$$\sin(\pi/4)+\cos(\pi/4)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}= \frac{2\sqrt{2}}{2}=\sqrt{2}$$

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?

  • $\begingroup$ Maybe some of the diagrams in this Wikipedia article might be interesting for you: en.wikipedia.org/wiki/Phasor#Addition $\endgroup$ – Martin Sleziak Jul 30 '15 at 8:48
  • $\begingroup$ @themhz Trying to understand your question ... Geometric representation in what context? In summing the coordinates, i.e., the abscissa and the ordinate together, are you really asking about the function $\sin { x } +\quad \cos { x }$? Or, are you wondering if summing the pairs in each coordinate has a special meaning in terms of the original graph of the unit circle? $\endgroup$ – oemb1905 Aug 2 '15 at 0:47

Well, in the case of adding sines and cosines, you can think of it algebraically:


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:$$\sin\theta+\cos\theta=\frac{\sqrt{2}+\sqrt{2}}{2}=\sqrt{2}$$


And the same for $\sin\theta+\tan\theta$.

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  • $\begingroup$ so the result is actually the length of the two lines right? $\endgroup$ – themhz May 25 '13 at 1:12
  • $\begingroup$ @themhz No; if it were the length of the two radii, then the result would be $2r$. Remember, trig functions represent ratios, not lengths. $\endgroup$ – user66698 May 25 '13 at 1:13
  • $\begingroup$ Ok but also, isn't cos projected on the x line and sine on the y line? $\endgroup$ – themhz May 25 '13 at 1:21
  • $\begingroup$ @themhz Yes, but cosine isn't the length of the x-axis; it's a ratio of the lengths of the x-axis and radius. $\endgroup$ – user66698 May 25 '13 at 1:28
  • $\begingroup$ Ok. So if I add them squared I get by pythagorian theory the length of the hypotenuse which is ρ. But in other case I can add the ratios and get that "abstract" number that I mentioned? $\endgroup$ – themhz May 25 '13 at 1:35

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 $\sqrt2$.

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  • $\begingroup$ Excuse me I am confused. a sinθ. "a" is an angle and "θ"? $\endgroup$ – themhz May 25 '13 at 1:08
  • $\begingroup$ $a$ is a real number, $\theta$ and $\alpha$ are angles. And $(\cos\theta,\,\sin\theta)$ is the unit vector which has angle $\theta$, measured from the positive half of $x$-axis. $\endgroup$ – Berci May 25 '13 at 1:26
  • $\begingroup$ aha ok, I notice the distinction on the font size. But that's the length of the hypotenuse right? $\endgroup$ – themhz May 25 '13 at 1:29
  • $\begingroup$ Ok I think I get it. But I am not on vectors yet. Thank you $\endgroup$ – themhz May 25 '13 at 1:42
  • $\begingroup$ Blue has some very nice constructions $\endgroup$ – Narasimham Jul 30 '15 at 5:07

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