# Find $\cos2\theta+\cos2\phi$, given $\sin\theta + \sin\phi = a$ and $\cos\theta+\cos\phi = b$

If $$\sin\theta + \sin\phi = a \quad\text{and}\quad \cos\theta+\cos\phi = b$$

then find the value of $$\cos2\theta+\cos2\phi$$

My attempt:

Squaring both sides of the second given equation:

$$\cos^2\theta+ \cos^2\phi + 2\cos\theta\cos\phi= b^2$$

Multiplying by 2 and subtracting 2 from both sides we obtain,

$$\cos2\theta+ \cos2\phi = 2b^2-2 - 4\cos\theta\cos\phi$$

How do I continue from here?

PS: I also found the value of $\sin(\theta+\phi)= \dfrac{2ab}{a^2+b^2}$

Edit: I had also tried to use $\cos2\theta + \cos2\phi= \cos(\theta+\phi)\cos(\theta-\phi)$ but that didn't seem to be of much use

## 4 Answers

$$\cos2\theta+\cos2\phi=2\cos(\theta+\phi)\cos(\theta-\phi),$$ $$a^2+b^2=2+2\cos(\theta-\phi)$$ and $$b^2-a^2=\cos2\theta+\cos2\phi+2\cos(\theta+\phi).$$ Thus, $$\cos2\theta+\cos2\phi=2\cdot\frac{b^2-a^2-(\cos2\theta+\cos2\phi)}{2}\cdot\frac{a^2+b^2-2}{2},$$ which gives $$\cos2\theta+\cos2\phi=\frac{(a^2+b^2-2)(b^2-a^2)}{a^2+b^2}.$$

• (+1) This is a slick solution without complex numbers, so I guess mine can rest in peace. – Jack D'Aurizio Aug 5 '17 at 21:19

HINT: use that $$\cos(2\theta)+\cos(2\phi)=\cos(\theta-\phi)\cos(\theta+\phi)$$ and $$\sin(\theta)+\sin(\phi)=2\cos\left(\frac{\theta-\phi}{2}\right)\sin\left(\frac{\theta+\phi}{2}\right)$$ and $$\cos(\theta)+\cos(\phi)=2\cos\left(\frac{\theta-\phi}{2}\right)\cos\left(\frac{\theta+\phi}{2}\right)$$ so another idea, and this works: use that $$\sin(\theta)+\sin(\phi)=2\,{\frac {\tan \left( \theta/2 \right) }{1+ \left( \tan \left( \theta /2 \right) \right) ^{2}}}+2\,{\frac {\tan \left( \phi/2 \right) }{1+ \left( \tan \left( \phi/2 \right) \right) ^{2}}}$$ and $$\cos(\theta)+\cos(\phi)={\frac {1- \left( \tan \left( \theta/2 \right) \right) ^{2}}{1+ \left( \tan \left( \theta/2 \right) \right) ^{2}}}+{\frac {1- \left( \tan \left( \phi/2 \right) \right) ^{2}}{1+ \left( \tan \left( \phi/2 \right) \right) ^{2}}}$$ and $$\cos(2\theta)+\cos(\phi)={\frac {1- \left( \tan \left( \theta \right) \right) ^{2}}{1+ \left( \tan \left( \theta \right) \right) ^{2}}}+{\frac {1- \left( \tan \left( \phi \right) \right) ^{2}}{1+ \left( \tan \left( \phi \right) \right) ^{2}}}$$ now convert $$\tan(x)$$ into $\tan(x/2)$ and solve the equations above for $$\tan(\phi/2)$$ respective $$\tan(\theta/2)$$

• and thank you for the $-1$!!!!!!!!!!!!!!! – Dr. Sonnhard Graubner Aug 5 '17 at 19:38
• I do not see any way for these identities alone to actual lead to a solution. That is probably the reason for the (-1), Doctor. – Jack D'Aurizio Aug 5 '17 at 19:43
• and now this leads to a solution – Dr. Sonnhard Graubner Aug 5 '17 at 19:56

Let $s= \cos^2(\theta)+\cos^2(\phi)$, we want $2(s-1)$. Square the first equation and use Pythagorus \begin{eqnarray*} a^2+s-2 =2 \sin( \theta) \sin (\phi) \\ (a^2+s-2)^2 = 4(1- \cos^2( \theta))(1- \cos^2 (\phi)) \\ (a^2-2)^2-4+2a^2 s+s^2 =4 \cos^2( \theta) \cos^2 (\phi) \end{eqnarray*} From the second equation \begin{eqnarray*} (b^2-s)^2=4 \cos^2( \theta) \cos^2 (\phi) \end{eqnarray*} Put these together & we have \begin{eqnarray*} s= \frac{b^4+4a^2-a^4}{2(a^2+b^2)} \\ \cos(2 \theta)+ \cos( 2\phi)=(b^2-a^2) \left(1- \frac{2}{a^2+b^2} \right). \end{eqnarray*}

Using Prosthaphaeresis Formula,

$$\cos2\theta+\cos2\phi=2\cos(\theta+\phi)\cos(\theta-\phi),$$ Now $$a^2+b^2=2+2\cos(\theta-\phi)\implies\cos(\theta-\phi)=?$$

Again, using Prosthaphaeresis formula, $$a=\sin\theta+\sin\phi=2\sin\dfrac{\theta+\phi}2\cos\dfrac{\theta-\phi}2$$

and $$b=\cdots=2\cos\dfrac{\theta+\phi}2\cos\dfrac{\theta-\phi}2$$

$$\implies\dfrac ab=\tan\dfrac{\theta+\phi}2$$

Now use $$\cos2A=\dfrac{1-\tan^2A}{1+\tan^2A}$$