Chapter 5.5, Problem 35E

To determine

Tofind:The exact value of trigonometric ratios $\sin {75}^0$ , $\cos {75}^0$ , $\tan {75}^0$ using half - angle formulae.

Answer to Problem 35E

Therequired values of trigonometric ratios are $\sin {75}^0=0.9659$ , $\cos {75}^0=0.2588$ , and $\tan {75}^0=3.7320$ .

Explanation of Solution

Given information:

The angle $A={75}^0$ .

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 $A={150}^0$ in result (1) we have

  $\begin{array}{l}\cos \frac{{150}^0}{2}=\sqrt{\frac{1+\cos {150}^0}{2}}\\ \cos {75}^0=\sqrt{\frac{1+\cos {150}^0}{2}}\\ \cos {75}^0=\sqrt{\frac{1-0.8660}{2}}=0.2588\end{array}$

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 $A={150}^0$ in result (2) we have

  $\begin{array}{l}\sin \frac{{150}^0}{2}=\sqrt{\frac{1-\cos {150}^0}{2}}\\ \sin {75}^0=\sqrt{\frac{1-\cos {150}^0}{2}}\\ \sin {75}^0=\sqrt{\frac{1-\left(-0.8660\right)}{2}}=\sqrt{\frac{1+0.8660}{2}}=0.9659\end{array}$

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

Now, substituting $A={150}^0$ in result (3) we have

  $\tan {150}^0=\frac{2\tan \left({150}^0/2\right)}{1-{\tan }^2\left({150}^0/2\right)}\Rightarrow \tan {150}^0=\frac{2\tan {75}^0}{1-{\tan }^2{75}^0}$

  $\Rightarrow -0.57735=\frac{2\tan {75}^0}{1-{\tan }^2{75}^0}$ . Suppose $x=\tan {75}^0$ , then $-0.57735=\frac{2x}{1-x^2}$

  $\Rightarrow 1-x^2=\frac{2x}{-0.57735}=-3.4641x\Rightarrow x^2-3.4641x-1=0$ ..................(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

  $x=\frac{-\left(-3.4641\right)\pm \sqrt{{\left(-3.4641\right)}^2-4\left(1\right)\left(-1\right)}}{2\cdot 1}=\frac{3.4641\pm \sqrt{12+4}}{2}$

  $x=\frac{3.4641\pm 4}{2}=3.7320,-0.26795$ , we accept positive value so $x=3.7320$ .

Thus, $\tan {75}^0=3.7320$ .

From direct formula $\tan A/2=\frac{\sin A/2}{\cos A/2}\Rightarrow \tan {75}^0=\frac{\sin {75}^0}{\cos {75}^0}=\frac{0.965926}{0.258819}=3.7320$

Conclusion:

Thevalues of trigonometric ratios are $\sin {75}^0=0.9659$ , $\cos {75}^0=0.2588$ , and $\tan {75}^0=3.7320$ .