Question

If sinA=1/root5 & sinB= 1/root10, find the values of cosA and cosB. Hence using the formula cos (A+B)=cosA .cosB- sinA.sinB, show that A+B=45

Answer 1

We know sin^2 x + cos^2 x = 1,
=> sin^2 x = 1 - cos^2 x
=> sin x = root of ( 1- cos^2 x)
given that sin A = 1/root5
=> sin^2 A = (1/root5)^2
=> sin^2 A = 1/5
=> 1 - cos^2 A = 1/5
=> cos^2 A = 1-(1/5)
=> cos^2 A = 4/5
=> cos A = root of (4/5)
=> cos A = 2/root 5
Following the similar way for sin B= 1/ root 10, we get cos B = 3/root 10
We know that cos(A+B) = cosA.cosB - sinA.sinB
=> cos(A+B) = [ 2/root 5 * 3/root 10 ] - [ 1/root 5 * 1/root 10 ]
=> cos(A+B) = 6/root 50 - 1/root 50
=> cos(A+B) = 5/root 50
but, root 50 = root 25 * root 2
=> root 50 = 5 * root 2   [ since square root of 25 is 5 ]
therefore,
cos(A+B) = 5/ ( 5*root 2 ) = 1/root 2
cos x will be 1/root2 only if x is 45
=> A+B = 45
hence proved