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(B)/(2)-1`

Substitute cos B by 7/8

`(7)/(8)=2\cos ^2(B)/(2)-1`

Add 1 to both sides

`\begin{gathered} (7)/(8)+1=2\cos ^2(B)/(2)-1+1 \\ (15)/(8)=2\cos ^2(B)/(2) \end{gathered}`

Divide both sides by 2

`\begin{gathered} ((15)/(8))/(2)=(2\cos ^2(B)/(2))/(2) \\ (15)/(16)=\cos ^2(B)/(2) \end{gathered}`

Take a square root for both sides

`\begin{gathered} \sqrt[]{(15)/(16)}=\cos (B)/(2) \\ \cos (B)/(2)=\frac{\sqrt[]{15}}{4} \end{gathered}`

let us find sin B

Since

`\sin ^2(B)/(2)+\cos ^2(B)/(2)=1`

Then

`\sin ^2(B)/(2)+(15)/(16)=1`

Subtract 15/16 from both sides

`\begin{gathered} \sin ^2(B)/(2)+(15)/(16)-(15)/(16)=1-(15)/(16) \\ \sin ^2(B)/(2)=(1)/(16) \end{gathered}`

Take a square root for both sides

`\begin{gathered} \sin (B)/(2)=\sqrt[]{(1)/(16)} \\ \sin (B)/(2)=(1)/(4) \end{gathered}`

Since

`\tan (B)/(2)=(\sin (B)/(2))/(\cos (B)/(2))`

Then

`\begin{gathered} \tan (B)/(2)=\frac{(1)/(4)}{\frac{\sqrt[]{15}}{4}} \\ \tan (B)/(2)=\frac{1}{\sqrt[]{15}} \end{gathered}`

The value of tan(1/2 B) is

`\frac{1}{\sqrt[]{15}}*\frac{\sqrt[]{15}}{\sqrt[]{15}}=\frac{\sqrt[]{15}}{15}`