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?
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Do you know the formula of sine of a sum? – J. W. Tanner Mar 19 '19 at 13:18
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Yep I do....... – Arkilo Mar 19 '19 at 13:19
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@Afzal Then what does that formula tell you about $\sin(x-\pi)$? What about $\sin(x + \pi)$? – Arthur Mar 19 '19 at 13:20
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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 – Arkilo Mar 19 '19 at 13:24
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1Do you know $\sin(\pi)$ and $\cos(\pi)$? – J. W. Tanner Mar 19 '19 at 13:25
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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 $\sin(x+\pi)=-\sin x$.
Then note that by "adding a full cirle", the same holds for the angle $x-\pi$: $$\sin(x-\pi)=\sin(x-\pi\color{blue}{+2\pi})=\sin(x+\pi)=-\sin x$$
StackTD
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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. – StackTD Mar 19 '19 at 14:37
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$$\sin(x)+\color{green}{\sin(x-\pi)}+\color{red}{\sin(x+\pi)}$$ $$=\sin(x)+\color{green}{\sin(x)\cos(-\pi)+\cos(x)\sin(-\pi)}+\color{red}{\sin(x)\cos(\pi)+\cos(x)\sin(\pi)}$$ $$=\sin(x)\color{green}{-\sin(x)}\color{red}{-\sin(x)}=-\sin(x)$$
using the formula for $\sin(x+\theta)$ and the facts that $\cos(\pm\pi)=-1$ and $\sin(\pm\pi)=0$
J. W. Tanner
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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. – Arkilo Mar 19 '19 at 13:39
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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 $-\sin(\theta)=\sin(-\theta)$ – J. W. Tanner Mar 19 '19 at 13:45
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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 $$\sin x - \sin x - \sin x$$
CiaPan
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Note that $\sin(\pi -x)=\sin x$ and $\sin(\pi+x)=-\sin x$, using which we get:
$$\begin{aligned}\lambda&=\sin x+\sin(x-\pi)+\sin(x+\pi)\\&= \sin x-\sin(\pi -x)+\sin(\pi+x)\\&=\sin x-\sin x-\sin x=-\sin x\end{aligned}$$
$$\sin(\pi -x)=\sin \pi \cos x-\sin x\cos\pi=+\sin x \\ \sin(\pi+x)=\sin\pi\cos x+\sin x\cos \pi =-\sin x$$
Paras Khosla
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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? – Arkilo Mar 19 '19 at 13:26
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@Afzal: Did you try to apply $\sin(\alpha+\beta)$ formula? Try and see that $\sin(\pi+x)=-\sin x$ – Vasili Mar 19 '19 at 13:37
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Yea I wasn't evaluating the value of π in sine so that was messing it all up. – Arkilo Mar 19 '19 at 13:40
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@Afzal: Good, now you may try to obtain other useful formulas for $\sin(\pi/2-x)$, $\sin(\pi/2+x)$ – Vasili Mar 19 '19 at 13:44
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Yes that is true @Arkilo Notice that $\cos x$ is an even function i.e. $\cos(-x)=\cos(x)$. – Paras Khosla Mar 19 '19 at 14:07
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@ParasKhosla Holy sheet man, I came here a little frustrated from hw, and now I have a math b*ner – Arkilo Mar 19 '19 at 14:20