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< Mentor Note -- The OP's question was in the thread title. They have been advised to please make more detailed and clear thread starters >

thanksthanks

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- Thread starter ah4p
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thanks

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- #2

Fredrik

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You're welcome.

(This isn't a question about a textbook-style problem, so I'm moving the thread to general math).

- #3

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$$\sin\theta = \sum_{n=0}^\infty \frac{(-1)^n \theta^{2n+1}}{(2n+1)!}\text{ and }\cos\theta = \sum_{n=0}^\infty \frac{(-1)^n \theta^{2n}}{(2n)!}$$

When [itex]\theta\approx 0[/itex], these approximations will get better and better for smaller numbers of terms. Where do they come from, you ask? Well, in many cases this is how mathematicians define sine and cosine. Otherwise it takes quite a bit of calculus to come up with these formulas, but you can check that they indeed work. Now, going back to [itex]\theta\approx 0[/itex], we write out the first term of sine and the second term of cosine (basically, we're taking terms with [itex]\theta^2[/itex] or lower exponents. We end up with, lo and behold

$$\sin\theta\approx\theta\text{ and }\cos\theta\approx 1-\frac{\theta^2}{2}.$$

Indeed, if you use a graphing program to graph this you get some very promising results around [itex]\theta = 0[/itex].

The first image plots sine in black and [itex]f(x)=x[/itex] in red, the second image plots cosine in black and its approximation in red. Amazingly, the approximations don't look so bad!

Of course, there are always geometric explanations too, but this is the slightly more sophisticated way to view the approximations.

- #4

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Where do they come from, you ask? Well, in many cases this is how mathematicians define sine and cosine. Otherwise it takes quite a bit of calculus to come up with these formulas,

I wouldn't say so - taking the derivative of sin(x) and cos(x) repeatedly and plugging into Taylor's equation is simple enough for anyone who asks this question!

Rationalising

- #5

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thank you very much for that. I noticed when I tried to find the point of intersection of sinx and tanx graphically they were x = 0, pi, 2piThe most obvious way to explain these approximations is truncating the Taylor series about zero. If you haven't seen these, they're basically infinite series for the trig functions. In the case of sine and cosine, they are

$$\sin\theta = \sum_{n=0}^\infty \frac{(-1)^n \theta^{2n+1}}{(2n+1)!}\text{ and }\cos\theta = \sum_{n=0}^\infty \frac{(-1)^n \theta^{2n}}{(2n)!}$$

When [itex]\theta\approx 0[/itex], these approximations will get better and better for smaller numbers of terms. Where do they come from, you ask? Well, in many cases this is how mathematicians define sine and cosine. Otherwise it takes quite a bit of calculus to come up with these formulas, but you can check that they indeed work. Now, going back to [itex]\theta\approx 0[/itex], we write out the first term of sine and the second term of cosine (basically, we're taking terms with [itex]\theta^2[/itex] or lower exponents. We end up with, lo and behold

$$\sin\theta\approx\theta\text{ and }\cos\theta\approx 1-\frac{\theta^2}{2}.$$

Indeed, if you use a graphing program to graph this you get some very promising results around [itex]\theta = 0[/itex].

View attachment 76762

View attachment 76763

The first image plots sine in black and [itex]f(x)=x[/itex] in red, the second image plots cosine in black and its approximation in red. Amazingly, the approximations don't look so bad!

Of course, there are always geometric explanations too, but this is the slightly more sophisticated way to view the approximations.

how does this mean sinx = tanx ?? I thought it meant sinx only equals tan x when x is 0 pi or 2pi??

tah again

- #6

Fredrik

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It doesn't, but when x is small, we have ##\cos x\approx1##, and thereforehow does this mean sinx = tanx ??

$$\tan x=\frac{\sin x}{\cos x}\approx \frac{\sin x}{1}=\sin x.$$

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- #7

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ah thank you very much that makes sense :DIt doesn't, but when x is small, we have ##\cos x=1##, and therefore

$$\tan x=\frac{\sin x}{\cos x}\approx \frac{\sin x}{1}=\sin x.$$

- #8

Fredrik

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$$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.$$ This implies that when ##h## is small, we have

$$f'(x)\approx \frac{f(x+h)-f(x)}{h}$$ for all ##x##, and therefore

$$f(x+h)\approx f(x)+f'(x)h$$ for all ##x##. This implies that when ##h## is small, we have

$$f(h)\approx f(0)+f'(0)h.$$ This result tells us in particular that when ##x## is small, we have

$$\sin x\approx\sin 0+\sin'(0)x =0+\cos 0\cdot x=x,$$ and

$$\cos x\approx\cos 0+\cos'(0)x= 1-\sin 0\cdot x =1.$$

- #9

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thank very much :D but why does that result in sin x = xIt doesn't, but when x is small, we have ##\cos x\approx1##, and therefore

$$\tan x=\frac{\sin x}{\cos x}\approx \frac{\sin x}{1}=\sin x.$$

Edit:I typed ##\cos x=1## above, but I meant ##\cos x\approx 1##. I fixed that after the reply below.

- #10

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cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! + x^8/8! + ...

For x<1, x^n<x(<1) for all n>1. For very small x, by extension, x^n is even smaller for all n>1; so for x close to 0, x^3 and x^5 and further is extremely close to 0 and can be called 0. So, for x close to 0,

sin(x) = x

cos(x) = 1

approximately.

(So tan(x) = sin(x)/cos(x) = sin(x)/1 (using cos(x)=1) = x/1 (using sin(x)=x) = x approximately, for x close to 0)

These relationships are all

- #11

Fredrik

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It doesn't. I explained why ##\sin x\approx x## for small ##x## in post #8, i.e. the post I wrote just before you asked the question I'm quoting now. It was also explained three times in post #2 (use the link), and twice in post #3. Now it's also explained in post #10.thank very much :D but why does that result in sin x = x

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