# Concavity of Trig Functions

1. Jun 3, 2008

### DMac

[SOLVED] Concavity of Trig Functions

I'm working out a problem, but I'm stuck at a spot.

The original equation is y = 2 cos x + sin 2x, and yes I did the second derivative, which came out to be

cos x (4 sin x + 1)

when i equate it to zero, i get
cos x = 0 and 4 sin x + 1 = 0

(Let's just say the domain was defined from 0 to 2 pi)

For SOME reason, I keep trying to evaluate x when 4 sin x + 1 = 0

by using inverse sin ....so x = inverse sin (-0.25)

My calculator is in radians, but it still gives me an answer of -0.25. I used a graphing calculator, and the answer SHOULD be 3.39. What am I doing wrong?

2. Jun 3, 2008

### Defennder

You forgot that the third quadrant anti-clockwise of the ASTC diagram is also negative. And that corresponds to pi + the basic angle. Basic angle is given by arcsin 1/4.

3. Jun 4, 2008

### konthelion

In your second derivative, I believe you are missing a constant

y'=2(cos(2x)-sin(x)), then y'' = -2(2sin(2x)+cos(x)) = -2cos(x)(4sin(x)+1)
By double-angle formula sin(2A)=2sin(A)cos(A)