How do you find the value of #sin(tan^-1 (7/24))#?

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Answer 1 · 2016-08-28T07:23:02

#7/25#

Let #theta# be the angle whose tan is #7/24#
Draw the right angle triangle with the two shorter sides are 7and 24

Pythagoras gives the hypotenuse as 25

Sin #theta# =#7/25#

Answer 2 · 2016-08-28T07:29:39

#sin(tan^-1(7/24))=7/25#.

Suppose that, #tan^-1(7/24)=theta#. Then, by Defn. of #tan^-1# function, #tan theta=7/24, theta in (-pi/2,pi/2)#

Since, #tan theta >0, theta !in (-pi/2,0), &, so, theta in (0,pi/2)#.

We know that, #csc^2 theta=1+cot^2 theta=1+1/tan^2 theta#.

#:. csc^2 theta=1+(24/7)^2=(7^2+24^2)/7^2=25^2/7^2#

#:. csc theta=+-24/7", but, as," theta in (0,pi/2), csc theta =+25/7, or, sin theta=7/25#.

Therefore, #"the reqd. value"=sin(tan^-1(7/24))#

#=sin theta=7/25#.

Alternatively , we can use the conversion formula :

# tan^-1x=sin^-1(x/sqrt(1+x^2)), where, x>0#, to give,

#tan^-1(7/24)=sin^-1(7/25)#, and hence,

#sin(tan^-1(7/24))=sin(sin^-1(7/25))=7/25#, as before!

Enjoy Maths.!