Question

Use the figure to evaluate the function given that g(x)=cos xg(theta/2)=

Answer 1

First, notice that since the point (a,1) belongs to the circumference x^2+y^2=5, then:

$\begin{gathered} a^2+1^2=5 \\ \Rightarrow a^2+1=5 \\ \Rightarrow a^2=4 \end{gathered}$

Since the point (a,1) is in the second quadrant, then:

$a=-2$

Remember that the coordinates of a point on a circle of radius r can be written in terms of the angle as:

$(r\cos (\theta),r\sin (\theta))$

The radius of the circle in this case is the square root of 5. Then:

$(-2,1)=(\sqrt[]{5}\cos (\theta),\sqrt[]{5}\sin (\theta))$

Therefore:

$\begin{gathered} \sqrt[]{5}\cos (\theta)=-2 \\ \Rightarrow\cos (\theta)=-\frac{2}{\sqrt[]{5}} \end{gathered}$

On the other hand, remember the following identity:

$\cos (\frac{\theta}{2})=\sqrt[]{\frac{1+\cos (\theta)}{2}}$

Substitute the value for the cosine of θ:

$\begin{gathered} \cos (\frac{\theta}{2})=\sqrt[]{\frac{1-\frac{2}{\sqrt[]{5}}}{2}} \\ =\sqrt[]{\frac{\sqrt[]{5}-2}{2\sqrt[]{5}}} \\ =\sqrt[]{\frac{5-2\sqrt[]{5}}{10}} \end{gathered}$

Therefore:

$g(\frac{\theta}{2})=\sqrt[]{\frac{5-2\sqrt[]{5}}{10}}$