4
$\begingroup$

So I have this trig question:

$ \sin(x)+\sin(x−π)+\sin(x+π) = $ _____

The answer is $- \sin(x)$

I can't figure out how to solve it.

Any help?

$\endgroup$
  • $\begingroup$ Do you know the formula of sine of a sum? $\endgroup$ – J. W. Tanner Mar 19 at 13:18
  • $\begingroup$ Yep I do....... $\endgroup$ – Arkilo Mar 19 at 13:19
  • $\begingroup$ @Afzal Then what does that formula tell you about $\sin(x-\pi)$? What about $\sin(x + \pi)$? $\endgroup$ – Arthur Mar 19 at 13:20
  • $\begingroup$ sin(x) cos (π) - sin (π) cos(x) + sin (x) cos(π) + sin(π)cos(x) , 2nd and fourth one cancels out and then you have sin(x) cos (π) + sin (x) cos(π), if I add em then I just get 2 sin(x) 2cos(π). How do I reduce it down from this point $\endgroup$ – Arkilo Mar 19 at 13:24
  • 1
    $\begingroup$ Do you know $\sin(\pi)$ and $\cos(\pi)$? $\endgroup$ – J. W. Tanner Mar 19 at 13:25
3
$\begingroup$

$$\sin(x)+\color{green}{\sin(x-\pi)}+\color{red}{\sin(x+\pi)}$$ $$=\sin(x)+\color{green}{\sin(x)\cos(-\pi)+\cos(x)\sin(-\pi)}+\color{red}{\sin(x)\cos(\pi)+\cos(x)\sin(\pi)}$$ $$=\sin(x)\color{green}{-\sin(x)}\color{red}{-\sin(x)}=-\sin(x)$$

using the formula for $\sin(x+\theta)$ and the facts that $\cos(\pm\pi)=-1$ and $\sin(\pm\pi)=0$

$\endgroup$
  • $\begingroup$ Aight this is v cool. You should have followed the signs of the general formula for expanding the second term tho, it makes it a little confusing to not do it. $\endgroup$ – Arkilo Mar 19 at 13:39
  • $\begingroup$ You could say $\sin(x-\theta)=\sin(x)\cos(\theta)-\cos(x)\sin(\theta)$ or $\sin(x+(-\theta))=\sin(x)\cos(-\theta)+\cos(x)\sin(-\theta);$ they're the same because $\cos(\theta)=\cos(-\theta)$ and $-\sin(\theta)=\sin(-\theta)$ $\endgroup$ – J. W. Tanner Mar 19 at 13:45
  • $\begingroup$ This is great, thanks man. $\endgroup$ – Arkilo Mar 19 at 14:23
4
$\begingroup$

As shown in some other answers, this is very simple if you know that: $$\sin(x-\pi)=-\sin x \quad\mbox{and}\quad \sin(x+\pi)=-\sin x$$ If you don't know these formulas or you have a hard time understanding why they are true, you should spend some time to carefully study the unit circle and how symmetry there leads to these simple relations.

The image below should help you understand why $\sin(x+\pi)=-\sin x$.

enter image description here

Then note that by "adding a full cirle", the same holds for the angle $x-\pi$: $$\sin(x-\pi)=\sin(x-\pi\color{blue}{+2\pi})=\sin(x+\pi)=-\sin x$$

$\endgroup$
  • $\begingroup$ Wow man, this was very neat $\endgroup$ – Arkilo Mar 19 at 13:57
  • $\begingroup$ You're welcome; it's worth being able to work and reason with the unit circle when it comes to these basic properties of the trig. functions. $\endgroup$ – StackTD Mar 19 at 14:37
1
$\begingroup$

You probably know, that $$ \sin(x−\pi) = -\sin(x).$$

Also $$\sin(x+\pi) = \sin(x-\pi + 2\pi) = \sin(x-\pi)$$ so your given expression reduces to $$\sin x - \sin x - \sin x$$

$\endgroup$
0
$\begingroup$

Note that $\sin(\pi -x)=\sin x$ and $\sin(\pi+x)=-\sin x$, using which we get:

$$\begin{aligned}\lambda&=\sin x+\sin(x-\pi)+\sin(x+\pi)\\&= \sin x-\sin(\pi -x)+\sin(\pi+x)\\&=\sin x-\sin x-\sin x=-\sin x\end{aligned}$$

$$\sin(\pi -x)=\sin \pi \cos x-\sin x\cos\pi=+\sin x \\ \sin(\pi+x)=\sin\pi\cos x+\sin x\cos \pi =-\sin x$$

$\endgroup$
  • $\begingroup$ Wait where did you get sin(π−x)=sinx and sin(π+x)=−sinx from? Aren't you supposed to apply the sin(alpha+beta) = sin(alpha)cos(beta) + cos(alpha)sin(beta) to it? $\endgroup$ – Arkilo Mar 19 at 13:26
  • $\begingroup$ @Afzal: Did you try to apply $\sin(\alpha+\beta)$ formula? Try and see that $\sin(\pi+x)=-\sin x$ $\endgroup$ – Vasya Mar 19 at 13:37
  • $\begingroup$ Yea I wasn't evaluating the value of π in sine so that was messing it all up. $\endgroup$ – Arkilo Mar 19 at 13:40
  • $\begingroup$ @Afzal: Good, now you may try to obtain other useful formulas for $\sin(\pi/2-x)$, $\sin(\pi/2+x)$ $\endgroup$ – Vasya Mar 19 at 13:44
  • $\begingroup$ sin(π/2−x) = cos (-x) and sin(π/2+x) = cos (x)? $\endgroup$ – Arkilo Mar 19 at 14:05

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.