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

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?

oemb1905
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themhz
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  • Maybe some of the diagrams in this Wikipedia article might be interesting for you: https://en.wikipedia.org/wiki/Phasor#Addition – Martin Sleziak Jul 30 '15 at 08:48
  • @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? – oemb1905 Aug 02 '15 at 00:47

2 Answers2

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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 $\sqrt2$.

Berci
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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 $\sin\theta+\tan\theta$.

  • so the result is actually the length of the two lines right? – themhz May 25 '13 at 01:12
  • @themhz No; if it were the length of the two radii, then the result would be $2r$. Remember, trig functions represent ratios, not lengths. –  May 25 '13 at 01:13
  • Ok but also, isn't cos projected on the x line and sine on the y line? – themhz May 25 '13 at 01:21
  • @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. –  May 25 '13 at 01:28
  • 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? – themhz May 25 '13 at 01:35
  • See my above answer - you can add any trig functions, so yes. You can find any of these ratios using the above method. –  May 25 '13 at 01:39
  • In the unit circle only, yes, they would be the line lengths would they not? – oemb1905 Aug 02 '15 at 00:49
  • @oemb I'm not sure I understand, but if I do, then yeah, since $r=1$, they are just the $x$ and $y$ lengths. –  Aug 02 '15 at 02:13
  • @Emrakul yeah, i just think that is @ themhz was thinking of, back in his first comment ... since the ratio simplifies in that case and since his examples had amplitudes of 1 ... – oemb1905 Aug 02 '15 at 02:15