Question 593965: Evaluate cos(Arcsin (3/5) + Arctan (12/5))
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! To solve this without a calculator we will use the formula:
cos(A+B) = cos(A)cos(B) - sin(A)sin(B)
with A = Arcsin(3/5) and B = Arctan(12/5):
cos(Arcsin (3/5) + Arctan (12/5))
cos(Arcsin(3/5)*cos(Arctan(12/5) - sin(Arcsin(3/5)*sin(Arctan(12/5)
Except for sin(Arcsin(3/5) (which, it should be obvious, is 3/5), it will help to draw a right triangle for Arcsin(3/5) and another right triangle for Arctan(12/5).
For one triangle, designate one of the acute angles as Arcsin(3/5). Since sin is opposite over hypotenuse, label the opposite side as 3 and the hypotenuse as 5. Using the Pythagorean Theorem (or your knowledge of 3/4/5 right triangles) you can find that the adjacent side is 4. Now we can use this to find cos(Arcsin(3/5): 4/5
For the other triangle, designate one of the acute angles as Arctan(12/5). Since tan is opposite over adjacent, label the opposite side as 12 and the adjacent side as 5. Using the Pythagorean Theorem (or your knowledge of 5/12/13 right triangles) you can find that the hypotenuse is 13. Now we can use this to find the sin(Arctan(12/5): 12/13 and the cos(Arctan(12/5): 5/13
Now we can substitute these values into
cos(Arcsin(3/5)*cos(Arctan(12/5) - sin(Arcsin(3/5)*sin(Arctan(12/5)
(4/5)(5/13) - (3/5)*(12/13)
Multiplying we get:
20/65 - 36/65
Subtracting we get:
-16/65
So cos(Arcsin (3/5) + Arctan (12/5)) = -16/65
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