Chapter 5.5, Problem 24E

To determine

To find the values of $\sin 2u,\cos 2u\&\tan 2u.$

Answer to Problem 24E

  $\sin 2u=\frac{\sqrt{3}}{2},\cos 2u=-\frac{1}{2}\&\tan 2u=-\sqrt{3}$

Explanation of Solution

Given:

  $\sec u=-2,\pi <u<\frac{3\pi }{2}$

Concept used:

Reciprocal identity:

  $\cos u=\frac{1}{\sec u}$

Pythagorean identity:

  ${\sin }^{2}u+{\cos }^{2}u=1$

Quotient identity:

  $\tan u=\frac{\sin u}{\cos u}$

Double angle formulas:

  $\begin{array}{l}\sin 2u=2\sin u\cos u\\ \cos 2u=1-2{\sin }^{2}u\end{array}$

Calculation:

Given, $\sec u=-2,\pi <u<\frac{3\pi }{2}$

B using reciprocal identity, we have

  $\cos u=\frac{1}{\sec u}\Rightarrow \cos u=\frac{1}{-2}\Rightarrow \cos u=-\frac{1}{2}$

Now by using Pythagorean identity, we have

  $\begin{array}{l}{\sin }^{2}u=1-{\cos }^{2}u\\ \Rightarrow {\sin }^{2}u=1-{\left(-\frac{1}{2}\right)}^2\\ \Rightarrow {\sin }^{2}u=1-\frac{1}{4}\\ \Rightarrow {\sin }^{2}u=\frac{3}{4}\\ \Rightarrow \sin u=\pm \frac{\sqrt{3}}{2}\end{array}$

Since $\pi <u<\frac{3\pi }{2}$ , it follows that $u$ lies in quadrant III. Moreover, $\sin u$ is negative in quadrant III. Therefore,

  $\sin u=-\frac{\sqrt{3}}{2}$

Now by using double angle formulas, we have

  $\begin{array}{l}\sin 2u=2\sin u\cos u\\ \Rightarrow \sin 2u=2\left(-\frac{\sqrt{3}}{2}\right)\left(-\frac{1}{2}\right)\\ \Rightarrow \sin 2u=\frac{\sqrt{3}}{2}\\ \&\\ \cos 2u=1-2{\sin }^{2}u\\ \Rightarrow \cos 2u=1-2{\left(-\frac{\sqrt{3}}{2}\right)}^2\\ \Rightarrow \cos 2u=1-2\left(\frac{3}{4}\right)\\ \Rightarrow \cos 2u=\frac{4-6}{4}\\ \Rightarrow \cos 2u=-\frac{1}{2}\end{array}$

Now by using quotient identity, we have

  $\begin{array}{l}\tan 2u=\frac{\sin 2u}{\cos 2u}\\ \Rightarrow \tan 2u=\frac{\raisebox{1ex}{$\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ \Rightarrow \tan 2u=-\sqrt{3}\end{array}$