What is Cos(arcsin(5/13) - arccos(7/25))?

2 Answers

\frac{204}{325}

Explanation:

\cos(\sin^{-1}(5/13)-\cos^{-1}(7/25))

=\cos(\sin^{-1}(5/13))\cos(\cos^{-1}(7/25))+\sin(\sin^{-1}(5/13))\sin(\cos^{-1}(7/25))

=\cos(\cos^{-1}(12/13))(7/25)+(5/13)\sin(\sin^{-1}(24/25))

=(12/13)(7/25)+(5/13)(24/25)

=\frac{84+120}{13\cdot 25}

=\frac{204}{325}

Jul 28, 2018

The answer is =218/325

Explanation:

Let theta_1=arcsin(5/13)

Then,

sintheta_1=5/13

And

costheta_1=sqrt(1-sin^2theta_1)

=sqrt(1-25/169)

=sqrt(144/169)

=14/13

Let theta_2=arccos(7/25)

costheta_2=7/25

sintheta_2=sqrt(1-cos^2theta_2)

=sqrt(1-49/625)

=sqrt(576/625)

=24/25

Therefore,

cos(arcsin(5/13)-arccos(7/25))

=cos(theta_1-theta_2)

=costheta_1costheta_2+sintheta_1sintheta_2

=14/13*7/25+5/13*24/25

=218/325