Exact Value of tan 72°

We will learn to find the exact value of tan 72 degrees using the formula of submultiple angles.

Let, A = 18°                        

Therefore, 5A = 90° 

⇒ 2A + 3A = 90˚

⇒ 2A = 90˚ - 3A

Taking sine on both sides, we get 

sin 2A = sin (90˚ - 3A) = cos 3A 

⇒ 2 sin A cos A = 4 cos\(^{3}\) A - 3 cos A

⇒ 2 sin A cos A - 4 cos\(^{3}\) A + 3 cos A = 0

⇒ cos A (2 sin A - 4 cos\(^{2}\) A + 3) = 0 

Dividing both sides by cos A = cos 18˚ ≠ 0, we get

⇒ 2 sin A - 4 (1 - sin\(^{2}\) A) + 3 = 0

⇒ 4 sin\(^{2}\) A + 2 sin A - 1 = 0, which is a quadratic in sin A

Therefore, sin A = \(\frac{-2 \pm \sqrt{- 4 (4)(-1)}}{2(4)}\)

⇒ sin A = \(\frac{-2 \pm \sqrt{4 + 16}}{8}\)

⇒ sin A = \(\frac{-2 \pm 2 \sqrt{5}}{8}\)

⇒ sin A = \(\frac{-1 \pm \sqrt{5}}{4}\)

Now sin 18° is positive, as 18° lies in first quadrant.

Therefore, sin 18° = sin A = \(\frac{√5 - 1}{4}\)

Now, cos 72° = cos (90° - 18°) = sin 18° = \(\frac{√5 - 1}{4}\)

And cos 18° = √(1 - sin\(^{2}\) 18°), [Taking positive value, cos 18° > 0]

⇒ cos 18° = \(\sqrt{1 - (\frac{\sqrt{5} - 1}{4})^{2}}\)

⇒ cos 18° = \(\sqrt{\frac{16 - (5 + 1 - 2\sqrt{5})}{16}}\)

⇒ cos 18° =  \(\sqrt{\frac{10 + 2\sqrt{5}}{16}}\)

Thus, sin 72° = sin (90° - 18°) = cos 18° = \(\frac{\sqrt{10 + 2\sqrt{5}}}{4}\)

Now, tan 72° = \(\frac{sin 72°}{cos 72°}\) = \(\frac{\frac{\sqrt{10 + 2\sqrt{5}}}{4}}{\frac{√5 - 1}{4}}\) = \(\frac{\sqrt{10 + 2√5}}{√5 - 1}\)

Therefore, tan 72° =\(\frac{\sqrt{10 + 2√5}}{√5 - 1}\)

 Submultiple Angles






11 and 12 Grade Math

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