$$\sin(\pi/4)+\cos(\pi/4)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}= \frac{2\sqrt{2}}{2}=\sqrt{2}$$
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?
Best Answer
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 $\sqrt2$.