I'm trying to prove this result $$\lim_{x\to 0} \frac{1 - \cos(x)}{x} = 0$$ In this process I have come across an identity $1-\cos^2x=\sin^2x$. Why should this hold ? Here are a few steps of my working: \begin{array}\\ \lim_{x\to 0} \dfrac{1 - \cos(x)}{x}\\ = \lim_{x\to 0} \left[\dfrac{1 - \cos(x)}{x} \times \dfrac{1 + \cos(x)}{1 + \cos(x)}\right] \\ =\lim_{x\to 0} \left[\dfrac{1 - \cos^2(x)}{x(1+\cos(x))}\right] \\ =\lim_{x\to 0} \left[\dfrac{\sin^2(x)}{x(1+\cos(x))}\right] \end{array}
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1$\begingroup$ What do you take as the definition of $\cos$ and $\sin$? $\endgroup$ – Element118 Dec 20 '15 at 7:09
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$\begingroup$ Google Pythagorean identity $\endgroup$ – user223391 Dec 20 '15 at 7:09
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$\begingroup$ $\sin^2(x) + \cos^2(x) = 1$ follows immediately from the Pythagorean theorem. $\endgroup$ – littleO Dec 20 '15 at 8:32
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It's a Pythagorean identity and comes from $$\sin^2 x + \cos ^2 x = 1$$
Just subtract $\cos ^2 x$ from both sides and you have your answer. Now, as to where $\sin^2 x + \cos ^2 x = 1$ comes from:
Let's say you have a right triangle with legs $a$ and $b$. By the Pythagorean theorem, the hypotenuse is $$\sqrt {a^2 + b^2}$$
Next, $\sin x$ is defined as $\dfrac{opposite}{hypotenuse}$, and $\cos x$ is defined as $\dfrac{adjacent}{hypotenuse}$
So in your triangle you have $\sin x = \dfrac{a}{\sqrt {a^2 + b^2}}$ and $\cos x = \dfrac{b}{\sqrt {a^2 + b^2}}$
$\sin^2 x = \dfrac{a^2}{a^2 + b^2}$
$\cos^2 x = \dfrac{b^2}{a^2 + b^2}$
$\sin^2 x + \cos^2 x = \dfrac{a^2}{a^2 + b^2} + \dfrac{b^2}{a^2 + b^2} = 1$
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This is just a basic property in Trigonometry. There are many ways to prove it . Here is one way
$\sin^2x + \cos^2x= \sin x \sin x+ \cos x \cos x = \cos(x-x) = \cos 0 = 1$
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1$\begingroup$ its wrong answer. To prove the statement $\sin x \sin x+ \cos x \cos x = \cos(x-x)$ we have to use that $\sin^2 x+ \cos^2 x=1.$ $\endgroup$ – Leox Dec 20 '15 at 8:45
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$\begingroup$ @Leox Where am i using $\sin^2x + \cos^2x=1$ in my proof ? $\endgroup$ – Dimenein Dec 20 '15 at 8:58
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$\begingroup$ I am using the formula $cos(x-y)=cos(x)cos(y)+sin(y)sin(x)$. And when we put $x=y$ we get $cos(x-x) = cos^2x+sin^2x = cos0 =1$ $\endgroup$ – Dimenein Dec 20 '15 at 9:05
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1$\begingroup$ yes, of cource.. but to prove the formula you need use the identity $\sin^2 x+\cos^2x=1$ $\endgroup$ – Leox Dec 20 '15 at 9:07
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Here is another simple way to prove the trig. identity, we know $i^2=-1$ so we have $$\sin^2 x+\cos^2x$$$$=\cos^2x-i^2\sin^2x$$ $$=(\cos x+i\sin x)(\cos x-i\sin x)$$ using Euler's theorem, $$=(e^{ix})(e^{-ix})=e^{0}=1$$
answered Dec 20 '15 at 8:12
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Let $F(x)=\sin ^2 x + \cos ^ 2 x$.
$$F'(x)=2 \sin x\cos x- 2 \cos x\sin x=0$$
Since $F(0)=1$ and $F$ is constant, we get
$$\sin ^2 x + \cos ^ 2 x=1$$
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3$\begingroup$ My guess is that, since the OP is looking for the limit of $\frac{1-\cos(x)}{x}$, he may be at a stage where derivatives of $\sin$ and $\cos$ are not proved yet/are being proved. So he may be looking for something very elementary... But proving identities using differentiation is a nice method, so you have my +1 anyway. $\endgroup$ – Taladris Dec 20 '15 at 8:27
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1$\begingroup$ it is wrong answer. To prove the formulas for the derivatives of $\sin x$ and $\cos x$ we have to use the identity $\sin^2 x+ \cos^2 x=1.$ $\endgroup$ – Leox Dec 20 '15 at 8:48
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1$\begingroup$ @Leox not at all. You can do it using power series expansions and it's easy. Power series is a standard way of defining sine and cosine functions in many older books. $\endgroup$ – luka5z Dec 20 '15 at 8:50
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$\begingroup$ @Leox but it does not mean it's a wrong answer... But I agree, in this case Ovi's answer is probably the best and the most accessible $\endgroup$ – luka5z Dec 20 '15 at 8:56
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If you define $\cos$ and $\sin$ as ratios of legs to hypotenuse of a right triangle, then $\cos^2 (x) +\sin^2 (x) =1$ follows directly from the Pythagorean identity.
