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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(cos,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 corresponds to the shadow projection on y axes(opposite) and the cosine to the shadow projection on x axes(adjacent).

At the previous operations I actually added those lines shadowed on the Cartesian axes. With other words, I added those sides of the triangle that is a 45,45,90.

What is the actual geometric meaning of trigonometric operations such as adding cos,sine,tan etc or subtracting them? I am just adding those sides,lines to get one new line with length $$\sqrt{2}$$ that is all?

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2 Answers 2

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

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so the result is actually the length of the two lines right? –  themhz May 25 '13 at 1:12
    
@themhz No; if it were the length of the two radii, then the result would be $2r$. Remember, trig functions represent ratios, not lengths. –  Emrakul May 25 '13 at 1:13
    
Ok but also, isn't cos projected on the x line and sine on the y line? –  themhz May 25 '13 at 1:21
    
@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. –  Emrakul May 25 '13 at 1:28
    
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 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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Excuse me I am confused. a sinθ. "a" is an angle and "θ"? –  themhz May 25 '13 at 1:08
    
$a$ 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. –  Berci May 25 '13 at 1:26
    
aha ok, I notice the distinction on the font size. But that's the length of the hypotenuse right? –  themhz May 25 '13 at 1:29
    
Ok I think I get it. But I am not on vectors yet. Thank you –  themhz May 25 '13 at 1:42

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