trig function

Let f(x) = 2 - cos(x) + sin(x)^2. Find the minimum and maximum values of f(x).

Hello Guest!

\(f(x) = 2 - cos(x) + sin^2(x)\\ \frac{df(x)}{dx}= sin(x)+2sin(x)cos(x) =0\)

\(sin(x)(1+2cos(x))=0\)

\(sin(x)=0\\ x=arcsin(0)\\ \color{blue}x_1=0\\ \color{blue}x_3=\pi =180^\circ\\ \color{blue}x_5=2\pi=360^\circ \)

\(1+2cos(x)=0\\ cos(x)=-\frac{1}{2}\\ x=arccos(-\frac{1}{2})\)

\(x_2=\frac{2}{3}\pi=120^\circ\\ x_4=\frac{4}{3}\pi=240^\circ\)

  !

\(\displaystyle f(x)=2-\cos(x)+\sin^{2}(x) \\=2-\cos(x)+1-\cos^{2}(x) \\ = 3-\{\cos^{2}(x)+\cos(x)+1/4-1/4\} \\ = 3-\{(\cos(x)+1/2)^{12}-1/4\} \\ =13/4-(\cos(x)+1/2)^{2}.\)

So, max will be 13/4 when cos(x) = -1/2, and min will be 13/4 - 9/4 = 1, when cos(x) = 1.