Half-Angle Trigonometric Identities: Questions and Solutions
This page shows how to compute the exact and approximate values of trigonometric functions involving
half-angles using the
half-angle formulas.
Worked examples are followed by
practice exercises and their
complete solutions.
Review of the Half-Angle Formulas
The half-angle identities are:
\[
\sin\left(\frac{\theta}{2}\right)=\pm\sqrt{\frac{1-\cos\theta}{2}}
\]
\[
\cos\left(\frac{\theta}{2}\right)=\pm\sqrt{\frac{1+\cos\theta}{2}}
\]
\[
\tan\left(\frac{\theta}{2}\right)=\pm\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}
\]
\[
\tan\left(\frac{\theta}{2}\right)=\frac{\sin\theta}{1+\cos\theta}
\qquad
\tan\left(\frac{\theta}{2}\right)=\frac{1-\cos\theta}{\sin\theta}
\]
Notes
- The sign ± depends on the quadrant in which \( \frac{\theta}{2} \) lies.
- Three equivalent formulas exist for \( \tan\left(\frac{\theta}{2}\right) \).
Examples with Solutions
Example 1
Given \( \cos\theta=-0.9 \) and \( \theta \) lies in Quadrant III:
- Determine the quadrant of \( \frac{\theta}{2} \) and the signs of
\( \sin\left(\frac{\theta}{2}\right) \) and
\( \cos\left(\frac{\theta}{2}\right) \).
- Find \( \sin\left(\frac{\theta}{2}\right) \).
- Find \( \cos\left(\frac{\theta}{2}\right) \).
Solution
Since \( \theta \) is in Quadrant III:
\[
\pi<\theta<\frac{3\pi}{2}
\quad\Rightarrow\quad
\frac{\pi}{2}<\frac{\theta}{2}<\frac{3\pi}{4}
\]
Thus \( \frac{\theta}{2} \) lies in Quadrant II, so
\( \sin\left(\frac{\theta}{2}\right)>0 \) and
\( \cos\left(\frac{\theta}{2}\right)<0 \).
b) Compute \( \sin\left(\frac{\theta}{2}\right) \)
\[
\sin\left(\frac{\theta}{2}\right)
=\sqrt{\frac{1-\cos\theta}{2}}
=\sqrt{\frac{1+0.9}{2}}
=\sqrt{\frac{19}{20}}
\approx0.97467
\]
c) Compute \( \cos\left(\frac{\theta}{2}\right) \)
\[
\cos\left(\frac{\theta}{2}\right)
=-\sqrt{\frac{1+\cos\theta}{2}}
=-\sqrt{\frac{0.1}{2}}
=-\frac{\sqrt{5}}{10}
\approx-0.22360
\]
Example 2
Given \( \tan\theta=-4 \) and \( \theta \) lies in Quadrant II,
find \( \tan\left(\frac{\theta}{2}\right) \).
Solution
\[
\tan\left(\frac{\theta}{2}\right)=\frac{\sin\theta}{1+\cos\theta}
\]
Using a reference triangle with opposite side 4 and adjacent side 1:
\[
\sin\theta=\frac{4}{\sqrt{17}},\quad
\cos\theta=-\frac{1}{\sqrt{17}}
\]
\[
\tan\left(\frac{\theta}{2}\right)
=\frac{4}{\sqrt{17}-1}
\approx1.28077
\]
Example 3
Given \( \sin\theta=-0.2 \) and \( \theta \) lies in Quadrant IV:
- Determine the signs of
\( \sec\left(\frac{\theta}{2}\right) \),
\( \csc\left(\frac{\theta}{2}\right) \),
and \( \cot\left(\frac{\theta}{2}\right) \).
- Find \( \sec\left(\frac{\theta}{2}\right) \).
- Find \( \csc\left(\frac{\theta}{2}\right) \).
- Find \( \cot\left(\frac{\theta}{2}\right) \).
