Half Angle Questions With Solutions
\( \) \( \)\( \)\( \)\( \)\( \)Example to calculate the exact value of trigonometric functions of a half angle using the half angle formulas are presented. Exercises and their solutions are also included.
Review of the Half Angle Formulas
The double angle formulas are given by
A - \( \qquad \sin \left(\dfrac{\theta}{2} \right) = \pm \sqrt {\dfrac{1 - \cos \theta} {2}} \)
B - \( \qquad \cos \left(\dfrac{\theta}{2} \right) = \pm \sqrt {\dfrac{1 + \cos \theta} {2}} \)
C 1 - \( \qquad \tan \left(\dfrac{\theta}{2} \right) = \pm \sqrt { \dfrac{1 - \cos \theta}{1 + \cos \theta} } \),
C 2 - \( \quad \tan \left(\dfrac{\theta}{2} \right) = \dfrac{\sin \theta}{1 + \cos \theta} \),
C 3 - \( \quad \tan \left(\dfrac{\theta}{2} \right) = \dfrac{1 - \cos \theta}{\sin \theta} \)
NOTES
1) Selecting the + or - in these formulas depends on the quadrant where \( \dfrac{\theta}{2} \) is located
2) There are three formulas for \( \quad \tan \left(\dfrac{\theta}{2} \right) \)
Examples With Solutions
Example 1
Given that \( \cos(\theta) = - 0.9 \) and \( \theta \) is in quadrant 3 ,
a) Find the quadrant of \( \dfrac{\theta }{2} \) and the signs of \( \sin \left(\dfrac{\theta }{2} \right) \) and \( \cos \left(\dfrac{\theta }{2} \right) \)
and find the exact and approximate values of
b) \( \quad \sin \left(\dfrac{\theta }{2} \right) \)
a) \( \quad \cos \left(\dfrac{\theta }{2} \right) \)
Solution to Example 1
a)
Given that \( \theta \) is in quadrant 3 and may be written in inequality form as follows
\( \qquad \pi \lt \theta \lt \dfrac{3 \pi}{2} \)
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 inequality tells us that \( \dfrac{ \theta}{2} \) is in quadrant 2 and therefore
\( \sin \left(\dfrac{\theta }{2} \right) \gt 0 \) and \( \cos \left(\dfrac{\theta }{2} \right) \lt 0 \)
b)
According to the above formulas in A, we have
\( \qquad \sin \left(\dfrac{\theta}{2} \right) = \pm \sqrt {\dfrac{1 - \cos \theta} {2}} \qquad (I) \)
\( \cos(\theta) \) is given.
We need to decide whether to use the \( + \) or the \(-\) in front of the formula in \( (I) \). In part a) we found the sign of \( \sin \left(\dfrac{\theta}{2} \right) \) to be positive, hence
\( \qquad \sin \left(\dfrac{\theta}{2} \right) = \color{red}{ + } \sqrt {\dfrac{1 - \cos \theta} {2}} \)
Substitute \( \cos(\theta) = - 0.9 \) (given) in the above to obtain
\( \qquad \sin \left(\dfrac{\theta}{2} \right) = \sqrt {\dfrac{1 + 0.9} {2}} \)
Simplify to obtain
\( \qquad \sin \left(\dfrac{\theta}{2} \right) = \sqrt {\dfrac{19}{20}} \approx 0.97467 \)
c)
According to the above formulas in B, we have
\( \qquad \cos \left(\dfrac{\theta}{2} \right) = \pm \sqrt {\dfrac{1 + \cos \theta} {2}} \qquad (II) \)
In part a) bove we determined that \( \cos \left(\dfrac{\theta}{2} \right) \) is negative , hence
\( \qquad \cos \left(\dfrac{\theta}{2} \right) = \color{red}{ - } \sqrt {\dfrac{1 + \cos \theta} {2}} \)
Substitute \( \cos(\theta) \) by its value in the above to obtain
\( \qquad \cos \left(\dfrac{\theta}{2} \right) = - \sqrt {\dfrac{1 - 0.9} {2}} = - \dfrac{\sqrt{5}}{10} \approx - 0.22360 \)
Example 2
Given that \( \quad \tan(\theta) = - 4 \) and \( \theta \) is in quadrant 2 , find the exact and approximate values of \( \tan \left(\dfrac{\theta }{2} \right) \) .
