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Mathematics Question on Trigonometric Functions

In ΔABC(a−b)2cos2C2+(a+b)2sin2C2=\Delta ABC \left(a-b\right)^{2} cos^{2} \frac{C}{2} + \left(a+b\right)^{2} sin^{2} \frac{C}{2} =

A

b2b^{2}

B

c2c^{2}

C

a2a^{2}

D

a2+b2+c2a^{2}+b^{2}+c^{2}

Answer

c2c^{2}

Explanation

Solution

(a−b)2cos2C2+(a+b)2sin2C2(a-b)^{2} cos ^{2} \frac{C}{2}+(a+b)^{2} sin ^{2} \frac{C}{2}
=(a2+b2−2ab)cos2C2+a2+b2+2absin2C2=\left(a^{2}+b^{2}-2 ab\right) cos ^{2} \frac{C}{2}+a^{2}+b^{2}+2ab sin ^{2} \frac{C}{2}
=a2(cos2C2+sin2C2)+b2(cos2C2+sin⁡2C2)=a^{2}\left(cos ^{2} \frac{C}{2}+sin ^{2} \frac{C}{2}\right)+b^{2}\left(cos ^{2} \frac{C}{2}+\sin ^{2} \frac{C}{2}\right)
−2ab(cos2C2−sin2C2)-2 ab\left(cos ^{2} \frac{C}{2}-sin ^{2} \frac{C}{2}\right)
=a2+b2−2ab cos C=a^{2}+b^{2}-2 a b\,cos\, C
[∵cos2C2−sin⁡2C2=cos⁡C]\left[\because cos ^{2} \frac{C}{2}-\sin ^{2} \frac{C}{2}=\cos C\right]
=a2+b2−2ab(a2+b2−c22ab)=c2=a^{2}+b^{2}-2 a b\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)=c^{2}
[∵cos⁡C=a2+b22ab]\left[\because \cos C=\frac{a^{2}+b^{2}}{2ab}\right]