Writing an Expression...1

[nycmathguy]

Section 4.7

Question 64

sin (arctan x)

Let u = arctan x

Then tan u = x.

tan u = opp/adj.

I draw a right triangle to find the hypotenuse.

Using the Pythagorean Theorem, the hypotenuse turns out to be sqrt{x^2 - 1}.

After the u-substitution, the given problem becomes sin u.

I know that sin u = opp/hyp.

My answer is sin u = x/sqrt{x^2 - 1}.

However, the correct answer is x/sqrt{1 + x^2}, according to Photomath.

Why is my answer wrong?

 

[HallsofIvy]

Section 4.7

View attachment 1125

Question 64

sin (arctan x)

Let u = arctan x

Then tan u = x.

tan u = opp/adj.

I draw a right triangle to find the hypotenuse.

Using the Pythagorean Theorem, the hypotenuse turns out to be sqrt{x^2 - 1}.

No, it's sqrt(x^2+ 1).
The hypotenuse is the longest side of a right triangle. It is always "+".

After the u-substitution, the given problem becomes sin u.

I know that sin u = opp/hyp.

My answer is sin u = x/sqrt{x^2 - 1}.

However, the correct answer is x/sqrt{1 + x^2}, according to Photomath.

Why is my answer wrong?

 

[nycmathguy]

No, it's sqrt(x^2+ 1).
The hypotenuse is the longest side of a right triangle. It is always "+".

In that case, sin u = x/sqrt{x^2 + 1}.