How do you calculate #sin(tan^-1(3/4))#?

Question

James Adkins · Accepted Answer

#3/5.# If we write #tan^-1(3/4)=theta#, then, by defn. of #tan^-1# fun., we get, #tantheta=3/4, theta in (-pi/2,pi/2).# Since, #tantheta >0, theta !in (-pi/2,0)#, but, #theta in (0,pi/2)# Now, desired value #=sin (tan^-1(3/4)) = sintheta,# where, #tantheta=3/4.# Now, #tantheta =3/4 rArr cottheta =4/3 rArr csc^2theta=1+cot^2theta=1+(4/3)^2=25/9 rArr csctheta=+-5/3 rArr sintheta=+-3/5.# As, #theta in (o,pi/2), sintheta >0, so, sintheta=+3/5# The reqd. value #=3/5.#

Brooklyn Anderson · Answer

undefined To calculate $ \sin(\tan^{-1}(3/4)) $, you can use the properties of trigonometric functions and the definition of the inverse tangent function.

Let $ \theta = \tan^{-1}(3/4) $. This means that $ \tan(\theta) = \frac{3}{4} $.

Using the definition of tangent, we can construct a right triangle with opposite side 3 and adjacent side 4. Then, using the Pythagorean theorem, we find that the hypotenuse is $ \sqrt{3^2 + 4^2} = 5 $.

So, in the right triangle, we have:

$$ \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{5} $$

Therefore, $ \sin(\tan^{-1}(3/4)) = \frac{3}{5} $.