Question

If cos B = 7/8, then what is the positive value of tan 1/2 B, in simplest radical formwith a rational denominator?

Answer 1

Let us use the rule of the double of the angle

Since

$\cos B=2\cos ^2\frac{B}{2}-1$

Substitute cos B by 7/8

$\frac{7}{8}=2\cos ^2\frac{B}{2}-1$

Add 1 to both sides

$\begin{gathered} \frac{7}{8}+1=2\cos ^2\frac{B}{2}-1+1 \\ \frac{15}{8}=2\cos ^2\frac{B}{2} \end{gathered}$

Divide both sides by 2

$\begin{gathered} \frac{\frac{15}{8}}{2}=\frac{2\cos ^2\frac{B}{2}}{2} \\ \frac{15}{16}=\cos ^2\frac{B}{2} \end{gathered}$

Take a square root for both sides

$\begin{gathered} \sqrt[]{\frac{15}{16}}=\cos \frac{B}{2} \\ \cos \frac{B}{2}=\frac{\sqrt[]{15}}{4} \end{gathered}$

let us find sin B

Since

$\sin ^2\frac{B}{2}+\cos ^2\frac{B}{2}=1$

Then

$\sin ^2\frac{B}{2}+\frac{15}{16}=1$

Subtract 15/16 from both sides

$\begin{gathered} \sin ^2\frac{B}{2}+\frac{15}{16}-\frac{15}{16}=1-\frac{15}{16} \\ \sin ^2\frac{B}{2}=\frac{1}{16} \end{gathered}$

Take a square root for both sides

$\begin{gathered} \sin \frac{B}{2}=\sqrt[]{\frac{1}{16}} \\ \sin \frac{B}{2}=\frac{1}{4} \end{gathered}$

Since

$\tan \frac{B}{2}=\frac{\sin \frac{B}{2}}{\cos \frac{B}{2}}$

Then

$\begin{gathered} \tan \frac{B}{2}=\frac{\frac{1}{4}}{\frac{\sqrt[]{15}}{4}} \\ \tan \frac{B}{2}=\frac{1}{\sqrt[]{15}} \end{gathered}$

The value of tan(1/2 B) is

$\frac{1}{\sqrt[]{15}}\times\frac{\sqrt[]{15}}{\sqrt[]{15}}=\frac{\sqrt[]{15}}{15}$