Why should we have $\sin^2(x) = \frac{1-\cos(2x)}{2}$ knowing that $\sin^2(x) = 1 - \cos^2(x)$?

Logically, can you not subtract $\cos^2(x)$ to the other side from this Pythagorean identity $\sin^2(x)+\cos^2(x)=1?$

When I look up trig identities, however, it says $\sin^2(x) = \frac{1-\cos(2x)}{2}$.

Why is this?

  • $\begingroup$ Welcome to StackExchange. Please use LaTeX. $\endgroup$ – user394255 Aug 5 '17 at 17:34
  • $\begingroup$ @A.Molendijk or you know MathJax $\endgroup$ – user451844 Aug 5 '17 at 17:35
  • $\begingroup$ The second formula is the double angle formula. $\endgroup$ – Donald Splutterwit Aug 5 '17 at 17:35
  • 6
    $\begingroup$ Both formulas are true, they don't contradict themselves. $\endgroup$ – Reiner Martin Aug 5 '17 at 17:39
  • $\begingroup$ Your proposed identity $\sin^2 x = 1-\cos^2 x$ is correct. And so is the one you "looked up", tha $\sin^2 x = \dfrac{1-\cos(2x)} 2. \qquad $ $\endgroup$ – Michael Hardy Aug 5 '17 at 18:20

Notice that $\cos^{2}(x):=(\cos(x))^{2}$ is not the same thing as $\cos(2x)$. It is indeed true that $\sin^{2}(x)=1-\cos^{2}(x)$ and that $\sin^{2}(x)=\frac{1-\cos(2x)}{2}$.


Both formulas are true, however, both are useful in different contexts (applications).

  • You use $\sin^2(x) = \frac{1-\cos(2x)}{2}$ for integrating $\sin^2(x)$.

  • You use $\sin^2(x) = 1 - \cos^2(x)$, for example, when solving $\sin^2(x) = 2\cos(x)$.

Note that it is just in some way more "natural" to write $\sin^2(x) + \cos^2(x)=1$, because this gives both $\sin^2(x) = 1 - \cos^2(x)$ and $\cos^2(x) = 1 - \sin^2(x)$ in one "natural looking" formula.


From the angle addition formula, we have

$$\begin{align} \cos(2x)&=\cos(x+x)\\\\ &=\cos(x)\cos(x)-\sin(x)\sin(x)\\\\ &=\cos^2(x)-\sin^2(x)\\\\ &=\left(1-\sin^2(x)\right)-\sin^2(x)\\\\ &=1-2\sin^2(x)\tag 1 \end{align}$$

Solving $(1)$ for $\sin^2(x)$ yields


as expected.


Let's address two things, the question in the post and the confusion others have pointed out.

Part 1: Proof

Here's a proof that $\cos^2(\theta)+\sin^2(\theta)=1$, from which you can show that $\sin^2(\theta)=1-\cos^2(\theta)$. It requires a bit of set up using the Euler identity and that $i=\sqrt{-1}$. If you are not familiar with complex numbers, find yourself a video on YouTube, you'll get the hang of them in next to no time.

$$e^{i\theta}=\cos(\theta)+i\sin(\theta)\tag{1}$$ $$e^{-i\theta}=\cos(\theta)-i\sin(\theta)\tag{2}$$ Now, add (1) and (2). $$e^{i\theta}+e^{-i\theta}=\cos(\theta)+i\sin(\theta)+\cos(\theta)-i\sin(\theta)=2\cos(\theta)\tag{3}$$

Subtract (2) from (1) $$e^{i\theta}-e^{-i\theta}=\cos(\theta)+i\sin(\theta)-\cos(\theta)+i\sin(\theta)=2i\sin(\theta)\tag{4}$$ From (3) and (4) we get (5) and (6) $$\cos(\theta)=\frac{e^{i\theta}+e^{-i\theta}}{2}\tag{5}$$ $$\sin(\theta)=\frac{e^{i\theta}-e^{-i\theta}}{2i}\tag{6}$$

now, add the squares of (5) and (6) $$\cos^2(\theta)=\frac{e^{2i\theta}+2e^{i\theta-i\theta}+e^{-2i\theta}}{4}=\frac{e^{2i\theta}+2+e^{-2i\theta}}{4}\tag{7}$$ $$\sin^2(\theta)=\frac{e^{2i\theta}-2e^{i\theta-i\theta}+e^{-2i\theta}}{-4}=\frac{e^{2i\theta}-2+e^{-2i\theta}}{-4}\tag{8}$$



Part 2: Addressing the confusion of $\cos(2x)$ and $\cos^2(x)$

$$\cos(2\theta)=\frac{e^{2i\theta}+e^{-2i\theta}}{2}= \frac{\left(e^{i\theta}\right)^2+\left(e^{-i\theta}\right)^2}{2} = \frac{(\cos(\theta)+i\sin(\theta))^2+(\cos(\theta)-i\sin(\theta))^2}{2}\tag{11}$$ Processing this result further, we obtain

$$\cos(2\theta)=\frac{\cos^2(\theta)+2i\cos(\theta)\sin(\theta) - \sin^2(\theta) + \cos^2(\theta) -2i\cos(\theta)\sin(\theta)-\sin^2(\theta) }{2}=\frac{2\cos^2(\theta)-2\sin^2(\theta)}{2}$$ Therefore $$\cos(2\theta)=\cos^2(\theta)-\sin^2(2\theta)\tag{12}$$.

Ok, you stated that $$\sin^2(x)=\frac{1-\cos(2x)}{2}\tag{13}$$ well, if you substitute the result from (12) into (13), you find that

$$\sin^2(\theta)=\frac{1-\cos^2(\theta)+\sin^2(2\theta)}{2}\tag{14}$$ Rearrange this (multiply both sides by 2 and then subtract the $sin^2(\theta)$ from the right side to the left) to get $$\sin^2(\theta)=1-\cos^2(\theta)\tag{15}$$


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