Solution
\[
\frac{3\pi}{4}<\frac{\theta}{2}<\pi
\]
So \( \frac{\theta}{2} \) is in Quadrant II.
\[
\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{0.96}
\]
\[
\sec\left(\frac{\theta}{2}\right)
=-\sqrt{\frac{2}{1+\sqrt{0.96}}}
\approx-1.00508
\]
\[
\csc\left(\frac{\theta}{2}\right)
=\sqrt{\frac{2}{1-\sqrt{0.96}}}
\approx9.94936
\]
\[
\cot\left(\frac{\theta}{2}\right)
=-\sqrt{\frac{1+\sqrt{0.96}}{1-\sqrt{0.96}}}
\approx-9.89897
\]
Exercises
- Given \( \tan\theta=4 \) in Quadrant III, find
\( \sin\left(\frac{\theta}{2}\right) \).
- Given \( \csc\theta=-5 \) in Quadrant IV, find
\( \tan\left(\frac{\theta}{2}\right) \).
Solutions to Exercises
Solutions to the Above Exercises
-
Use the formula given in A above to write
\[ \qquad \sin \left(\dfrac{\theta}{2} \right) = \pm \sqrt {\dfrac{1 - \cos \theta} {2}} \]
We need to find the quadrant of \( \dfrac{\theta}{2} \) to select the + or - in the formula and we also need to determine \( \cos \theta \).
\( \theta \) is in quadrant 3 (given) may be written using an inequality as
\[ \qquad \pi \lt \theta \lt \dfrac{3 \pi}{2} \]
Divide all terms in the above inequality by \( 2 \) to obtain
\[ \qquad \dfrac{\pi}{2} \lt \dfrac{ \theta}{2} \lt \dfrac{3 \pi}{4} \]
The above means that \( \dfrac{ \theta}{2} \) is in quadrant 2 and therefore \( \sin \left(\dfrac{\theta}{2} \right) \) is positive and given by
\[ \qquad \sin \left(\dfrac{\theta}{2} \right) = \sqrt {\dfrac{1 - \cos \theta} {2}} \quad (I) \]
Given that \( \qquad \tan \theta = 4 \), we can write
\[ \qquad \tan(\theta) = 4 = \dfrac{4}{1} = \dfrac{\text{ Opposite Side}}{\text{ Adjacent Side }} \] and write that
\[ \text{ Opposite Side} = 4 \) and \( \text{ Adjacent Side } = 1 \]
Calculate the Hypotenuse of the right triangle using the Pythagorean theorem
\[ \qquad \text{Hypotenuse} = \sqrt { \text{Adjacent Side }^2 + \text{Opposite Side}^2 } = \sqrt{1^2+4^2} = \sqrt{17} \]
We now determine \( \cos \theta \)
\[ \qquad |\cos \theta| = \dfrac{\text{ Adjacent Side }}{\text{ Hypotenuse }} = \dfrac{1}{ \sqrt{17} } \]
Since \( \theta \) is in quadrant 3, \( \cos \theta \) is negatine and therefore
\[ \cos \theta = - \dfrac{1}{ \sqrt{17}} \]
Substitute the value of \( \cos \theta \) in \( (I) \) to obtain
\[ \qquad \sin \left(\dfrac{\theta}{2} \right) = \sqrt {\dfrac{1 + \dfrac{1}{ \sqrt{17}}} {2}} \approx 0.78820 \]
-
Given \( \csc \theta = - 5 \), we write
\[ \csc \theta = \dfrac{1}{\sin \theta} = - 5 \] hence
\[ \sin \theta = - \dfrac{1}{5} \]
Use the identity \( \cos^2 \theta = 1 - \sin^2 \theta \) to claculate \( \cos \theta \)
\[ \cos \theta = \pm \sqrt {1 - \sin^2 \theta} \]
Since \( \theta \) is in quadrant 4 (given) , \( \cos \theta \) is positive and therefore
\[ \cos \theta = \sqrt {1 - \left( - \dfrac{1}{5} \right)^2 } = \dfrac{2\sqrt{6}}{5} \]
We now use formula C2 given above to write
\[ \quad \tan \left(\dfrac{\theta}{2} \right) = \dfrac{\sin \theta}{1 + \cos \theta} \]
Substitute \( \sin \theta \) and \( \cos \theta \) by their values to obtain
\[ \quad \tan \left(\dfrac{\theta}{2} \right) = \dfrac{- \dfrac{1}{5}}{1 + \dfrac{2\sqrt{6}}{5}} \]
Which may be simplified to
\[ \quad \tan \left(\dfrac{\theta}{2} \right) = -\dfrac{1}{5+2\sqrt{6}} \approx -0.10102 \]
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