In -ABC prove that - a - b2cos 2C2 - a - b2sin 2C2 - c2-

Now, #160; ( a b)^2cos ^2C2 + ( a + b)^2sin ^2C2 =(a^2+b^2)(cos^2 C2+sin^2 C2) 2ab(cos^2C2 sin^2 C2) =(a^2+b^2) 2ab(cos C) =c^2 . [ Since ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Properties of Determinants

In ΔABC prov...

Question

In $Δ ABC$ prove that , ${(a - b)}^{2} {cos}^{2} \frac{C}{2} + {(a + b)}^{2} {sin}^{2} \frac{C}{2} = c^{2}$ .

Open in App

Solution

Suggest Corrections

Similar questions

△ A B C

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

Δ A B C

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

{sin}^{2} (c / 2) = c^{2}

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

Δ A B C

\frac{a^{2} - b^{2}}{c^{2}} = \frac{sin (A - B)}{sin (A + B)}

In any $Δ A B C$ , prove that

$\frac{(b^{2} - c^{2})}{a^{2}} s i n 2 A + \frac{(c^{2} - a^{2})}{b^{2}} s i n 2 B + \frac{(a^{2} - b^{2})}{c^{2}} s i n 2 C = 0$

Join BYJU'S Learning Program