Solve $\cos(\theta) + \sin(\theta) = x$ for known $x$, unknown $\theta$?

Question

After looking at the list of trigonometric identities, I can't seem to find a way to solve this. Is it solvable?

$$\cos(\theta) + \sin(\theta) = x.$$

What if I added another equation to the problem:

$$-\sin(\theta) + \cos(\theta) = y,$$ where $\theta$ is the same and $y$ is also known?

Thanks.

EDIT:

OK, so using the linear combinations I was able to whip out:

$$a \sin(\theta) + b \cos(\theta) = x = \sqrt{a^2 + b^2} \sin(\theta + \phi),$$ where $\phi = \arcsin \left( \frac{b}{\sqrt{a^2 + b^2}} \right) = \frac{\pi}{4}$ (as long as $a\geq 0$)

Giving me:

$$x = \sin(\theta + \frac{\pi}{4}) \text{ and } \arcsin(x) - \frac{\pi}{4} = \theta.$$

All set! Thanks!

Answer 1

Yeah you can write $\frac{1}{\sqrt{2}}\Bigl[\cos{\theta} + \sin{\theta}\Bigl]$ as $\sin\Bigl(\frac{\pi}{4}+\theta\Bigr)$ and solve for $x$.

Multiply both sides by $\frac{1}{\sqrt{2}}$ and then try something.

Answer 2

Another method goes by noting that $\cos^2\theta+\sin^2\theta=1$. We have $\cos\theta+\sin\theta=x$, so $$\cos^2\theta+2\cos\theta\sin\theta+\sin^2\theta=x^2,$$ or $2\cos\theta\sin\theta=x^2-1$. But $2\cos\theta\sin\theta=\sin(2\theta)$, so $2\theta=\sin^{-1}(x^2-1)$, or $$ \theta=\frac12\sin^{-1}(x^2-1).$$

Answer 3

Linear equations in $\sin \theta $ and $\cos \theta $ can be solved by a resolvent quadratic equation, using the two identities (also here):

$$\cos \theta =\frac{1-\tan ^{2}\frac{\theta }{2}}{1+\tan ^{2}\frac{\theta }{2% }}$$

and

$$\sin \theta =\frac{2\tan \frac{\theta }{2}}{1+\tan ^{2}\frac{\theta }{2}}.$$

In this case, we have

$$\begin{eqnarray*} \cos \theta +\sin \theta &=&x \\ &\Leftrightarrow &\left( 1-\tan ^{2}\frac{\theta }{2}\right) +2\tan \frac{% \theta }{2}=x\left( 1+\tan ^{2}\frac{\theta }{2}\right) \\ &\Leftrightarrow &(x+1)\tan ^{2}\frac{\theta }{2}-2\tan \frac{\theta }{2}% +x-1=0 \\ &\Leftrightarrow &\tan \frac{\theta }{2}=\frac{2\pm \sqrt{4-4(x+1)(x-1)}}{% 2(x+1)} \\ &\Leftrightarrow &\tan \frac{\theta }{2}=\frac{1\pm \sqrt{2-x^{2}}}{x+1} \\ &\Leftrightarrow &\theta =2\arctan \frac{1\pm \sqrt{2-x^{2}}}{x+1}. \end{eqnarray*}$$

Answer 4

If you know

$$ \cos(\theta) + \sin(\theta) = x $$

and

$$ -\sin(\theta) + \cos(\theta) = y $$

then you have a system of two linear equations in the 'unknowns' $\cos(\theta)$ and $\sin(\theta)$, and thus can solve for the values of $\cos(\theta)$ and $\sin(\theta)$:

$$ \cos(\theta) = \frac{x+y}{2} $$ $$ \sin(\theta) = \frac{x-y}{2} $$

and then obtain $\theta$ in your favorite manner.