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Exercise 3.2 · Q9

Q.Find the value of the trigonometric function sin⁡(−11π3)\sin\left(-\frac{11\pi}{3}\right).

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The key idea is to use the periodicity of sine (sin⁡(θ+2π)=sin⁡θ\sin(\theta + 2\pi) = \sin\theta) and the odd property (sin⁡(−θ)=−sin⁡θ\sin(-\theta) = -\sin\theta) to reduce the angle to a standard one. The value is 32\boxed{\frac{\sqrt{3}}{2}}.

Why This Approach Works

Trigonometric functions repeat every 2π2\pi radians. That means sin⁡(θ+2πn)=sin⁡θ\sin(\theta + 2\pi n) = \sin\theta for any integer nn. For a negative angle like −11π3-\frac{11\pi}{3}, we can add multiples of 2π2\pi until we land in a familiar range — usually between 00 and 2π2\pi, or between −π-\pi and π\pi. Then we use the quadrant signs to get the exact value.

The angle −11π3-\frac{11\pi}{3} is large in magnitude, but the periodicity lets us shrink it down.

Step-by-Step Solution

  1. Rewrite the angle by adding 2π2\pi repeatedly. Since sin⁡(θ+2π)=sin⁡θ\sin(\theta + 2\pi) = \sin\theta, we can add 2π2\pi (which is 6π3\frac{6\pi}{3}) to −11π3-\frac{11\pi}{3} until the angle becomes positive and small.

−11π3+6π3=−5π3-\frac{11\pi}{3} + \frac{6\pi}{3} = -\frac{5\pi}{3}

Still negative. Add another 2π2\pi:

−5π3+6π3=π3-\frac{5\pi}{3} + \frac{6\pi}{3} = \frac{\pi}{3}

So sin⁡(−11π3)=sin⁡(π3)\sin\left(-\frac{11\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right).

Tip

A faster way: divide the angle by 2π2\pi to see how many full cycles to remove. Here −11π3÷2π=−116-\frac{11\pi}{3} \div 2\pi = -\frac{11}{6}, so we add 22 full cycles (+2×2π+2 \times 2\pi) to get −11π3+4π=−11π3+12π3=π3-\frac{11\pi}{3} + 4\pi = -\frac{11\pi}{3} + \frac{12\pi}{3} = \frac{\pi}{3}. …

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