# If $\tan\alpha$, $\tan\beta$ are roots of $x^2+px+q=0$, evaluate: $\sin^2(\alpha+\beta)+p\sin(\alpha+\beta)\cos(\alpha+\beta)+q\cos^2(\alpha+\beta)$

Question : Knowing that $$\tan\alpha$$ , $$\tan\beta$$ are roots of the quadratic equation $$x^2+px+q=0$$ ;

Compute the expression $$\sin^2(\alpha +\beta) +p\sin(\alpha +\beta) \cos(\alpha +\beta)+q\cos^2(\alpha +\beta$$)

My Working :

Sum of the roots are : $$\tan\alpha +\tan\beta = -p;$$ product of the roots $$\tan\alpha \tan\beta = q;$$

After putting these values of roots in the given equation I got :

$$x^2-(\tan\alpha + \tan\beta) x + ( \tan\alpha \tan\beta) =0$$

Please suggest whether is it correct method of approaching this or some other better method. Thanks..

• Why do you have that last $q$ in your equation? You could just write $(x - \tan \alpha)(x - \tan \beta)$ and then multiply out to figure out $p$ and $q$. Make sense? Did you mean to have an $x^2$ in the expression to compute? Jul 8, 2013 at 17:29
• Look at the formula for $\tan(\alpha+\beta)$.
– OR.
Jul 8, 2013 at 17:29

We have $-(\tan\alpha +\tan \beta) = p$ and $\tan\alpha\tan\beta=q$.

Note that $\tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-q}$ and thus $p = -\tan(\alpha+\beta)(1-q)$.

Substitute this for $p$ to obtain,

$\sin^2(\alpha +\beta) -\tan(\alpha+\beta)(1-q)\sin(\alpha +\beta) \cos(\alpha +\beta)+q\cos^2(\alpha +\beta)$

or $\sin^2(\alpha +\beta) -(1-q)\sin^2(\alpha +\beta) +q\cos^2(\alpha +\beta)$

= $q\sin^2(\alpha +\beta) +q\cos^2(\alpha +\beta) = q$

• NIce minimalist calcultion. Jul 8, 2013 at 17:54

Here goes ugly.

Note that $$\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=-\frac{p}{1-q}.$$

Multiply and divide the expression we were given by $\cos^2(\alpha+\beta)$. We get $$\cos^2(\alpha+\beta)\left(\tan^2(\alpha+\beta)+p\tan(\alpha+\beta)+q\right).$$

Almost finished, since $\cos^2(\alpha+\beta)=\frac{1}{\tan^2(\alpha+\beta)+1}$.

• @DonAntonio: Thanks for the TeX fix. Jul 8, 2013 at 17:40

Hints and ideas:

$$1+\tan^2x=\frac1{\cos^2x}\;,\;\;1+\cot^2x=\frac1{\sin^2x}$$

$$\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}$$

Thus, for example:

$$\sin^2(\alpha+\beta)=\frac1{1+\cot^2(\alpha+\beta)}=\frac1{1+\left(\frac{1-\tan\alpha\tan\beta}{\tan\alpha+\tan\beta}\right)^2}=\frac{(\tan\alpha+\tan\beta)^2}{(\tan^2\alpha+1)(\tan^2\beta+1)}\ldots$$

and etc.