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Exercise 3.3 · Q4

Q.Prove that 2sin⁡23π4+2cos⁡2π4+2sec⁡2π3=102\sin^2\frac{3\pi}{4} + 2\cos^2\frac{\pi}{4} + 2\sec^2\frac{\pi}{3} = 10.

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Evaluate each trigonometric term at its special angle, then sum: the left side simplifies to 1+1+8=101 + 1 + 8 = 10.

The heart of this problem is recognizing that 3π4\frac{3\pi}{4}, π4\frac{\pi}{4}, and π3\frac{\pi}{3} are special angles whose trigonometric values we know exactly. Once we substitute these values, the algebra becomes straightforward arithmetic.

Special angles are the backbone of trigonometry problems. The angles π4\frac{\pi}{4} (45°), π3\frac{\pi}{3} (60°), and 3π4\frac{3\pi}{4} (135°) appear constantly in exams because their sine, cosine, and tangent values are either simple fractions or involve 2\sqrt{2} and 3\sqrt{3} in predictable ways.

Let me evaluate each term separately, then combine them.

Step-by-step evaluation

  1. Find sin⁡3π4\sin\frac{3\pi}{4}

    The angle 3π4\frac{3\pi}{4} lies in the second quadrant (between π2\frac{\pi}{2} and π\pi). We can write it as π−π4\pi - \frac{\pi}{4}, so:

sin⁡3π4=sin⁡(π−π4)=sin⁡π4=12\sin\frac{3\pi}{4} = \sin\left(\pi - \frac{\pi}{4}\right) = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}

Therefore:

sin⁡23π4=(12)2=12\sin^2\frac{3\pi}{4} = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}

  1. Find cos⁡π4\cos\frac{\pi}{4}

    This is one of the most basic special angles:

cos⁡π4=12\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}

So:

cos⁡2π4=(12)2=12\cos^2\frac{\pi}{4} = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}

  1. Find sec⁡π3\sec\frac{\pi}{3}

    Since sec⁡θ=1cos⁡θ\sec\theta = \frac{1}{\cos\theta}, we need cos⁡π3\cos\frac{\pi}{3} first. The angle π3\frac{\pi}{3} (60°) has:

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

Therefore:

sec⁡π3=1cos⁡π3=112=2\sec\frac{\pi}{3} = \frac{1}{\cos\frac{\pi}{3}} = \frac{1}{\frac{1}{2}} = 2

And:

sec⁡2π3=22=4\sec^2\frac{\pi}{3} = 2^2 = 4

  1. Combine all terms …

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