12
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ It's definitely solvable. You may have overlooked the identity though. Just look for identities that involve sums of trigonometric functions. $\endgroup$ – Raskolnikov Feb 5 '11 at 16:46
  • $\begingroup$ Hint: use en.wikipedia.org/wiki/… $\endgroup$ – Dan Petersen Feb 5 '11 at 16:46
  • $\begingroup$ As you can solve the first equation, adding the second to the mix may well make it inconsistent. The second is solvable alone, using a very similar technique. $\endgroup$ – Ross Millikan Feb 5 '11 at 16:58
  • $\begingroup$ Hmmm... $x$ is not the sine you write, there is is still a factor missing in your solution (equivalently, try $x=1$ and see what happens). $\endgroup$ – Did Aug 12 '11 at 14:03
13
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I thought the identity was: $cos(\theta) = sin(\theta + \frac{\pi}{2})$. If I'm not mistaken, that would give me $sin(\theta) + sin(\theta + \frac{\pi}{2}) = x$. $\endgroup$ – levesque Feb 5 '11 at 16:48
  • $\begingroup$ @Jcl: yeah sorry $\endgroup$ – anonymous Feb 5 '11 at 16:49
  • $\begingroup$ @JCL: You could try this. $\endgroup$ – anonymous Feb 5 '11 at 16:53
  • $\begingroup$ Both identities are true. But the one Chandru1 suggested is more useful for the problem at hand. $\endgroup$ – Ross Millikan Feb 5 '11 at 16:56
8
$\begingroup$

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 $$ \theta=\frac12\sin^{-1}(x^2-1).$$

$\endgroup$
  • $\begingroup$ I think squaring introduces an extraneous solution $\pi + \theta$ (where $\theta$ is the correct solution) in this problem. $\endgroup$ – Srivatsan Aug 12 '11 at 11:24
7
$\begingroup$

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

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Wow.. I am ashamed of myself for even asking this question :) $\endgroup$ – levesque Jun 25 '12 at 15:47
  • $\begingroup$ Don't worry too much. While this is a neat trick and obvious in retrospect, there is something to be said about recognizing that you can solve for $\theta$ from just one equation. I've gotten a lot of good results over the years by recognizing I can use some method to solve a problem and then doing so, rather than spending more time looking for a 'simpler' solution that might not exist. $\endgroup$ – Hurkyl Jun 25 '12 at 16:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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