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?
 A: 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 $\pi$.
A: 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.
A: @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  $-\sin 4> \sin 3\implies \sin 3+\sin 4<0$
