Question

Given that sinX=(3/8) and X is obtuse, what is the value of tan2X? In the textbook, the correct answer is (-3√55)/23. How did they get this answer? I appreciate any help.

Answer 1

Answer:

Step-by-step explanation:

$\sin(X) = 3/8 \\ \\\sin^2(X) + \cos^2(X) = 1 \\ \implies (3/8)^2 + \cos^2(X) = 1 \\ \implies \cos^2(X) = 1-(3/8)^2 = 1-9/64 = 55/64. \\\\\implies \cos(X) = \pm\sqrt{55/64} = \pm\sqrt{55}/8.\\\\\text{Now, since X is obtuse, it is in the second quadrant and so we know that the cosine} \\\text{of X must be negative. Hence we pick:}\\\\\cos(X) = -\sqrt{55}/8.\\\implies \tan(X) = \sin(X)/\cos(X) = \frac{3/8}{-\sqrt{55}/8} = -3/\sqrt{55}.$

$\therefore \tan(2X) = \frac{2\tan(X)}{1-\tan^2(X)} = \frac{-6/\sqrt{55}}{1-(-3/\sqrt{55})^2} = \frac{-6/\sqrt{55}}{1-9/55} = \frac{-6/\sqrt{55}}{46/55} \\\\\implies \tan(2X) = \frac{-6}{\sqrt{55}} \times \frac{55}{46} = \frac{-3}{23} \times \frac{55}{\sqrt{55}} = -\frac{3}{23} \times \sqrt{55} = \underline{\underline{-\frac{3\sqrt{55}}{23}}}$