1. ## Calculate !

Hii !

$\displaystyle \tan(a)$ and $\displaystyle \tan(b)$ are the solutions of the equation :

$\displaystyle x^2+\pi x+\sqrt{2}=0$

calculate: $\displaystyle sin^2(a+b)+\pi\sin(a+b)\cos(a+b)+\sqrt{2}cos^2(a+b )$

2. Originally Posted by Perelman
Hii !

$\displaystyle \tan(a)$ and $\displaystyle \tan(b)$ are the solutions of the equation :

$\displaystyle x^2+\pi x+\sqrt{2}=0$

calculate: $\displaystyle \sin^2(a+b)+\pi\sin(a+b)\cos(a+b)+\sqrt{2}\cos^2(a +b)$
If $\displaystyle \tan a$ and $\displaystyle \tan b$ are the solutions of the equation $\displaystyle x^2+px+q=0$ then $\displaystyle \tan a + \tan b = -p$ and $\displaystyle \tan a\tan b = q$. Thus $\displaystyle \tan(a+b) = \frac{\tan a + \tan b}{1-\tan a\tan b} = \frac{-p}{1-q}$. Also, $\displaystyle \sec^2(a+b) = 1+\tan^2(a+b) = \frac{p^2+(1-q)^2}{(1-q)^2}$, and so $\displaystyle \cos^2(a+b) = \bigl(\sec^2(a+b)\bigr)^{-1} = \frac{(1-q)^2}{p^2+(1-q)^2}$.

Then $\displaystyle \sin^2(a+b)+p\sin(a+b)\cos(a+b)+q\cos^2(a+b) = \cos^2(a+b)\bigl(\tan^2(a+b)+p\tan(a+b)+q\bigr)$, and you can check by using the results of the previous paragraph that this simplifies to $\displaystyle q$ (which in this case is $\displaystyle \sqrt2$).