cos^{2}\theta+sin^{2}\theta=cos2\thetacos
2
θ+sin
2
θ=cos2θ
(a^{2} +b^{2})(1)-2ab(cos c)(a
2
+b
2
)(1)−2ab(cosc)
a^{2} +b^{2}-2ab(cos c)a
2
+b
2
−2ab(cosc)
by the law of cosines (a^{2} +b^{2}-2ab(cos c)=c^{2})(a
2
+b
2
−2ab(cosc)=c
2
)
c^{2}c
2
=Right hand side
Hence, proved