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

Consider the equation $$ \cos^2\phi + \alpha\sin\phi\cos\phi-\beta=0\;, $$ where $\alpha,\beta\in\mathbb{R}$. I need to find an explicit expression for $\phi$. I have tried completing the square, but that did not go far. Any ideas are welcome.

share|cite|improve this question

Use the two trigonometric identities $$ \cos^2\varphi = \frac 1 2 + \frac 1 2 \cos(2\varphi)\qquad \text{ and } \qquad 2\sin\varphi\cos\varphi = \sin(2\varphi). $$ This transforms your equation to $$ \frac 1 2 + \frac 1 2 \cos(2\varphi) + \frac 1 2 \alpha\sin(2\varphi) - \beta = 0, $$ or $$ \cos(2\varphi)+ \alpha\sin(2\varphi) = 2\beta - 1. $$ Now you have a linear combination of sine and cosine functions that both have the same period. The coefficients are $1$ and $\alpha$, and you want a linear combination in which the sum of square of the coefficients is $1$, so that one of them will be a sine and the other a cosine. So you can write $$ \sqrt{1+\alpha^2}\left( \frac{1}{\sqrt{1+\alpha^2}}\cos(2\alpha)+ \frac{\alpha}{\sqrt{1+\alpha^2}} \sin(2\alpha) \right) = 2\beta - 1. $$ This becomes $$ \sqrt{1+\alpha^2}\Big( \sin(\chi)\cos(2\alpha)+ \cos(\chi) \sin(2\alpha) \Big) = 2\beta - 1 $$ and then $$ \sqrt{1+\alpha^2} \sin(2\varphi+\chi) = 2\beta - 1. $$

Can you take it from there?

share|cite|improve this answer

Multiply by $2$. Then note that $2 \cos ^2 x = \cos 2x +1$. Let $\cos 2 \phi = y$. Then we have $y+\alpha \sqrt{1-y^2}+c = 0 ,(c = 1- \beta)$ Wolfram Alpha solves this as $\cos 2 \phi = \frac{c \pm \sqrt{\alpha4- \alpha^2 c^2+\alpha^2}}{\alpha^2+1}$.

share|cite|improve this answer

My answer is a little tedious, which involves the discussion of the values of $\alpha$ and $\beta$.

First, you can use trigonometric identities to change your equation into: $$ \frac{1+ \cos{2 \phi}}{2} +\frac{\alpha}{2} \sin{2 \phi} - \beta = 0$$ Moving $\cos{2\phi}$ to the right hand side and taking squares of both side gives you $$(1+\alpha^2)\cos^2{2\phi} - 2(2\beta -1) \cos{2\phi} + (2\beta-1)^2 - \alpha^2=0$$ which can be solved as a quadratic function. Note that you need to discuss the existence of solution, since $|\cos{2\phi}| \leq 1$. Once you get the solution to $\cos{2\phi}$, it is easy to get the expression of $\phi$.

share|cite|improve this answer

Divide either sides by $\cos^2\phi$ to get

$$1+a\tan\phi=\frac b{\cos^2\phi}=b\sec^2\phi=b(1+\tan^2\phi)$$

$$\iff b\tan^2\phi-a\tan\phi+b-1=0$$ which is a Quadratic Equation in $\tan\phi$

share|cite|improve this answer

Hint: $\cos^2 \phi=\dfrac{\cos 2\phi +1}2$ and $\sin \phi \cos\phi=\dfrac{\sin 2\phi}2$.

Then your expression is a linear function of $\cos 2\phi$ and $\sin 2\phi$.

Do you know how to solve $a\sin x +b\cos x=c$ for $x$?

share|cite|improve this answer

Another way is to use the trigonometric identity:- $1 = sin^2\phi + \cos^2\phi$.

Then, the original equation can be re-written as $\cos^2\phi + \alpha\sin\phi\cos\phi-\beta (sin^2\phi + \cos^2\phi)=0$

After re-arranging terms, we have $(1 - \beta)\cos^2\phi + \alpha\sin\phi\cos\phi-\beta sin^2\phi=0$

Which is a (homogeneous) quadratic equation solvable by (1) applying the quadratic formula appropriately or (2) factorization (if $\alpha$ and $\beta$ are 'well behave' known numerical quantities.

The identity:- $\frac {sin \theta} {cos \theta} = tan \theta$ is also needed to get the value(s) of $\theta$.

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.