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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.

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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.

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    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?2012-08-20

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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".

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    Ok, so function that this does work with. Can they said to "preserve addition"? It's an expression I've heard.2012-08-20
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    @Algific - only *linear functions* preserve this property. In 1D, this is true only for function of the form: $f(x) = ax+b$2012-08-20
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    @nbubis: only if $b=0$, of course!2012-08-20
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    @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.2012-08-20
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    @tomasz - You're right of course, but for functions over the reals this is the same.2012-08-20
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    @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*.)2012-08-20
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    @nbubis Take care, it's not true for `f(x)=1` or `f(x)=x+1`, as these are "affine" rather than "linear".2012-08-20
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    @haydoni - I'm familiar with the terminology you are using , but in other contexts affine functions are also called linear.2012-08-20
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    @nbubis In which case linear does not imply additive, see examples above. :)2012-08-20
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    @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"*.2012-08-20
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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.

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    (+1) for the 'In maths it's the single most important thing to stick to given rules and not accidentally "invent" new ones.'2012-08-20
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    @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”.2012-08-20
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    @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!2012-08-20
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    @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.2012-08-20
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    @tomasz, "The thing is, before applying a „rule”, you should verify if it is actually true." that is exactly my point. I don't know why you disagree :)2012-08-20
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    @tomasz: You like to disagree even if things aren't wrong, huh?! :) Anyway, you might do it right, but if you teach students your way, the average student will fail exams. The option of intuition is poison to the average student.2012-08-20
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    @Gerenuk: I primarily disagree with the wording. And mathematics is about mathematics, the art, the reasoning, and yes, the intuition, not passing (or failing) exams. Of course, you must be wary not to trust your intuition too much, since it *is* often misleading. In mathematics as much as anywhere else.2012-08-20
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    @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.2012-08-20
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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)