Chapter 5.5, Problem 37E

To determine

Tofind:The exact value of trigonometric ratios $\sin {112}^{0}30^{\prime }$ , $\cos {112}^{0}30^{\prime }$ , $\tan {112}^{0}30^{\prime }$ using half - angle formulae.

Answer to Problem 37E

Therequired values of trigonometric ratios are $\sin {112}^{0}30^{\prime }=0.9239$ , $\cos {112}^{0}30^{\prime }=-0.3827$ , and $\tan {112}^{0}30^{\prime }=-2.4142$ .

Explanation of Solution

Given information:

The angle $A={112}^{0}30^{\prime }$ .

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 $A={225}^0$ in result (1) we have

  $\cos \frac{{225}^0}{2}=\sqrt{\frac{1+\cos {225}^0}{2}}\Rightarrow \cos {112}^{0}30^{\prime }=\sqrt{\frac{1+\cos {225}^0}{2}}$

Now, $\cos {225}^0=\cos \left({180}^0+{45}^0\right)=-\cos \left({45}^0\right)=-\frac{1}{\sqrt{2}}=-0.7071$

  $\cos {112}^{0}30^{\prime }=\sqrt{\frac{1+\cos {225}^0}{2}}=\pm \sqrt{\frac{1-0.7071}{2}}=\pm 0.3827$

Since angle ${112}^{0}30^{\prime }$ falls in second quadrant where $\cos {112}^{0}30^{\prime }$ is negative, therefore, we accept $\cos {112}^{0}30^{\prime }=-0.3827$ .

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 $A={225}^0$ in result (2) we have

  $\sin \frac{{225}^0}{2}=\sqrt{\frac{1-\cos {225}^0}{2}}\Rightarrow \sin {112}^{0}30^{\prime }=\sqrt{\frac{1-\cos {225}^0}{2}}$

Now, $\sin {225}^0=\sin \left({180}^0+{45}^0\right)=-\sin \left({45}^0\right)=-\frac{1}{\sqrt{2}}=-0.7071$

  $\sin {112}^{0}30^{\prime }=\sqrt{\frac{1-\left(-0.7071\right)}{2}}=\sqrt{\frac{1+0.7071}{2}}=\pm 0.9239$

Since angle ${112}^{0}30^{\prime }$ falls in second quadrant where $\sin {112}^{0}30^{\prime }$ is positive, therefore, we accept $\sin {112}^{0}30^{\prime }=0.9239$ .

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

Now, substituting $A={225}^0$ in result (3) we have

  $\tan {225}^0=\frac{2\tan \left({225}^0/2\right)}{1-{\tan }^2\left({225}^0/2\right)}\Rightarrow \tan {225}^0=\frac{2\tan {1112}^{0}30^{\prime }}{1-{\tan }^2{112}^{0}30^{\prime }}$

Now, $\tan {225}^0=\tan \left({180}^0+{45}^0\right)=\tan \left({45}^0\right)=1$

  $\Rightarrow 1=\frac{2\tan {112}^{0}30^{\prime }}{1-{\tan }^2{112}^{0}30^{\prime }}$ . Suppose $x=\tan {112}^{0}30^{\prime }$ , then $1=\frac{2x}{1-x^2}$

  $\Rightarrow 1-x^2=\frac{2x}{1}\Rightarrow x^2+2x-1=0$..................(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 ${112}^{0}30^{\prime }$ falls in second quadrant where $\tan {112}^{0}30^{\prime }$ is negative, therefore, we accept $x=-2.4142$ .Thus, $\tan {112}^{0}30^{\prime }=-2.4142$ .

Conclusion:

Thevalues of trigonometric ratios are $\sin {112}^{0}30^{\prime }=0.9239$ , $\cos {112}^{0}30^{\prime }=-0.3827$ , and $\tan {112}^{0}30^{\prime }=-2.4142$ .