Question

If sin a = 1/√10, sin b = 1/√5, prove that a+b=π/4, a and b being positive acute angles.

Answer 1

sin(a+b)= sinacosb+ cosasinb
Sina =1/root10
cosa= 3/root10(from pythogreas rule)
sinb= 1/root5
cosb= 2/root5
sin(a+b)= sinpi/4
sin(a+b)= 1/root2
now sin(a+b)= 1/root10 * 2/root5 + 3/root10 * 1/root5
= 2/5root2 + 3/5 root2
= 5/ 5root2
= 1/root2
sin(a+b) = sinpi/4 = 1/root2
therefore a+b = pi/4 is proved

Answer 2

Answer:

sin(a+b)= sinacosb+ cosasinb

Sina =1/root10

cosa= 3/root10(from pythogreas rule)

sinb= 1/root5

cosb= 2/root5

sin(a+b)= sinpi/4

sin(a+b)= 1/root2

now sin(a+b)= 1/root10 * 2/root5 + 3/root10 * 1/root5

= 2/5root2 + 3/5 root2

= 5/ 5root2

= 1/root2

sin(a+b) = sinpi/4 = 1/root2

therefore a+b = pi/4 is proved