Solution to Example 2
According to the formula C2 given above in A, we have
\( \qquad \tan \left(\dfrac{\theta}{2} \right) = \dfrac{\sin \theta}{1 + \cos \theta} \qquad (I) \)
We are given \( \tan(\theta) \) and the quadrant of \( \theta \).
Use the definition of \( \tan(\theta) \) in a right triangle
\( \qquad \tan(\theta) = \dfrac{\text{Opposite Side}}{\text{Adjacent Side}} \)
Given that \( \qquad \tan(\theta) = - 4\), we can write
We use right triangle and use \( |\tan(\theta)| = 4 \)
\( \qquad |\tan(\theta)| = \dfrac{\text{Opposite Side}}{\text{Adjacent Side}} = 4 = \dfrac{4}{1} \),
and say that
\( \text{Opposite Side} = 4 \) and \( \text{Adjacent Side} = 1 \)
Calculate the Hypotenuse of the right triangle using the Pythagorean theorem
\( \qquad \text{Hypotenuse} = \sqrt { \text{Opposite Side}^2 + \text{Adjacent Side}^2 } \)
Substitute the \( \text{Opposite Side} \) and \( \text{Adjacent Side} \) by their values to obtain
\( \qquad \text{Hypotenuse} = \sqrt{4^2 + 1^2} = \sqrt {17} \)
We now calculate \( \sin(\theta) \) and \( \cos(\theta) \) taking into account the fact that \( \sin(\theta) \) is positive and \( \cos(\theta) \) is negative in quadrant 2.
\( \qquad \sin(\theta) = \dfrac{\text{Opposite Side}}{\text{Hypotenuse}} = \dfrac{4}{\sqrt {17}}\)
\( \qquad \cos(\theta) = - \dfrac{\text{Adjacent Side}}{\text{Hypotenuse}} = - \dfrac{1}{\sqrt {17}}\)
Substitute \( \sin(\theta) \) and \( \cos(\theta) \) by their values in \( (I) \), to obtain
\( \qquad \tan \left(\dfrac{\theta}{2} \right) = \dfrac{\dfrac{4}{\sqrt {17}}}{1 - \dfrac{1}{\sqrt {17}}} \)
Simplify
\( \qquad \tan \left(\dfrac{\theta}{2} \right) = \dfrac{4}{\sqrt {17} - 1} \approx 1.28077\)
Example 3
Given that \( \sin(\theta) = - 0.2 \) and \( \theta \) is in quadrant 4 , find the exact and approximate values of
a) Find the quadrant of angle \( \dfrac{\theta}{2} \) and hence the signs of \( \sec \left(\dfrac{\theta}{2} \right) \), \( \csc \left(\dfrac{\theta}{2} \right) \) and \( \cot \left(\dfrac{\theta}{2} \right) \)
b) \( \quad \sec \left(\dfrac{\theta}{2} \right) \)
c) \( \quad \csc \left(\dfrac{\theta}{2} \right) \)
d) \( \quad \cot \left(\dfrac{\theta}{2} \right) \).
