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
ANSWERED
asked 2021-03-12

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.

Answers (1)

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 \(\displaystyle \cos{{\left({112.5}^{\circ}\right)}}= \cos{{\left(\frac{{225}^{\circ}}{{2}}\right)}}{i}{s},\)
\(\displaystyle \cos{{\left(\frac{{225}^{\circ}}{{2}}\right)}}=-\sqrt{{\frac{{{1}+ \cos{{225}}^{\circ}}}{{2}}}}\)
\(\displaystyle=\sqrt{{\frac{{{1}+{\left(-\frac{\sqrt{{2}}}{{2}}\right)}}}{{2}}}}\)
\(\displaystyle=-\sqrt{{{\left(\frac{{{2}-\sqrt{{2}}}}{{2}}\right)}}}\)

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