Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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..

share|cite|improve this question
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? – Amzoti Jul 8 '13 at 17:29
Look at the formula for $\tan(\alpha+\beta)$. – Mlazhinka Shung Gronzalez LeWy Jul 8 '13 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 $\sin^2(\alpha +\beta) -(1-q)\sin^2(\alpha +\beta) +q\cos^2(\alpha +\beta)$

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

share|cite|improve this answer
NIce minimalist calcultion. – André Nicolas Jul 8 '13 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 $\cos^2(\alpha+\beta)=\frac{1}{\tan^2(\alpha+\beta)+1}$.

share|cite|improve this answer
@DonAntonio: Thanks for the TeX fix. – André Nicolas Jul 8 '13 at 17:40

Hints and ideas:


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

Thus, for example:


and etc.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.