If #sintheta=1/3# and #theta# is in quadrant I, how do you evaluate #tan2theta#?

1 Answer
Apr 4, 2017

Answer:

#tan(2theta)=(4sqrt2)/7#

Explanation:

Use the equation #tan(2theta)=(2tantheta)/(1-tan^2theta)#.

The given information tells us that #sintheta=1/3#. Since #theta# is in the first quadrant, we know that all its trig functions will be positive.

The formula for #tan(2theta)# uses #tantheta#. All we have is #sintheta#.

From #sintheta#, we can figure out #costheta# using the Pythagorean identity:

#sin^2theta+cos^2theta=1#

Using #sintheta=1/3#, we see that:

#1/9+cos^2theta=1#

#costheta=sqrt(8/9)=(2sqrt2)/3#

So now we can figure out #tantheta#:

#tantheta=sintheta/costheta=(1/3)/((2sqrt2)/3)=1/3(3/(2sqrt2))=1/(2sqrt2)#

Applying the formula for #tan(2theta)#:

#tan(2theta)=(2tantheta)/(1-tan^2theta)=(2(1/(2sqrt2)))/(1-(1/(2sqrt2))^2)#

Simplifying:

#tan(2theta)=(1/sqrt2)/(1-1/8)=1/sqrt2(8/7)=(4sqrt2)/7#