Chapter 5.5, Problem 40E

To determine

Tofind:The exact value of trigonometric ratios $\sin \frac{7\pi }{12}$ , $\cos \frac{7\pi }{12}$ , $\tan \frac{7\pi }{12}$ using half - angle formulae.

Answer to Problem 40E

Therequired values of trigonometric ratios are $\sin \frac{7\pi }{12}=0.9659$ , $\cos \frac{7\pi }{12}=-0.2588$ , and $\tan \frac{7\pi }{12}=-3.7321$ .

Explanation of Solution

Given information:

The angle $A=\frac{7\pi }{12}=\frac{7\times {180}^0}{12}={105}^0$ .

Formulaused:

  $\cos A=2{\cos }^2\frac{A}{2}-1$ , $\cos A=1-2{\sin }^2\frac{A}{2}$ , and $\tan A=\frac{2\tan \left(A/2\right)}{1-{\tan }^2\left(A/2\right)}$ .

Calculations:

We have $\cos A=2{\cos }^2\frac{A}{2}-1\Rightarrow \cos \frac{A}{2}=\sqrt{\frac{1+\cos A}{2}}$ ..................(1)

Now, substituting $A=\frac{7\pi }{6}={210}^0$ in result (1) we have

  $\cos \frac{7\pi }{12}=\sqrt{\frac{1+\cos 7\pi /6}{2}}\Rightarrow \cos \frac{7\pi }{12}=\sqrt{\frac{1+\cos 210}{2}}$

Now,

  $\cos \frac{7\pi }{6}=\cos {210}^0=\cos \left({180}^0+{30}^0\right)=-\cos {30}^0=-\frac{\sqrt{3}}{2}=-0.8660$

  $\Rightarrow \cos \frac{7\pi }{12}=\sqrt{\frac{1-0.8660}{2}}=\pm 0.2588$

Since angle $\frac{7\pi }{8}={105}^0$ falls in second quadrant where $\cos {105}^0$ is negative, therefore, we accept $\cos {105}^0=-0.2588$ .

Thus, $\cos \frac{7\pi }{12}=-0.2588$ .

We have $\cos A=1-2{\sin }^2\frac{A}{2}\Rightarrow \sin \frac{A}{2}=\sqrt{\frac{1-\cos A}{2}}$ ..................(2)

Now, substituting $A=\frac{\pi }{4}$ in result (2) we have

  $\begin{array}{l}\sin \frac{7\pi }{12}=\sqrt{\frac{1-\cos 7\pi /6}{2}}\\ \Rightarrow \sin \frac{7\pi }{12}=\sqrt{\frac{1-\cos {210}^0}{2}}=\sqrt{\frac{1-(-0.8660)}{2}}\\ \Rightarrow \sin \frac{7\pi }{12}=\sqrt{\frac{1+0.8660}{2}}=0.9659\end{array}$

Thus, $\sin \frac{7\pi }{12}=0.9659$ .

We have $\tan A=\frac{2\tan \left(A/2\right)}{1-{\tan }^2\left(A/2\right)}$..................(3)

Now, substituting $A=\frac{7\pi }{6}$ in result (3) we have

  $\begin{array}{l}\tan \frac{7\pi }{6}=\frac{2\tan \left(7\pi /12\right)}{1-{\tan }^2\left(7\pi /12\right)}\\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{2\tan \left(7\pi /12\right)}{1-{\tan }^2\left(7\pi /12\right)}\left(\begin{array}{l}\because \tan \frac{7\pi }{16}=\tan {210}^0\\ =\tan \left({180}^0+{30}^0\right)\\ =\tan \left({30}^0\right)=\frac{1}{\sqrt{3}}\end{array}\right)\end{array}$ .

Suppose $x=\tan \frac{7\pi }{12}$ , then $\frac{1}{\sqrt{3}}=\frac{2x}{1-x^2}$

  $\Rightarrow 1-x^2=2\sqrt{3}x\Rightarrow x^2+2\sqrt{3}x-1=0$..................(4)

Now, we solve quadratic equation (4) with formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ for quadratic polynomial $a{x}^2+bx+c=0$ as

  $x=\frac{-2\sqrt{3}\pm \sqrt{{\left(2\sqrt{3}\right)}^2-4\left(1\right)\left(-1\right)}}{2\cdot 1}=\frac{-2\sqrt{3}\pm \sqrt{12+4}}{2}$ ,

  $\Rightarrow x=\frac{-2\sqrt{3}\pm 4}{2}=-3.7321,0.2680$

Since angle $\frac{7\pi }{12}$ falls in second quadrant where $\tan \frac{\text{7π}}{\text{12}}$ isnegative, therefore, we accept $x=-3.7321$ .Thus, $\tan \frac{7\pi }{12}=-3.7321$ .

Conclusion:

Thevalues of trigonometric ratios are $\sin \frac{7\pi }{12}=0.9659$ , $\cos \frac{7\pi }{12}=-0.2588$ , and $\tan \frac{7\pi }{12}=-3.7321$ .