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Exercise 3.3 · Q1

Q.Prove that sin⁡2π6+cos⁡2π3−tan⁡2π4=−12\sin^2\frac{\pi}{6} + \cos^2\frac{\pi}{3} - \tan^2\frac{\pi}{4} = -\frac{1}{2}.

Uttar Pradesh UpmspTextbookSubjective· 2mImportance★★★★★est
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✓ Free question

This problem is a direct application of standard trigonometric values at special angles. By substituting sin⁡π6=12\sin\frac{\pi}{6} = \frac{1}{2}, cos⁡π3=12\cos\frac{\pi}{3} = \frac{1}{2}, and tan⁡π4=1\tan\frac{\pi}{4} = 1, the expression simplifies to 14+14−1=−12\frac{1}{4} + \frac{1}{4} - 1 = -\frac{1}{2}.

The key insight here is that you don't need to manipulate identities or solve equations — you just need to recall the exact values of sine, cosine, and tangent at the standard angles π6\frac{\pi}{6}, π3\frac{\pi}{3}, and π4\frac{\pi}{4}. These angles appear constantly in trigonometry, and their values are worth memorising cold.

Let’s walk through it step by step.

  1. Recall the value of sin⁡π6\sin\frac{\pi}{6}. The angle π6\frac{\pi}{6} is 30∘30^\circ. In a 30-60-90 triangle, the side opposite 30∘30^\circ is half the hypotenuse. So

sin⁡π6=12.\sin\frac{\pi}{6} = \frac{1}{2}.

Squaring gives

sin⁡2π6=(12)2=14.\sin^2\frac{\pi}{6} = \left(\frac{1}{2}\right)^2 = \frac{1}{4}.

  1. Recall the value of cos⁡π3\cos\frac{\pi}{3}. The angle π3\frac{\pi}{3} is 60∘60^\circ. In the same 30-60-90 triangle, the side adjacent to 60∘60^\circ is half the hypotenuse. So

cos⁡π3=12.\cos\frac{\pi}{3} = \frac{1}{2}.

Squaring gives

cos⁡2π3=(12)2=14.\cos^2\frac{\pi}{3} = \left(\frac{1}{2}\right)^2 = \frac{1}{4}.

  1. Recall the value of tan⁡π4\tan\frac{\pi}{4}. The angle π4\frac{\pi}{4} is 45∘45^\circ. In a 45-45-90 triangle, the legs are equal, so tan⁡45∘=1\tan 45^\circ = 1. Thus

tan⁡π4=1.\tan\frac{\pi}{4} = 1.

Squaring gives

tan⁡2π4=12=1.\tan^2\frac{\pi}{4} = 1^2 = 1.

  1. Substitute into the expression. The original expression is

sin⁡2π6+cos⁡2π3−tan⁡2π4.\sin^2\frac{\pi}{6} + \cos^2\frac{\pi}{3} - \tan^2\frac{\pi}{4}.

Replacing each term:

14+14−1.\frac{1}{4} + \frac{1}{4} - 1.

  1. Simplify.

14+14=24=12.\frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}.

Then

12−1=−12.\frac{1}{2} - 1 = -\frac{1}{2}.

Watch out

A common mistake is to confuse sin⁡π6\sin\frac{\pi}{6} with sin⁡π3\sin\frac{\pi}{3} or to misremember tan⁡π4\tan\frac{\pi}{4} as something other than 1. Always double-check which angle corresponds to which standard value — a quick sketch of the triangles helps.

Tip

If you ever forget these values, draw the unit circle or the special triangles. For π6\frac{\pi}{6} and π3\frac{\pi}{3}, the 30-60-90 triangle has sides 11, 3\sqrt{3}, 22; for π4\frac{\pi}{4}, the 45-45-90 triangle has sides 11, 11, 2\sqrt{2}.

✓Final answer

The value of the expression is −12\boxed{-\frac{1}{2}}.

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