# why $1 - \cos^2x = \sin^2x$? [duplicate]

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}

• What do you take as the definition of $\cos$ and $\sin$? – Element118 Dec 20 '15 at 7:09
• Google Pythagorean identity – user223391 Dec 20 '15 at 7:09
• $\sin^2(x) + \cos^2(x) = 1$ follows immediately from the Pythagorean theorem. – littleO Dec 20 '15 at 8:32

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 $\cos x = \dfrac{b}{\sqrt {a^2 + b^2}}$

$\sin^2 x = \dfrac{a^2}{a^2 + b^2}$

$\cos^2 x = \dfrac{b^2}{a^2 + b^2}$

$\sin^2 x + \cos^2 x = \dfrac{a^2}{a^2 + b^2} + \dfrac{b^2}{a^2 + b^2} = 1$

This is just a basic property in Trigonometry. There are many ways to prove it . Here is one way

$\sin^2x + \cos^2x= \sin x \sin x+ \cos x \cos x = \cos(x-x) = \cos 0 = 1$

• its wrong answer. To prove the statement $\sin x \sin x+ \cos x \cos x = \cos(x-x)$ we have to use that $\sin^2 x+ \cos^2 x=1.$ – Leox Dec 20 '15 at 8:45
• @Leox Where am i using $\sin^2x + \cos^2x=1$ in my proof ? – Dimenein Dec 20 '15 at 8:58
• Why $\sin x \sin x+ \cos x \cos x = \cos(x-x)?$ – Leox Dec 20 '15 at 9:00
• I am using the formula $cos(x-y)=cos(x)cos(y)+sin(y)sin(x)$. And when we put $x=y$ we get $cos(x-x) = cos^2x+sin^2x = cos0 =1$ – Dimenein Dec 20 '15 at 9:05
• yes, of cource.. but to prove the formula you need use the identity $\sin^2 x+\cos^2x=1$ – Leox Dec 20 '15 at 9:07

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, $$=(e^{ix})(e^{-ix})=e^{0}=1$$

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

$$\sin ^2 x + \cos ^ 2 x=1$$

• 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. – Taladris Dec 20 '15 at 8:27
• it is wrong answer. To prove the formulas for the derivatives of $\sin x$ and $\cos x$ we have to use the identity $\sin^2 x+ \cos^2 x=1.$ – Leox Dec 20 '15 at 8:48
• @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. – luka5z Dec 20 '15 at 8:50
• yes, but power series are not basic things – Leox Dec 20 '15 at 8:54
• @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 – luka5z Dec 20 '15 at 8:56

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.