What is the actual geometric meaning of trigonometric operations such as adding cos,sine,tan $$\sin(\pi/4)+\cos(\pi/4)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}= \frac{2\sqrt{2}}{2}=\sqrt{2}$$
Thinking of trig components (cosine, sine) that I used to produce the result using the mechanics of algebra, makes me wonder what is the geometric representation of 
  $$\sin(π/4)+\cos(π/4)$$ 
The sine function corresponds to the shadow projection on $y$-axis (opposite) and the cosine function to the shadow projection on $x$-axis (adjacent). 
At the previous operations I actually added those lines shadowed on the Cartesian axes. In other words, I added those sides of the triangle that form a $45-45-90$.
What is the actual geometric meaning of trigonometric operations such as adding cosine, sine, tangent, etc., or subtracting them? Am I just adding those sides and lines in order to get one new line with length $$\sqrt{2}$$ Is that all?  
 A: We have
$$a\,\sin\theta+b\,\cos\theta=\sqrt{a^2+b^2}\,\sin(\theta+\alpha)$$
where $\alpha$ is the unique angle such that $\cos\alpha=a/\sqrt{a^2+b^2}$ and $\sin\alpha=b/\sqrt{a^2+b^2}$, in case $a^2+b^2 >0$.
Note that $\alpha$ is the angle of the vector $(a,b)$, measured from the positive half of $x$-axis, and $r:=\sqrt{a^2+b^2}$ is its length.
It means that, $a\,\sin\theta+b\,\cos\theta$ is the $y$ coordinate of the rotation of $(\cos\theta,\,\sin\theta)$ by $\alpha$, multiplied by $r$.
In your example $a=b=1$ so $r=\sqrt2$ and $\alpha=\pi/4$. Then the rotated and stretched vector will have angle $\pi/4+\pi/4=\pi/2$ and length $\sqrt2$. Its $y$ coordinate is indeed $\sqrt2$.
A: Well, in the case of adding sines and cosines, you can think of it algebraically:
$$\sin\theta=\frac{y}{r};\cos\theta=\frac{x}{r}$$
Therefore, when you're taking $\sin\theta+\cos\theta$ you're really finding an expression for $\displaystyle\frac{x+y}{r}$. Similarly:
$$\cos\theta+\tan\theta=\frac{x}{r}+\frac{y}{x}=\frac{x^2+ry}{rx}$$
$$\sin\theta+\tan\theta=\frac{y}{r}+\frac{y}{x}=\frac{xy+ry}{rx}$$
In the case of a 45-45-90 triangle, represented by $\theta=\frac{\pi}{4}$, you have $x=y=\sqrt{2}$ and $r=2$, therefore:$$\sin\theta+\cos\theta=\frac{\sqrt{2}+\sqrt{2}}{2}=\sqrt{2}$$
$$\cos\theta+\tan\theta=\frac{2+2\sqrt{2}}{2\sqrt{2}}=\frac{2+\sqrt{2}}{2}$$
And the same for $\sin\theta+\tan\theta$.
