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I just started trigonometry, and I came across this problem. I know that this problem would be very simple with a calculator, but without one, I'm lost. How would you determine if $\sin 4$ is bigger than $\sin 3$ in negativity or positivity? How would you even determine if they are positive or negative in the first place, and how would you know which one is bigger?

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    $\begingroup$ Recall that by definition $\sin(x)$ is positive for $x \in (0, \pi)$ and negative for $x\in (\pi, 2\pi)$. Also note that because of continuity, since $\sin(\pi)=0$, for some small $\varepsilon$ you can also say that $\sin(\pi \pm \varepsilon)\approx 0$. Lastly notice that $ 3 = \pi -0.1415...$" and that $4\in (\pi, 2\pi)$. $\endgroup$
    – Robert Lee
    Jun 26, 2021 at 16:50
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    $\begingroup$ pi=3.14. sine(pi)=0. sin(3)>0, sin(4)<0, but 3 is close to pi, so sin(3) is not very strongly positive .... $\endgroup$
    – mathse
    Jun 26, 2021 at 16:51
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    $\begingroup$ Radians or degrees? $\endgroup$
    – fleablood
    Jun 26, 2021 at 17:19
  • $\begingroup$ $\sin 3$ is much closer to $\sin \pi$ than $\sin 4$. Knowing that $\sin n$ is the $y$-component of a point on the unit circle angle $n$ from the $x$-axis, this tells me that $\sin 3 + \sin 4 < 0$. (This answer assumed radians. Degrees is trivial to see) $\endgroup$
    – William
    Jun 26, 2021 at 17:20

3 Answers 3

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Here's a possible method: $$\sin 3+\sin 4=2\sin\left(\frac {3+4}{2}\right) \cos\left(\frac {3-4}{2}\right)=2\sin\left(\frac 72\right)\cos\left(\frac 12\right)$$ Now, we know that $0<\frac 12<\frac {\pi}{2}$, hence the cosine is positive. Also, $\frac {3\pi}{2}>\frac 72>\pi$, so the sine part is negative. Hence the product is negative, and consequently, the sum is negative too.

Notes:

●Here the formula I've used is: $$\sin x+\sin y=2\sin\left(\frac {x+y}{2}\right) \cos \left(\frac {x-y}{2}\right)$$ ●Also note how I replaced $\cos \left(-\frac 12\right)$ with $\cos\left( \frac 12\right)$, this is due to the fact that $\cos$ is an even function, and hence, $\cos (-x)=\cos x$.

●Observe how this method gives you the general process for such questions, where it is possible that neither angle is very close to multiples of $\pi$.

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    $\begingroup$ ... and if we don't already know that $\frac72>\pi$, that is because $\frac7{12}>\frac{\pi}6$ because $\sin\frac{7}{12}>\frac12$ -- and just the first two terms of the power series for $\sin\frac{7}{12}$ show the latter fact. $\endgroup$ Jun 26, 2021 at 18:15
  • $\begingroup$ Indeed, but since OP is a beginner with trig, I suspect he/she doesn't know calculus yet, so $\pi \approx 3.14$ will probably be more useful. $\endgroup$ Jun 26, 2021 at 19:24
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For values close to $\pi$ (or close to $0$) we have $|\sin (\pi \pm k)| = |\sin k|$.

And if $|k|$ is closer to $0$ (but less than $\frac \pi 2$) than $|j|$ then $|\sin(\pi \pm k)| < |\sin (\pi \pm j)|$.

.......

So bearing in mind that $4-\pi \approx 4- 3.14 = 0.86$ and $\pi -3 \approx 3.14 - 3 = 0.14$ we have

So for $\frac \pi 2 < 3 < \pi$ we have $\sin 3 > 0$ and $\sin 3=\sin (\pi -3) < \sin (4 - \pi)$

And for $\pi < 4 < \frac {3\pi}2$ we have $\sin 4 < 0$ and $|\sin 4| = \sin (4-\pi)> \sin (\pi -3)$.

so $\sin 3 + \sin 4 = |\sin 3| - |\sin 4|= \sin(\pi -3) - \sin (4-\pi) < 0$.

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Alternatively (but this is more coincidental than general.

$\frac {3\pi}4 < 3 < \pi$ so $\frac 12 > \sin 3 > 0$.

And $\frac{5\pi} 4 < 4 < \frac {3\pi}2$ so $-\frac 12 > \sin 4 > -1$.

So $0 =\frac 12 -\frac 12 > \sin 3 + \sin 4 > 0 -1=-1$

But that would be harder to generalize for something like $\sin 2.8 + \sin 3.7$ something like that.

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  • $\begingroup$ Your 4th line is the key: $0<\pi - 3<4-\pi<\pi /2,\,$ so $\,0<\sin (\pi -3)<\sin (4-\pi).$ $\endgroup$ Jun 26, 2021 at 21:38
  • $\begingroup$ Well.... that and $\sin 4 < \sin \pi = 0$ and $\sin 3 > \sin \pi =0$...... $\endgroup$
    – fleablood
    Jun 26, 2021 at 23:32
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@Ritam_Dasgupta perfectly answered your question.

I will help you with other question i.e. which out of $\sin 3$ and $\sin 4$ is greater in magnitude

For this I will assume $\pi\approx3.14$

Now $\sin 3= \sin(3.14-3)\approx\sin(\pi-0.14)\approx \sin(0.14)$

Now $\sin 4= \sin(3.14+0.86)\approx\sin(\pi+0.86)\approx -\sin(0.86)$

Clearly both $0.86,0.14<1.57\approx0.5\pi$

Therefore their $\sin()$ is positive

Now $\sin x$ is increasing function on the interval $(0,\frac{\pi}{2})$

Therefore $\sin(0.86)>\sin(0.14)$

$|\sin 4| > \sin 3$

And since $|\sin 4|=-\sin 4$ as $\sin 4$ is negative

therefore $-\sin 4> \sin 3\implies \sin 3+\sin 4<0$

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