Sin Rule

• Jul 16th 2013, 06:42 AM
srirahulan
Sin Rule
Use only Sin rule and Cos rule for a triangle to proof this>>>>>>> $\displaystyle \frac{Cos2A}{a^2}-\frac{Cos2B}{b^2}=\frac{1}{a^2}-\frac{1}{b^2}$
• Jul 16th 2013, 07:30 AM
Prove It
Re: Sin Rule
What have you tried so far?
• Jul 16th 2013, 04:28 PM
Soroban
Re: Sin Rule
Hello, srirahulan!

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

Quote:

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

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

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

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

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

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

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