# How do you determine the sign of $\sin 3 + \sin 4$ without using a calculator? Is it even possible to have a sensible answer?

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?

• 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 $4\in (\pi, 2\pi)$. Jun 26 at 16:50
• 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 .... Jun 26 at 16:51
• Radians or degrees? Jun 26 at 17:19
• $\sin 3$ 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) Jun 26 at 17:20

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 $$\pi$$.

• ... 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. Jun 26 at 18:15
• 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. Jun 26 at 19:24

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.

• Your 4th line is the key: $0<\pi - 3<4-\pi<\pi /2,\,$ so $\,0<\sin (\pi -3)<\sin (4-\pi).$ Jun 26 at 21:38
• Well.... that and $\sin 4 < \sin \pi = 0$ and $\sin 3 > \sin \pi =0$...... Jun 26 at 23:32

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