Chapter 5.5, Problem 38E

To determine

Tofind:The exact value of trigonometric ratios $\sin {67}^{0}30^{\prime }$ , $\cos {67}^{0}30^{\prime }$ , $\tan {67}^{0}30^{\prime }$ using half - angle formulae.

Answer to Problem 38E

Therequired values of trigonometric ratios are $\sin {67}^{0}30^{\prime }=0.9239$ , $\cos {67}^{0}30^{\prime }=0.3827$ , and $\tan {67}^{0}30^{\prime }=2.4142$ .

Explanation of Solution

Given information:

The angle $A={67}^{0}30^{\prime }$ .

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 $A={135}^0$ in result (1) we have

  $\cos 135=\sqrt{\frac{1+\cos {135}^0}{2}}\Rightarrow \cos {67}^{0}30^{\prime }=\sqrt{\frac{1+\cos {135}^0}{2}}$

Now, $\cos {135}^0=\cos \left({90}^0+{45}^0\right)=-\sin \left({45}^0\right)=-\frac{1}{\sqrt{2}}=-0.7071$

  $\cos {67}^{0}30^{\prime }=\sqrt{\frac{1+\cos {135}^0}{2}}=\pm \sqrt{\frac{1-0.7071}{2}}=\pm 0.3827$

Since angle ${67}^{0}30^{\prime }$ falls in first quadrant where $\cos {67}^{0}30^{\prime }$ is positive, therefore, we accept $\cos {67}^{0}30^{\prime }=0.3827$ .

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 $A={135}^0$ in result (2) we have

  $\begin{array}{l}\sin \frac{{135}^0}{2}=\sqrt{\frac{1-\cos {135}^0}{2}}\\ \sin {67}^{0}30^{\prime }=\sqrt{\frac{1-\cos {135}^0}{2}}\\ \sin {67}^{0}30^{\prime }=\sqrt{\frac{1-\left(-0.7071\right)}{2}}=\sqrt{\frac{1+0.7071}{2}}=\pm 0.9239\end{array}$

Since angle ${67}^{0}30^{\prime }$ falls in first quadrant where $\sin {67}^{0}30^{\prime }$ is positive, therefore, we accept $\sin {67}^{0}30^{\prime }=0.9239$ .

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

Now, substituting $A={135}^0$ in result (3) we have

  $\tan {135}^0=\frac{2\tan \left({135}^0/2\right)}{1-{\tan }^2\left({135}^0/2\right)}\Rightarrow \tan {135}^0=\frac{2\tan {67}^{0}30^{\prime }}{1-{\tan }^2{67}^{0}30^{\prime }}$

Now, $\tan {135}^0=\tan \left({90}^0+{45}^0\right)=-\cot \left({45}^0\right)=-1$

  $\Rightarrow -1=\frac{2\tan {67}^{0}30^{\prime }}{1-{\tan }^2{67}^{0}30^{\prime }}$ . Suppose $x=\tan {67}^{0}30^{\prime }$ , then $-1=\frac{2x}{1-x^2}$

  $\Rightarrow -1+x^2=2x\Rightarrow x^2-2x-1=0$..................(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

  $x=\frac{2\pm \sqrt{{\left(2\right)}^2-4\left(1\right)\left(-1\right)}}{2\cdot 1}=\frac{2\pm \sqrt{4+4}}{2}=2.4142,-0.4142$ ,

Since angle ${67}^{0}30^{\prime }$ falls in first quadrant where $\tan {67}^{0}30^{\prime }$ is positive, therefore, we accept $x=2.4142$ .Thus, $\tan {67}^{0}30^{\prime }=2.4142$ .

Conclusion:

Thevalues of trigonometric ratios are $\sin {67}^{0}30^{\prime }=0.9239$ , $\cos {67}^{0}30^{\prime }=0.3827$ , and $\tan {67}^{0}30^{\prime }=2.4142$ .