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Exercise 3.3 · Q41

Q.Find sin⁡2x,cos⁡2x,tan⁡2x\sin2x,\cos2x,\tan2x if sec⁡x=−135\sec x=-\dfrac{13}{5}, π2<x<π\dfrac{\pi}{2}<x<\pi

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Step 1: Since π2<x<π\dfrac{\pi}{2}<x<\pi (quadrant II), cos⁡x<0\cos x<0: from sec⁡x=−135\sec x=-\dfrac{13}{5}, cos⁡x=−513\cos x=-\dfrac{5}{13}.

Step 2: sin⁡x=+1−cos⁡2x=1−25169=1213\sin x=+\sqrt{1-\cos^2x}=\sqrt{1-\frac{25}{169}}=\dfrac{12}{13} (sine positive in quadrant II).

Step 3: sin⁡2x=2sin⁡xcos⁡x=2(1213)(−513)=−120169\sin2x=2\sin x\cos x=2\left(\dfrac{12}{13}\right)\left(-\dfrac{5}{13}\right)=-\dfrac{120}{169}.

Step 4: cos⁡2x=2cos⁡2x−1=2(25169)−1=50−169169=−119169\cos2x=2\cos^2x-1=2\left(\dfrac{25}{169}\right)-1=\dfrac{50-169}{169}=-\dfrac{119}{169}.

Step 5: tan⁡2x=sin⁡2xcos⁡2x=−120/169−119/169=120119\tan2x=\dfrac{\sin2x}{\cos2x}=\dfrac{-120/169}{-119/169}=\dfrac{120}{119}.

✓Final answer

sin⁡2x=−120169\sin2x=-\dfrac{120}{169}, cos⁡2x=−119169\cos2x=-\dfrac{119}{169}, tan⁡2x=120119\tan2x=\dfrac{120}{119}

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