28.7k views
0 votes
If cos theta = 4/ square root 23 and angle theta is in quadrant 1, what is the exact value of tan2 theta in simplest radical form?

User JD Allen
by
4.5k points

1 Answer

2 votes

Solution

For this case we know that an angle theta is in the I quadrant and additionally:


\cos \theta=\frac{4}{\sqrt[]{23}}

We can find the sin of the angle usingt this identity:


\sin ^2\theta=1-\cos ^2\theta
\sin \theta=\sqrt[]{1-(\frac{4}{\sqrt[]{23}})^2}=\sqrt[]{(7)/(23)}

We can find tangent:


\tan \theta=(\sin\theta)/(\cos\theta)=\frac{\sqrt[]{(7)/(23)}}{\frac{4}{\sqrt[]{23}}}=\frac{\sqrt[]{7}}{4}

Finally we can find tan 2 theta and we have:


\tan 2\theta=(2\tan\theta)/(1-\tan^2\theta)=\frac{2\cdot\frac{\sqrt[]{7}}{4}}{1-(7)/(16)}=\frac{\frac{\sqrt[]{7}}{2}}{(9)/(16)}=\frac{8\sqrt[]{7}}{9}

User Walter Lockhart
by
3.4k points