If tan theta = 7/2 find sin theta ?

1 Answer
May 16, 2018

#sin arctan(a/b) = pm a/sqrt{a^2+b^2} = pm 7/sqrt{7^2+2^2 }= pm 7/sqrt{53}#

Explanation:

We won't known the sign of #sin theta#. There are two angles #180^circ # apart whose tangent (slope) is #7/2# and their sines will be negatives of each other.

This problem can be written with the multvalued inverse tangent:

#sin arctan( 7/2)#

If #7/2# is the tangent, that's a right triangle with opposite #7# and adjacent #2# so hypotenuse #sqrt{7^2+2^2}=sqrt{53}# so a sine of opposite over hypotenuse or

#\pm 7/sqrt{53} #