Find cos (x+y) when both x and y are acute angles and sin x=3/5 and sin y=5/13?

Answer 1

sinx = 3/5 so cos x = 4/5 (Use Pythagoras' rule)
siny = 5/13 so cos y = 12/13 (Use Pythagoras' rule)

cos (x + y) = cosx cosy - sinx siny
= 4/5*12/13 - 3/5*5/13

= (48 - 15)/65

= 33/65

Answer 2

cos (x+y) = cosxcosy - sinxsiny

x and y are acute angles :
sinx=3/5 -> cos²x=1-sin²x = 16/25 -> cosx=4/5
siny=5/13 -> cos²y=1-sin²y = 144/16 -> cosy=12/13

cos (x+y) = 4*12/(5*13) - 3*5/(5*13) = (48-15)/65 = 33/65

Answer 3

Sin x = 3/5....... gives us Cos x = 4/5
Sin y = 5/13 ...... therefore, cos y = 12/13
Cos (x+y) = Cos x Cos y - Sin x Sin y = 4/5 x 12/13 - 3/5 x 5/13
= 48/65 - 3/13 = 33/65

Answer 4

sinx = 3/5 => cosx = 4/5 ( x is acute so cosx >0)
siny = 5/13 => cosy = 12/13 (y is acute so cosy > 0)

cos(x+y) = cosxcosy - sinxsiny = 4/5 * 12/13 - 5/13 * 3/5 = 33/65

Answer 5

cos(x + y) = cosxcosy + sinxsiny and...
if sinx = 3/5, then the triangle is a 3-4-5 right triangle so
cosx = 4/5
also if siny = 5/13 then it is a 5-12-13 triangle so cosy = 12/13
so...
cos(x + y)
= cosxcosy + sinxsiny
= (4/5)(12/13) + (3/5)(5/13)
= (48/65) + (15/65)
= 63/65

Answer 6

omg stop that hurts my brain
man its friday
take a break