# 9sinA+40CosA= 41, Prove 41CosA = 40

sciencesolve | Certified Educator

You need to prove that `41CosA = 40` , meaning that substituting 40/41 for cos A in the relation `9sinA+40CosA= 41 ` holds such that:

`9sin A + 40*40/41 = 41`

`369 sin A + 40^2 = 41^2 => 369 sin A = 41^2 - 40^2`

`369 sin A = (41-40)(41+40)`

`369 sin A = 81 => sin A = 81/369 => sin A = 9/41`

You need to use the basic formula of trigonometry for `cos A = 40/41`  to check if `sin A = 9/41`  such that:

`sin^2 A = 1 - cos^2 A => sin^2 A = 1 - 40^2/41^2`

`sin^2 A = (41^2 - 40^2)/41^2`

`sin^2 A = 81/41^2 => sin A = 9/41`

Notice that if `cosA= 40/41` , hence `sin A = 9/41`  and for these two values of `sin A ` and `cos A` , the relation `9sinA+40CosA= 41`  is verified.

jeew-m | Certified Educator

Consider a triangle with leg lengths 40,9 and 41.

Then `40^2+9^2 = 41^2` is satisfied Pythagoras theorem.

So the triangle have a 90 deg angle.

So we can say either cosA = 40/41 or cosA = 9/40

But we know;

9sinA+40CosA= 41-----(1)

if cosA = 9/41 then sinA=40/41

9sinA+40CosA = (40*9+40*9)/41 = 17.56 ;not satisfied (1)

if cosA = 40/41 then sinA = 9/41

9sinA+40CosA = (9*9/41+40*40/41)/41 = 41 ;satisfied (1)

Therefore cosA = 41/40

cosA = 40/41

41cosA = 40