As an example, the definition you're for $\sin$ using is:
$$\sin\theta=\frac{o}{h}$$
I think your confusion stems from the fact that the angle $\theta$ isn't mentioned in the right hand side of this equation, which means if you're just given the angle, it's unclear how to calculate the $\sin$ (because what's $o$ and $h$?). The idea is that you imagine a right triangle where one corner has an angle $\theta$, and then $\sin\theta$ is equal to $\frac{o}{h}$ applied to the sides of that triangle. "But wait, how do I know I'm imagining the 'correct' triangle?". That's the thing - it doesn't matter. As long as the angle $\theta$ is the same, the ratio $\frac{o}{h}$ will always be the same. That's what justifies defining $\sin$ in terms of a triangle, rather than directly in terms of an angle.
That said, the triangular definitions that you're given in middle school are terrible definitions for the $\sin$ and $\cos$ functions. Some of the problems are:
- They only work for angles $< 90^\circ$. There can't be any oblique angles in a right triangle, so you can't talk about a hypotenuse.
- When working with polar coordinates, you have to perform complicated visualizations, superposing triangles over everything in your head in order to understand where the formulae are coming from.
- They obfuscate the relationship between $\sin\theta$ and the angle $\theta$.
- It's less clear why they're worth studying. Who cares about ratios of sides in right angle triangles?
These definitions caused a lot of confusion for me up until late high school when I suddenly realized other definitions were available. The best definition for high school mathematics is as follows (it's almost the same as the one given by Shahab, just explained in a different way).
Let's say you've got an angle of $\theta$ jutting up from the horizon towards the sky. You extend a segment along this angle, of length $r$.
And now you want to know what the height and width of this segment are ($y$ and $x$ in the diagram). It turns out there's no easy formula for these, so mathematicians simply define new symbols $\sin\theta$ for the height, and $\cos\theta$ for the width. Actually, I was simplifying a tiny bit. You can imagine that the values of $y$ and $x$ don't just depend on the angle $\theta$. If you make $r$ longer, both $x$ and $y$ will get bigger, but $\theta$ won't change. So to account for that, $\sin$ is actually defined as the value $\frac{y}{r}$, and similarly for $\cos$. Hopefully you can see that these values only depend on $\theta$. If you keep $\theta$ the same but, say, double $r$, then $y$ will get doubled as well, so the ratio $\frac{y}{r}$ will still be the same.
If you think of the triangle formed by the segments $x$, $y$ and $r$ in the diagram, you can see that $r$ is the hypotenuse, $x$ is the adjacent side and $y$ is the opposite side, so you get back the old triangle definition. The difference is that now, you have a definition that also applies to angles greater than $90รง\circ$
You can now easily explain why the trig functions have a part below the $x$-axis. When the angle is greater than $180^\circ$ but smaller than $360^\circ$, where will the angle be pointing? It'll be pointing downwards, so the "height" of the angle will be negative, hence $\sin$ will be negative. Why are they periodic? Because when the angle hits $360^\circ$, it's like you're back to an angle $0$ again. See the animation in robjohn's answer for a great visualization.
Too long for a comment...
Consider that this may be, at least in part, be due to a historical bias: since trigonometric functions were available since the earliest days of mathematics, mathematicians tried to frame their results in familiar terms instead of in other less popular functions. This effect compounds itself, as more and more results are found that involve trigonometric functions.
So perhaps it's not that sin and cos have fundamental properties that make them pervade modern mathematics, but rather that the mathematics we have developed as humans is one which is historically biased towards geometrical descriptions.
In another world where drum acoustics is paramount, perhaps expanding functions in Fourier-Bessel Series is the norm.
Best Answer
Rewrite it like this $$\underbrace{2(\sin a+\cos a)\sin b+\cos b}_{E}=3$$
Since we have by Cauchy inequality $$ \sin a+\cos a \leq \sqrt{(\sin ^2a+\cos ^2a)(1^2+1^2)}=\sqrt{2}$$ we have always $$E \leq \sqrt{8}\sin b+ \cos b $$
Now again by Cauchy inequality $$\sqrt{8}\sin b+ \cos b \leq 3$$
Since we have equality case we have $\sin b :\cos b = \sqrt{8}:1$ so $\tan b = \sqrt{8}$ . Similary we have $\tan a =1$. So the result is $35$.