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Miscellaneous Exercise 2 · Q57

Q.If α is a root of 25cos⁡2θ+5cos⁡θ−12=025\cos^2\theta + 5\cos\theta − 12 = 0, π2<α<π\dfrac{\pi}{2} < \alpha < \pi, then sin⁡2α\sin 2\alpha is equal to: A) −2425−\dfrac{24}{25} B) −1318−\dfrac{13}{18} C) 1318\dfrac{13}{18} D) 2425\dfrac{24}{25}

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✓ Free question

Step 1. 25cos⁡2θ+5cos⁡θ−12=025\cos^2\theta+5\cos\theta-12=0; by the quadratic formula, cos⁡θ=−5±25+120050=−5±3550\cos\theta=\dfrac{-5\pm\sqrt{25+1200}}{50}=\dfrac{-5\pm35}{50}, giving cos⁡θ=35\cos\theta=\tfrac35 or cos⁡θ=−45\cos\theta=-\tfrac45.

Step 2. Since π/2<α<π\pi/2<\alpha<\pi (Q2), cos⁡α\cos\alpha must be negative, so cos⁡α=−45\cos\alpha=-\tfrac45.

Step 3. sin⁡2α=1−1625=925\sin^2\alpha=1-\tfrac{16}{25}=\tfrac9{25}, and sinα is positive in Q2, so sin⁡α=35\sin\alpha=\tfrac35.

Step 4. sin⁡2α=2sin⁡αcos⁡α=2(35)(−45)=−2425\sin2\alpha=2\sin\alpha\cos\alpha=2\left(\tfrac35\right)\left(-\tfrac45\right)=-\tfrac{24}{25}.

✓Final answer

A) −24/25.

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