Math Help - Sin Rule

1. Sin Rule

Use only Sin rule and Cos rule for a triangle to proof this>>>>>>> $\frac{Cos2A}{a^2}-\frac{Cos2B}{b^2}=\frac{1}{a^2}-\frac{1}{b^2}$

2. Re: Sin Rule

What have you tried so far?

3. Re: Sin Rule

Hello, srirahulan!

We need this identity: . $\sin^2\theta \:=\:\frac{1-\cos2\theta}{2}$

$\text{Use only Sin rule and Cos rule for a triangle to prove:}$

. . $\frac{\cos2A}{a^2}-\frac{\cos2B}{b^2}\:=\:\frac{1}{a^2}-\frac{1}{b^2}$

From the Sine Rule:. . . . $\frac{\sin A}{a} \;=\;\frac{\sin B}{b}$

Square both sides:. . . . $\frac{\sin^2A}{a^2} \;=\;\frac{\sin^2B}{b^2}$

Apply the identity:. $\frac{1-\cos2A}{a^2} \;=\;\frac{1-\cos2B}{b^2}$

. . . . . . . . . . . . . $\frac{1}{a^2} - \frac{\cos2A}{a^2} \;=\;\frac{1}{b^2} - \frac{\cos2B}{b^2}$

Therefore: . . $\frac{\cos2A}{a^2} - \frac{\cos2B}{b^2} \;=\;\frac{1}{a^2} - \frac{1}{b^2}$