Chapter 7.3, Problem 6E

To determine

To find: the valuesof $\sin 2x,\cos 2x,\tan 2x$ .

Answer to Problem 6E

  $\begin{array}{l}\sin 2x=-\frac{\sqrt{15}}{8}\\ \cos 2x=\frac{7}{8}\\ \tan 2x=-\frac{\sqrt{15}}{7}\end{array}$ .

Explanation of Solution

Given information:

Given condition,

  $\csc x=4$

And

  $\tan x<0$

Formula used:

Double angle formula of sine

  $\sin 2x=2\sin x\cos x=\frac{2\tan x}{1+{\tan }^{2}x}$

Double angle formula of cosine

  $\cos 2x=1-2{\sin }^{2}x=2{\cos }^{2}x-1={\cos }^{2}x-{\sin }^{2}x=\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}$

Double angle formula of cosine

  $\tan 2x=\frac{2\tan x}{1-{\tan }^{2}x}$

Calculation:

Consider given conditions.

Given condition,

  $\csc x=4$

And

  $\tan x<0$

So, $\sin x$ must be positive, $\cos x$ must be negative and $\tan x$ must be negative.

So,

  $\begin{array}{c}\sin x=\frac{1}{\csc x}\\ =\frac{1}{4}\end{array}$

  $\begin{array}{c}\cos x=-\sqrt{1-{\sin }^{2}x}\\ =-\sqrt{1-{\left(\frac{1}{4}\right)}^2}\\ =-\frac{\sqrt{15}}{4}\\ \tan x=\frac{\sin x}{\cos x}\\ =\frac{\frac{1}{4}}{-\frac{\sqrt{15}}{4}}\\ =-\frac{1}{4}\times \frac{4}{\sqrt{15}}\\ =-\frac{1}{\sqrt{15}}\end{array}$

Calculate the value of $\sin 2x$ by using the double angle formula as follows:

  $\begin{array}{c}\sin 2x=2\sin x\cos x\\ =2\times \frac{1}{4}\times \left(-\frac{\sqrt{15}}{4}\right)\\ =-\frac{2\sqrt{15}}{16}\\ =-\frac{\sqrt{15}}{8}\end{array}$

Hence, $\sin 2x=-\frac{\sqrt{15}}{8}$ .

Calculate the value of $\cos 2x$ by using the double angle formula as follows:

  $\begin{array}{c}\cos 2x=1-2{\sin }^{2}x\\ =1-2\times {\left(\frac{1}{4}\right)}^2\\ =1-2\times \frac{1}{16}\\ =1-\frac{1}{8}\\ =\frac{7}{8}\end{array}$

Hence, $\cos 2x=\frac{7}{8}$ .

Calculate the value of $\tan 2x$ by using the double angle formula as follows:

  $\begin{array}{c}\tan 2x=\frac{2\tan x}{1-{\tan }^{2}x}\\ =\frac{2\times \left(-\frac{1}{\sqrt{15}}\right)}{1-{\left(-\frac{1}{\sqrt{15}}\right)}^2}\\ =\frac{-\frac{2}{\sqrt{15}}}{1-\frac{1}{15}}\\ =\frac{-\frac{2}{\sqrt{15}}}{\frac{14}{15}}\\ =-\frac{\sqrt{15}}{7}\end{array}$

Hence, $\tan 2x=-\frac{\sqrt{15}}{7}$ .