Solution to Example 3
a)
\( \theta \) in quadrant 4 could be written as an inequality
\( \dfrac{3 \pi}{2} \lt \theta \lt 2\pi \)
Divide all terms of the above inequality by 2 to obtain
\( \dfrac{3 \pi}{4} \lt \dfrac{\theta}{2} \lt \pi \)
The above inequality indicates that \( \dfrac{\theta}{2} \) is in quadrant 2 and therefore the signs of \( \sec \left(\dfrac{\theta}{2} \right) \), \( \csc \left(\dfrac{\theta}{2} \right) \) and \( \cot \left(\dfrac{\theta}{2} \right) \) are as follows:
\( \sec \left(\dfrac{\theta}{2} \right) \lt 0 \),
\( \csc \left(\dfrac{\theta}{2} \right) \gt 0 \)
\( \cot \left(\dfrac{\theta}{2} \right) \lt 0\)
b)
\( \sec \left(\dfrac{\theta}{2} \right) \) is given by
\( \qquad \sec \left(\dfrac{\theta}{2} \right) = \dfrac{1}{\cos \left(\dfrac{\theta}{2} \right)} \quad (I) \)
Use the formula of \( \cos \left(\dfrac{\theta}{2} \right) \) given in B above,
\( \qquad \cos \left( \dfrac{\theta}{2} \right) = \pm \sqrt {\dfrac{1 + \cos \theta} {2}} \quad (II) \)
We now need to find \( \cos \theta \) knowing \( \sin(\theta) = - 0.2 \). Use the identity \( \cos \theta = \pm \sqrt{1 - \sin^2 \theta} \). \( \theta \) is in quadrant 4 and therefore \( \cos \theta \) is positive, hence
\( \qquad \cos \theta = \sqrt {1 - (-0.2)^2 } = \sqrt {0.96} \)
In part a) above, we found that \( \cos \left( \dfrac{\theta}{2} \right) \) is negative, hence from \( (II) \), we have
\( \qquad \cos \left( \dfrac{\theta}{2} \right) = - \sqrt {\dfrac{1 + \cos \theta} {2}} \)
Substitute \( \cos \theta = \sqrt {0.96} \) in the above to obtain
\( \qquad \cos \left( \dfrac{\theta}{2} \right) = - \sqrt {\dfrac{1 + \sqrt {0.96} } {2}} \)
Substitute the above in \( (I) \) to obtain
\( \qquad \sec \left(\dfrac{\theta}{2} \right) = \dfrac{1}{ -\sqrt {\dfrac{1 + \sqrt {0.96} } {2}} } = - \sqrt {\dfrac{2} {1 + \sqrt {0.96} }} \approx - 1.00508 \)
c)
\( \csc \left(\dfrac{\theta}{2} \right) \) is given by
\( \qquad \csc \left(\dfrac{\theta}{2} \right) = \dfrac{1}{\sin \left(\dfrac{\theta}{2} \right)} \quad (III) \)
Use the formula of \( \sin \left(\dfrac{\theta}{2} \right) \) given in A above and the signs of \( \sin \left( \dfrac{\theta}{2} \right) \) was determined in part a), hence
\( \qquad \sin \left( \dfrac{\theta}{2} \right) = \sqrt {\dfrac{1 - \sqrt {0.96}} {2}} \quad (IV) \)
Substitute the above in \( (III) \) to obtain
\( \qquad \csc \left(\dfrac{\theta}{2} \right) = \sqrt {\dfrac{2}{1 - \sqrt {0.96}}} \approx 9.94936 \)
d)
We use the identity
\( \qquad \cot \left(\dfrac{\theta}{2} \right) = \dfrac{\cos \left(\dfrac{\theta}{2} \right) }{\sin \left(\dfrac{\theta}{2} \right)} = \dfrac{\csc \left(\dfrac{\theta}{2} \right) }{\sec \left(\dfrac{\theta}{2} \right) } \)
Substitute \( \csc \left(\dfrac{\theta}{2} \right) \) and \( \sec \left(\dfrac{\theta}{2} \right) \) by their exact values calculated above in parts a) and b)
\( \qquad \cot \left(\dfrac{\theta}{2} \right) = \dfrac{ \sqrt {\dfrac{2}{1 - \sqrt {0.96}}} }{ - \sqrt {\dfrac{2} {1 + \sqrt {0.96} }} } = - \sqrt{\dfrac{1+\sqrt{0.96}}{1-\sqrt{0.96}}} \approx - 9.89897\)
Exercises With Solutions
- Given \( \tan \theta = 4 \) and \( \theta \) is in quadrant 3, find the exact and approximate values of \( \sin \left(\dfrac{\theta}{2} \right) \).
- Given \( \csc \theta = - 5 \) and \( \theta \) is in quadrant 4, find the exact and approximate values of \( \tan \left(\dfrac{\theta}{2} \right) \).
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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