Ching-Kuang Shene (冼鏡光), Professor Emeritus
Department of Computer Science
Michigan Technological University
Houghton, MI 49931
USA
Created October 6, 2025
Updated April 4, 2026
It is extremely easy to prove the Half-Angle identities without using the Pythagorean Theorem and the Pythagorean Identity. What we need is the Product-to-Sum Identities. Here is the proof: $$ \begin{eqnarray*} \sin^2(x) &=& \sin(x)\sin(x) = -\frac{1}{2}\left( \cos(x+x)-\cos(x-x)\right) = \frac{1}{2}\left( 1-\cos(2x) \right), \\ \cos^2(x) &=& \cos(x)\cos(x) = \frac{1}{2}\left( \cos(x+x)-\cos(x-x)\right) = \frac{1}{2}\left( 1+\cos(2x) \right). \end{eqnarray*} $$ Adding the above together yields the Pythagorean Identity. Or, replacing \( x \) by \( \frac{x}{2} \) and taking square root yields \[ \sin\left( \frac{x}{2}\right) = \pm\sqrt{\frac{1-\cos(x)}{2}} \quad\mbox{and}\quad \cos\left( \frac{x}{2}\right) = \pm\sqrt{\frac{1+\cos(x)}{2}}. \]
Thus, we proved the Pythagorean Identity without using the Pythagorean Theorem and the Pythagorean Identity.
Luzia also proved the same without using the Pythagorean Identity and the Pythagorean Theorem. Please visit this page for more details: Luzia's 2015 Proof.
Consider a unit circle with center \( O \). Let \( \overleftrightarrow{OC} \) and \( \overleftrightarrow{OA} \) make an angle of \( x \), where \( A \) and \( C \) are on the circle and \( \overleftrightarrow{OC} \) is the x-axis. Without loss of generality, we assume \( 0 \leq x \leq 90^{\circ} \). Let \( E \) be the mid-point of \( \overline{AC} \). Let \( \varphi = \angle ACO \). From right triangle \( \bigtriangleup OCE \) we have \( \varphi = 90^{\circ} - \frac{x}{2} \) and hence \( \sin(\varphi) = \cos\left(\frac{x}{2}\right)\), \( \cos(\varphi) = \sin\left(\frac{x}{2}\right) \) and \( \overline{EC} = \sin\left( \frac{x}{2}\right) \). Therefore, \( \overline{AC}=2\overline{EC}=2\sin\left(\frac{x}{2}\right) \). From right triangle \( \bigtriangleup AOB \) we have \( \overline{OB} = \cos(x) \) and \( \overline{BC} = 1 - \cos(x) \). From right triangle \( \bigtriangleup ACB \) we have \[ \overline{BC}=\overline{AC}\cos(\varphi) =\overline{AC}\sin\left(\frac{x}{2}\right) = 2\sin^2\left(\frac{x}{2} \right). \] Because we have \[ 1-\cos(x) = \overline{BC} = 2\sin^2\left( \frac{x}{2}\right). \] rearranging yields the half-angle identity for \( \sin() \): \[ \sin^2\left( \frac{x}{2} \right) = \frac{1-\cos(x)}{2} \mbox{\ \ \ \ \ \ \ \ \ \ or \ \ \ \ \ \ \ \ \ \ } \sin\left(\frac{x}{2}\right) = \pm\sqrt{\frac{1-\cos(x)}{2}}. \]
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We have seen the double angle identities Double ngle Identities for \( \cos(x) \): \[ \cos(x) = \cos^2\left( \frac{x}{2} \right) - \sin^2\left( \frac{x}{2} \right). \] Plugging in the \( \sin^2(x/2) \) obtained earlier, rearranging and simplifying yields: \[ \cos^2\left( \frac{x}{2} \right) = \frac{1+\cos(x)}{2} \mbox{\ \ \ \ \ \ \ \ \ \ or \ \ \ \ \ \ \ \ \ \ } \cos\left(\frac{x}{2}\right) = \pm\sqrt{\frac{1+\cos(x)}{2}}. \]
Neither the Pythagorean Theorem nor the Pythagorean Identity was used in deriving the half-angle identities. Therefore, the half-angle identities are independent of the Pythagorean Theorem and the Pythagorean Identity.
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