Hi! I need to find the range of the function y = cos(x) on the interval -pi/6<x<pi/4. Help would be appreciated!
If you want a kind of graphical intepretation of the problem then draw the goniometric circle and also an angle $\displaystyle \alpha=\frac{\pi}{4}$ and an angle $\displaystyle \beta=\frac{-\pi}{6}$.
You can see that $\displaystyle \cos\left(\alpha)<\cos\left(\beta)$.
If $\displaystyle \beta \rightarrow 0$ then $\displaystyle \cos(\beta) \rightarrow 1$
If $\displaystyle \alpha \rightarrow 0$ then $\displaystyle \cos(\alpha) \rightarrow 1$
And offcourse if $\displaystyle \alpha=\beta=0$ then $\displaystyle \cos(\alpha)=\cos(\beta)=1$
That means that $\displaystyle \cos(\alpha)=\cos\left(\frac{\pi}{4}\right)=\frac{ \sqrt{2}}{2}$
is the lower limit of the range and 1 the upper limit of the range, therefore the range is:
$\displaystyle \[\frac{\sqrt{2}}{2},1\]$
Cos(x) decreases, from 1 to 0, for x from 0 to $\displaystyle \pi/2$. $\displaystyle \pi/4$ is in that range so one end point is $\displaystyle cos(\pi/4)$. One way to remember that is to remember that $\displaystyle \pi/4+ \pi/4= \pi/2$ so if a right triangle has one angle $\displaystyle \pi/4$ radians, the other is also. Of course, if two angles are equal, the sides opposite them are also equal. That is, we have an isosceles right triangle. If one leg has length s, so has the other and the length of the hypotenuse, c, is given by $\displaystyle c^2= s^2+ s^2= 2s^2$ so $\displaystyle c= s\sqrt{2}$. The cosine is "opposite side over they hypotenuse" so $\displaystyle cos(\pi/4)= \frac{s}{s\sqrt{2}}= \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2}$.