8
$\begingroup$

I noticed something just now. This is probably a stupid question, but I'm going to ask it anyway. Because when I discover that my understanding of a topic is fundamentally flawed, I get nervous.

enter image description here

Basically I'm suppose to show that the angle marked in red is $sin \space \alpha = \frac{3}{5}$. Note that this task is on the part of the test without calculator. My first thought was that the whole thing is 90 degrees. And the other to angles I can fine easily. AB is 1 and BE is 0.5. And the length of AE is $\frac{\sqrt 5}{2}$. So I calculate the angle for the bottom triangle.

$sin \space = \frac{BE}{AE} = \frac{\frac{1}{2}}{\frac{\sqrt 5}{2}} = \frac{1}{\sqrt 5}$

I know that sine of 90 is 1, right. Now it all falls apart, the following is wrong. The angle on the top side of the red angle is equal to the one just calculated. So I did this.

$1 - 2 \times \frac{1}{\sqrt 5}$
And expected to get $\frac{3}{5}$, which I didn't. The following is correct.

$\arcsin(1) - 2 \times \arcsin(\frac{1}{\sqrt 5}) = 36.86$
$\arcsin(\frac{3}{5}) = 36.86$

Why won't the expression without arcsin give me $\frac{3}{5}$ ? Hope this makes sense, I'll be right here pressing F5 and updating if more info is needed. Thank you for input.

$\endgroup$
1
  • 7
    $\begingroup$ A similar thing happens with $f(x) = x^2$. We have $f(1) = 1$ but $f(2) = 4 \neq 2 f(1)$. What's with that? $\endgroup$
    – Sean Eberhard
    Aug 20, 2012 at 8:45

3 Answers 3

Reset to default
14
$\begingroup$

A $sin$ of an angle is not the same as an angle, it's a function of the angle. You can add angles: $$\alpha = \alpha_1+\alpha_2$$ But not the $sin$'s: $$\sin(\alpha) \neq \sin(\alpha_1)+\sin(\alpha_2)$$ In fact, this is true for most functions, and this property is called "non-additivity".

$\endgroup$
10
  • $\begingroup$ Ok, so function that this does work with. Can they said to "preserve addition"? It's an expression I've heard. $\endgroup$
    – Algific
    Aug 20, 2012 at 8:51
  • 3
    $\begingroup$ @nbubis: only if $b=0$, of course! $\endgroup$
    – yatima2975
    Aug 20, 2012 at 11:48
  • 4
    $\begingroup$ @nbubis: Not exactly. This is called additivity. Linear functions, in addition to being additive, are also homogeneous. A function $f(x+y)=x+25y$, where $y$ is rational and $x$ is from a fixed complement subspace of $\bf Q$ in $\bf R$ (as a space over $\bf Q$) is additive, but not linear over reals. $\endgroup$
    – tomasz
    Aug 20, 2012 at 12:08
  • 2
    $\begingroup$ @nbubis: It isn't: the function I'm referring to is a function over the reals. (It is the same for measurable functions over the reals, for example, but not arbitrary.) $\endgroup$
    – tomasz
    Aug 20, 2012 at 12:20
  • 1
    $\begingroup$ @nbubis: A linear function must satisfy both $f(x+y) = f(x) + f(y)$ and $f(cx) = c*f(x)$ (with c a constant). So while it's true that not satisfying the first condition would make the function non-linear, saying "this property is called non-linearity" is incorrect, because it's possible for the first condition $f(x+y) = f(x) + f(y)$ to be satisfied but the function to still be non-linear (by not satisfying the second condition). The name of the first condition is "additivity", so the correct name for your property above would be "non-additive". $\endgroup$
    – BlueRaja - Danny Pflughoeft
    Aug 20, 2012 at 16:32
2
$\begingroup$

You know that $\alpha+2\beta=90^\circ$. There is no rule that $\sin\alpha+2\sin\beta=1$. In maths it's the single most important thing to stick to given rules and not accidentally "invent" new ones.

For your problem: Maybe writing $\sin\alpha=x/y$ with some auxiliarly lines (e.g. the height on the triangle) and successively derive $x$ and $y$ by a chain of Pythagoras applications.

EDIT: Or better use http://en.wikipedia.org/wiki/Law_of_cosines and derive the sine from cos.

$\endgroup$
8
  • $\begingroup$ (+1) for the 'In maths it's the single most important thing to stick to given rules and not accidentally "invent" new ones.' $\endgroup$
    – user11470
    Aug 20, 2012 at 11:42
  • $\begingroup$ @BiditAcharya: I disagree. I think inventing new „rules” is what mathematics is about. For example, this exercise can be easily done as soon as you see that $\sin(\alpha)=\cos(\pi/2-\alpha)$ and $\cos(2\beta)=\cos(\beta)^2-\sin(\beta)^2$ which are both „new rules”. $\endgroup$
    – tomasz
    Aug 20, 2012 at 12:12
  • $\begingroup$ @tomasz: It's about erroneously inventing rules. The main reason for pupils doing so many mistakes in maths is because teachers fail to explain that is no "oh this equations vaguely look like that or that rule", but there is only "it's exactly like" or "it's not applicable". One single incorrect rule in maths is like a virus that spoils all results. So don't guess rules! $\endgroup$
    – Gere
    Aug 20, 2012 at 15:25
  • 1
    $\begingroup$ @Gerenuk: Again, I disagree. You should guess „rules”, and when you do, you will guess wrong often. The thing is, before applying a „rule”, you should verify if it is actually true. If you never guess and invent things, you will never learn anything new. $\endgroup$
    – tomasz
    Aug 20, 2012 at 15:28
  • 1
    $\begingroup$ @Garenuk: The point is, if you make people afraid of making mistakes and force them to just follow same-old algorithms over and over, you're reducing „mathematics” to mindless rote learning. Which is why many people find mathematics incomprehensible. The best way to learn is to make mistakes and understand why they're mistakes, not avoiding mistakes at all costs. $\endgroup$
    – tomasz
    Aug 20, 2012 at 16:09
1
$\begingroup$

Let $\angle EAB = \alpha_1$. Then $\sin(\alpha+\alpha_1)=2/\sqrt{5}$.

$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$

i.e. $\sin\alpha\cos\alpha_1+\cos\alpha\sin\alpha_1=2/\sqrt{5}$

$\cos\alpha_1=2/\sqrt{5}$ and $\sin\alpha_1=1/\sqrt{5}$ (from figure)

Therefore

$$2\sin\alpha+\cos\alpha=2\tag1$$

(Canceling denominator $\sqrt 5$)

If $\sin\alpha=3/5$ then $\cos\alpha=4/5$, i.e.

$$2\sin\alpha+\cos\alpha=2\frac{3}{5}+\frac{4}{5}=2\tag2$$

Eq. $(1)$ = Eq. $(2)$

Hence proved...

Also $\sin(\alpha+\beta)\ne\sin\alpha+\sin\beta$.
(Which is called non-linearity, probably the answer to the question is this)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .