Evaluating a Composition of Functions...2

[nycmathguy]

Section 4.7

Question 54

sin (arccos sqrt{5})

Let u = arccos sqrt{5}.
This means that cos u = sqrt{5}/1.

The range of the arccos function is [0, pi]. I know that cos u is positive in quadrant 1.

cos u = adj/hyp. I draw a right triangle in quadrant 1 to find the opposite side.

I will use a^2 + b^2 = c^2 to find the opposite side.

Let b = opposite side of right triangle.

(sqrt{5})^2 + b^2 = 1^2

5 + b^2 = 1

b^2 = 1 - 5

b^2 = -4

Can't take the square of a negative numbers outside the complex number system.

So, I cannot calculate the opposite side of the right triangle formed in quadrant 1

My conclusion is as follows:

sin (arccos sqrt{5}) is undefined.

You say?

 

[HallsofIvy]

You are making the same mistake over and over again! cosine is sqrt(5). cosine is "near side over hypotenuse" and the hypotenuse is the longest side of a right triangle! The hypotenuse is sqrt(5) and the near side is 1. if b is the length of the opposite side, 1+ b^2= 5.

 

[nycmathguy]

You are making the same mistake over and over again! cosine is sqrt(5). cosine is "near side over hypotenuse" and the hypotenuse is the longest side of a right triangle! The hypotenuse is sqrt(5) and the near side is 1. if b is the length of the opposite side, 1+ b^2= 5.

I will do the remaining even number questions when time allows.