Question 49 of 50intermediate🔧 ApplyLong Answer4 marks

Find sin⁡x2\sin \frac{x}{2}, cos⁡x2\cos \frac{x}{2} and tan⁡x2\tan \frac{x}{2} in the following: cos⁡x=−13\cos x = -\frac{1}{3}, xx in quadrant III.

Correct Answer
sin⁡x2=63,cos⁡x2=−33,tan⁡x2=−2\sin\frac{x}{2} = \frac{\sqrt{6}}{3}, \quad \cos\frac{x}{2} = -\frac{\sqrt{3}}{3}, \quad \tan\frac{x}{2} = -\sqrt{2}
Exercise: MISCELLANEOUS EXERCISE ON CHAPTER 3 | Q: 9 | (Page 29)
For More Understanding

Explanation

This question tests half-angle formulas and quadrant analysis. The key insight is determining which quadrant x/2x/2 lies in when xx is in quadrant III.

Since π<x<3π2\pi < x < \frac{3\pi}{2}, dividing by 2 gives π2<x2<3π4\frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}, placing x/2x/2 in quadrant II where sine is positive and cosine is negative. This determines the signs in the half-angle formulas.

Solution Steps

  1. Step 1: Given cos⁡x=−13\cos x = -\frac{1}{3} and xx lies in quadrant III. Since π<x<3π2\pi < x < \frac{3\pi}{2}, we have π2<x2<3π4\frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}. Therefore, x/2x/2 lies in quadrant II where sin⁡(x/2)\sin(x/2) is positive and cos⁡(x/2)\cos(x/2) is negative.

  2. Step 2: Using the half-angle formula:

    sin⁡x2=±1−cos⁡x2=±1−(−13)2=±432=±23=±63\sin\frac{x}{2} = \pm\sqrt{\frac{1 - \cos x}{2}} = \pm\sqrt{\frac{1 - \left(-\frac{1}{3}\right)}{2}} = \pm\sqrt{\frac{\frac{4}{3}}{2}} = \pm\sqrt{\frac{2}{3}} = \pm\frac{\sqrt{6}}{3}

    Since x/2x/2 is in quadrant II, sin⁡(x/2)>0\sin(x/2) > 0. Therefore, sin⁡x2=63\sin\frac{x}{2} = \frac{\sqrt{6}}{3}.

  3. Step 3: Using the half-angle formula:

    cos⁡x2=±1+cos⁡x2=±1+(−13)2=±232=±13=±33\cos\frac{x}{2} = \pm\sqrt{\frac{1 + \cos x}{2}} = \pm\sqrt{\frac{1 + \left(-\frac{1}{3}\right)}{2}} = \pm\sqrt{\frac{\frac{2}{3}}{2}} = \pm\sqrt{\frac{1}{3}} = \pm\frac{\sqrt{3}}{3}

    Since x/2x/2 is in quadrant II, cos⁡(x/2)<0\cos(x/2) < 0. Therefore, cos⁡x2=−33\cos\frac{x}{2} = -\frac{\sqrt{3}}{3}.

  4. Step 4: Now,

    tan⁡x2=sin⁡x2cos⁡x2=63−33=−63=−2\tan\frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \frac{\frac{\sqrt{6}}{3}}{-\frac{\sqrt{3}}{3}} = -\frac{\sqrt{6}}{\sqrt{3}} = -\sqrt{2}

    Therefore, tan⁡x2=−2\tan\frac{x}{2} = -\sqrt{2}.

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