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Miscellaneous Examples · Example 20

Q.Find the value of tan⁡π8\tan\frac{\pi}{8}.

Chhattisgarh CgbseTextbookSubjective· 2mImportance★★★★★est
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The key idea is to use the half-angle identity for tangent, starting from tan⁡π4=1\tan\frac{\pi}{4} = 1. Solving the resulting quadratic gives tan⁡π8=2−1\tan\frac{\pi}{8} = \sqrt{2} - 1.

We want tan⁡π8\tan\frac{\pi}{8}. Since π8\frac{\pi}{8} is half of π4\frac{\pi}{4}, the half-angle formula for tangent is the natural tool. The formula comes in three forms, but the one that avoids square roots inside square roots is:

tan⁡θ2=sin⁡θ1+cos⁡θ=1−cos⁡θsin⁡θ\tan\frac{\theta}{2} = \frac{\sin\theta}{1 + \cos\theta} = \frac{1 - \cos\theta}{\sin\theta}

For θ=π4\theta = \frac{\pi}{4}, both sin⁡π4\sin\frac{\pi}{4} and cos⁡π4\cos\frac{\pi}{4} are 22\frac{\sqrt{2}}{2}, so the arithmetic will be clean.

  1. Set up the half-angle identity. Let θ=π4\theta = \frac{\pi}{4}. Then θ2=π8\frac{\theta}{2} = \frac{\pi}{8}. Using the first form:

tan⁡π8=sin⁡π41+cos⁡π4\tan\frac{\pi}{8} = \frac{\sin\frac{\pi}{4}}{1 + \cos\frac{\pi}{4}}

  1. Substitute the known values. sin⁡π4=22\sin\frac{\pi}{4} = \frac{\sqrt{2}}{2} and cos⁡π4=22\cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}. So:

tan⁡π8=221+22\tan\frac{\pi}{8} = \frac{\frac{\sqrt{2}}{2}}{1 + \frac{\sqrt{2}}{2}}

  1. Simplify the denominator. Write 11 as 22\frac{2}{2}:

1+22=2+221 + \frac{\sqrt{2}}{2} = \frac{2 + \sqrt{2}}{2}

  1. Divide the fractions.

tan⁡π8=222+22=22×22+2=22+2\tan\frac{\pi}{8} = \frac{\frac{\sqrt{2}}{2}}{\frac{2 + \sqrt{2}}{2}} = \frac{\sqrt{2}}{2} \times \frac{2}{2 + \sqrt{2}} = \frac{\sqrt{2}}{2 + \sqrt{2}}

  1. Rationalise the denominator. Multiply numerator and denominator by the conjugate 2−22 - \sqrt{2}:

tan⁡π8=2(2−2)(2+2)(2−2)\tan\frac{\pi}{8} = \frac{\sqrt{2}(2 - \sqrt{2})}{(2 + \sqrt{2})(2 - \sqrt{2})}

The denominator is a difference of squares: (2)2−(2)2=4−2=2(2)^2 - (\sqrt{2})^2 = 4 - 2 = 2.

  1. Simplify the numerator. …

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