30
°
0.52359878
rad
0.5
0.8660254
0.57735027
0.8660254
0.5
0.57735027
-1.110223024625e-16
30
°
0.52359878
rad
0.5
0.8660254
0.57735027
0.8660254
0.5
0.57735027
-1.110223024625e-16
The Half Angle Calculator evaluates the trigonometric functions of half a given angle using the half-angle identities. These formulas express $$\sin\!\left(\frac{\theta}{2}\right)$$, $$\cos\!\left(\frac{\theta}{2}\right)$$, and $$\tan\!\left(\frac{\theta}{2}\right)$$ in terms of $$\cos\theta$$, allowing you to find exact trig values of angles that are half of known angles.
The three primary half-angle identities are:
$$\sin\!\left(\frac{\theta}{2}\right) = \pm\sqrt{\frac{1 - \cos\theta}{2}}$$
$$\cos\!\left(\frac{\theta}{2}\right) = \pm\sqrt{\frac{1 + \cos\theta}{2}}$$
$$\tan\!\left(\frac{\theta}{2}\right) = \pm\sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}}$$
The $$\pm$$ sign depends on the quadrant in which $$\frac{\theta}{2}$$ lies. There are also two rationalized forms for the tangent half-angle that avoid the square root:
$$\tan\!\left(\frac{\theta}{2}\right) = \frac{\sin\theta}{1 + \cos\theta} = \frac{1 - \cos\theta}{\sin\theta}$$
The half-angle formulas are derived from the double-angle identity for cosine. Starting with $$\cos(2\alpha) = 1 - 2\sin^2\alpha$$, substitute $$\alpha = \theta/2$$:
$$\cos\theta = 1 - 2\sin^2\!\left(\frac{\theta}{2}\right)$$
Solving for $$\sin(\theta/2)$$ gives the half-angle sine formula. Similarly, using $$\cos(2\alpha) = 2\cos^2\alpha - 1$$ yields the half-angle cosine formula.
Half-angle formulas let you compute exact trigonometric values for many angles. For example, since $$\cos(60°) = \frac{1}{2}$$, you can find exact values at 30°. More usefully, knowing $$\cos(45°) = \frac{\sqrt{2}}{2}$$ lets you find exact values for 22.5°, and $$\cos(30°)$$ gives exact values for 15°.
The tangent half-angle substitution $$t = \tan(\theta/2)$$ is one of the most powerful techniques in calculus. Under this substitution:
$$\sin\theta = \frac{2t}{1+t^2}, \quad \cos\theta = \frac{1-t^2}{1+t^2}, \quad d\theta = \frac{2\,dt}{1+t^2}$$
This converts any rational expression of trigonometric functions into a rational function of $$t$$, which can always be integrated using partial fractions.
Enter angle $$\theta$$ in degrees. The calculator computes $$\theta/2$$ and evaluates $$\sin(\theta/2)$$, $$\cos(\theta/2)$$, and $$\tan(\theta/2)$$ both directly and using the half-angle identity forms. The alternative tangent calculation via $$\sin\theta/(1+\cos\theta)$$ serves as a cross-check.
The calculator converts $$\theta$$ and $$\theta/2$$ to radians, then evaluates trigonometric functions directly using JavaScript's Math library. It also computes $$\tan(\theta/2)$$ using the rationalized form $$\sin\theta/(1+\cos\theta)$$ as verification. Both methods should agree within floating-point precision.
The primary outputs show the trig values at $$\theta/2$$. The alternative tangent value (via sin/cos identity) should match the direct tangent calculation. If they differ by more than a tiny amount, it indicates a boundary case. NaN appears when the function is undefined (e.g., $$\tan(90°)$$ when $$\theta = 180°$$).
Inputs
Results
Half of 60° is 30°, giving the well-known values sin(30°) = 0.5 and cos(30°) = √3/2 ≈ 0.866. Both tan(θ/2) methods agree at 1/√3 ≈ 0.5774.
Inputs
Results
Half of 90° is 45°. At 45°, sine and cosine are equal (√2/2 ≈ 0.7071) and tangent is exactly 1. The identity tan(45°) = sin(90°)/(1+cos(90°)) = 1/1 = 1 checks out.
The half-angle formula for sine is $$\sin\!\left(\frac{\theta}{2}\right) = \pm\sqrt{\frac{1 - \cos\theta}{2}}$$. The sign is positive when $$\theta/2$$ is in the first or second quadrant, and negative in the third or fourth quadrant.
The sign depends on which quadrant $$\theta/2$$ falls in. For sine, the result is positive in quadrants I and II. For cosine, it is positive in quadrants I and IV. You must determine the quadrant of $$\theta/2$$ (not $$\theta$$) to choose correctly.
The Weierstrass (or tangent half-angle) substitution sets $$t = \tan(\theta/2)$$, then expresses $$\sin\theta$$, $$\cos\theta$$, and $$d\theta$$ in terms of $$t$$. This transforms trigonometric integrals into rational function integrals that can be solved with partial fractions.
Yes. Since $$15° = 30°/2$$ and $$\cos(30°) = \frac{\sqrt{3}}{2}$$, the half-angle formula gives $$\sin(15°) = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} = \frac{\sqrt{6} - \sqrt{2}}{4}$$.
The calculator shows $$\tan(\theta/2)$$ computed directly and also via the identity $$\frac{\sin\theta}{1 + \cos\theta}$$. Both should match, serving as a cross-verification. The rationalized form avoids square roots and sign ambiguity.
Yes, they are inverses. The half-angle formulas are derived by solving the double-angle cosine identity for $$\sin^2(\alpha)$$ or $$\cos^2(\alpha)$$ and substituting $$\alpha = \theta/2$$. Every half-angle identity can be traced back to a double-angle identity.
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