Question 50 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: sin⁡x=14\sin x = \frac{1}{4}, xx in quadrant II

Correct Answer
sin⁡(x2)=5+34,cos⁡(x2)=5−34,tan⁡(x2)=4+15\sin\left(\frac{x}{2}\right) = \frac{\sqrt{5} + \sqrt{3}}{4}, \quad \cos\left(\frac{x}{2}\right) = \frac{\sqrt{5} - \sqrt{3}}{4}, \quad \tan\left(\frac{x}{2}\right) = 4 + \sqrt{15}
Exercise: MISCELLANEOUS EXERCISE ON CHAPTER 3 | Q: 10 | (Page 29)
For More Understanding

Explanation

This question tests the application of half-angle formulas and quadrant analysis. The student must first determine cos⁡x\cos x using the Pythagorean identity, then identify the quadrant of x/2x/2 to determine the signs of the half-angle functions.

The key insight is that when xx lies in quadrant II (π/2<x<π\pi/2 < x < \pi), then x/2x/2 lies in quadrant I (π/4<x/2<π/2\pi/4 < x/2 < \pi/2), making all three functions positive.

Solution Steps

  1. Step 1: Given sin⁡x=14\sin x = \frac{1}{4}, xx in quadrant II. Find cos⁡x\cos x using:

    cos⁡2x=1−sin⁡2x=1−116=1516\cos^2 x = 1 - \sin^2 x = 1 - \frac{1}{16} = \frac{15}{16}

    Since xx is in quadrant II, cos⁡x\cos x is negative. Therefore, cos⁡x=−154\cos x = -\frac{\sqrt{15}}{4}.

  2. Step 2: Determine the quadrant of x/2x/2. Since xx is in quadrant II, π/2<x<π\pi/2 < x < \pi. Therefore, π/4<x/2<π/2\pi/4 < x/2 < \pi/2. So x/2x/2 lies in quadrant I, where sin⁡(x/2)>0\sin(x/2) > 0, cos⁡(x/2)>0\cos(x/2) > 0, and tan⁡(x/2)>0\tan(x/2) > 0.

  3. Step 3: Find sin⁡(x/2)\sin(x/2) using the formula:

    sin⁡2(x2)=1−cos⁡x2=1+1542=4+158\sin^2\left(\frac{x}{2}\right) = \frac{1 - \cos x}{2} = \frac{1 + \frac{\sqrt{15}}{4}}{2} = \frac{4 + \sqrt{15}}{8}

    Since x/2x/2 is in quadrant I,

    sin⁡(x2)=4+158=5+34\sin\left(\frac{x}{2}\right) = \sqrt{\frac{4 + \sqrt{15}}{8}} = \mathbf{\frac{\sqrt{5} + \sqrt{3}}{4}}

  4. Step 4: Find cos⁡(x/2)\cos(x/2) using the formula:

    cos⁡2(x2)=1+cos⁡x2=1−1542=4−158\cos^2\left(\frac{x}{2}\right) = \frac{1 + \cos x}{2} = \frac{1 - \frac{\sqrt{15}}{4}}{2} = \frac{4 - \sqrt{15}}{8}

    Since x/2x/2 is in quadrant I,

    cos⁡(x2)=4−158=5−34\cos\left(\frac{x}{2}\right) = \sqrt{\frac{4 - \sqrt{15}}{8}} = \mathbf{\frac{\sqrt{5} - \sqrt{3}}{4}}

  5. Step 5: Find tan⁡(x/2)\tan(x/2):

    tan⁡(x2)=sin⁡(x/2)cos⁡(x/2)=5+35−3\tan\left(\frac{x}{2}\right) = \frac{\sin(x/2)}{\cos(x/2)} = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}

    Rationalizing:

    tan⁡(x2)=(5+3)25−3=8+2152=4+15\tan\left(\frac{x}{2}\right) = \frac{(\sqrt{5} + \sqrt{3})^2}{5 - 3} = \frac{8 + 2\sqrt{15}}{2} = \mathbf{4 + \sqrt{15}}
Previous