What is Cos(arcsin(5/13) - arccos(7/25))cos(arcsin(513)arccos(725))?

2 Answers

\frac{204}{325}204325

Explanation:

\cos(\sin^{-1}(5/13)-\cos^{-1}(7/25))cos(sin1(513)cos1(725))

=\cos(\sin^{-1}(5/13))\cos(\cos^{-1}(7/25))+\sin(\sin^{-1}(5/13))\sin(\cos^{-1}(7/25))=cos(sin1(513))cos(cos1(725))+sin(sin1(513))sin(cos1(725))

=\cos(\cos^{-1}(12/13))(7/25)+(5/13)\sin(\sin^{-1}(24/25))=cos(cos1(1213))(725)+(513)sin(sin1(2425))

=(12/13)(7/25)+(5/13)(24/25)=(1213)(725)+(513)(2425)

=\frac{84+120}{13\cdot 25}=84+1201325

=\frac{204}{325}=204325

Jul 28, 2018

The answer is =218/325=218325

Explanation:

Let theta_1=arcsin(5/13)θ1=arcsin(513)

Then,

sintheta_1=5/13sinθ1=513

And

costheta_1=sqrt(1-sin^2theta_1)cosθ1=1sin2θ1

=sqrt(1-25/169)=125169

=sqrt(144/169)=144169

=14/13=1413

Let theta_2=arccos(7/25)θ2=arccos(725)

costheta_2=7/25cosθ2=725

sintheta_2=sqrt(1-cos^2theta_2)sinθ2=1cos2θ2

=sqrt(1-49/625)=149625

=sqrt(576/625)=576625

=24/25=2425

Therefore,

cos(arcsin(5/13)-arccos(7/25))cos(arcsin(513)arccos(725))

=cos(theta_1-theta_2)=cos(θ1θ2)

=costheta_1costheta_2+sintheta_1sintheta_2=cosθ1cosθ2+sinθ1sinθ2

=14/13*7/25+5/13*24/25=1413725+5132425

=218/325=218325