Evaluate cos0 if sin0= 5/3
Mathematics · College · Tue Nov 03 2020
Answered on
Solution [note question have a mistake in sin(thita) is sin^-1(thita) ]
Given, sin^-1(thita)= 5/3
We know that sin(thita) = L/K
And cos(thita)=A/K by Pythagoras theorem
So K^2=L^2+A^2…..(1)
By comparison, sin^-1(thita)= 5/3 here sin is in an inverse form that's why values will be
K=5, L=3, by equation (1)
(5)^2=(3)^2+A^2
25=9+A^2
A^2= 25-9
A^2= 16
A=4, so the cos(thita) = 4/3.