Question

When using the half-angle formulas for trigonometric functions of alpha/2, I determine the sign based on the quadrant in which alpha lies.Determine whether the statement makes sense or does not make sense, and explain your reasoning.

Trigonometric Functions

When using the half-angle formulas for trigonometric functions of $$\displaystyle\frac{\alpha}{{2}}$$, I determine the sign based on the quadrant in which $$\alpha$$ lies.Determine whether the statement makes sense or does not make sense, and explain your reasoning.

2021-03-13

For example:
The trigonometric function is given as, $$\displaystyle \cos{{\left({112.5}^{\circ}\right)}}$$
$$\displaystyle \cos{{\left({112.5}^{\circ}\right)}}= \cos{{\left(\frac{{225}^{\circ}}{{2}}\right)}}$$
Here, $$\alpha=225^{\circ}$$
Apply the half angle formula for the given trigonometric function, $$\cos(112.5^{\circ})$$ in which the angle $$112.5^{\circ}$$ lies in the quadrant (II) where sine and cosecant is positive. We have to determine the value of $$\cos(112.5^{\circ})$$ which is negative in second quadrant.
Therefore in half angle formula , negative sign has to be put.
$$\displaystyle \cos{{\left(\frac{\alpha}{{2}}\right)}}=-\sqrt{{\frac{{{1}+ \cos{\alpha}}}{{2}}}}$$
The value of $$\displaystyle \cos{{\left({112.5}^{\circ}\right)}}= \cos{{\left(\frac{{225}^{\circ}}{{2}}\right)}}{i}{s},$$
$$\displaystyle \cos{{\left(\frac{{225}^{\circ}}{{2}}\right)}}=-\sqrt{{\frac{{{1}+ \cos{{225}}^{\circ}}}{{2}}}}$$
$$\displaystyle=\sqrt{{\frac{{{1}+{\left(-\frac{\sqrt{{2}}}{{2}}\right)}}}{{2}}}}$$
$$\displaystyle=-\sqrt{{{\left(\frac{{{2}-\sqrt{{2}}}}{{2}}\right)}}}$$