Let a and b be acute angles with sina = 4/5 and sinb = 5/13.

Find cos(a - b)

 

Answer by AnlytcPhil(1806)   (Show Source): You can put this solution on YOUR website!
 
Let a and b be acute angles with sina = 4/5 and sinb = 5/13.
Find cos(a - b)

cos(a - b) = cosa·cosb + sina·sinb

We need the cosines of a and b, so 
we draw right triangles:

Since sinA = 4/5, and since sine = opposite/hypotenuse, we
place the numerator 4 on the opposite side to angle a and
the denominator 5 on the hypotenuse

    /|
 5 / |4
  /  |
 /a  |
 
Now we calculate the adjacent side using the Pythagorean
theorem. 

a² + b² = c²
a² + 4² = 5²
a² + 16 = 25
     a² = 9
      a = 3
     
    /|
 5 / |4
  /  |
 /a  |
   3

So cosa = 3/5

Since sin = 5/13, and since sine = opposite/hypotenuse, we
place the numerator 5 on the opposite side to angle a and
the denominator 13 on the hypotenuse

    /|
13 / |5
  /  |
 /b  |
 
Now we calculate the adjacent side using the Pythagorean
theorem. 

a² + b² = c²
a² + 5² = 13²
a² + 25 = 169
     a² = 144
      a = 12
     
    /|
13 / |5
  /  |
 /b  |
  12
  
So cosb = 12/13

Now we have

sina = 4/5, sinb = 5/13, cosa = 3/5, cosb = 12/13

Substitute into

cos(a - b) = cosa·cosb + sina·sinb

cos(a - b) = (3/5)(12/13) + (4/5)(5/13)

cos(a - b) = 36/65 + 20/65

cos(a - b) = 56/65         

Edwin
AnlytcPhil@aol.com
     

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