The steps to calculate the exact value of trigonometric functions of a double angle using the double angle formulas are presented. Exercises and their solutions are also included.
The double angle formulas are given by
A - \( \qquad \sin(2 \theta) = 2 \sin(\theta) \cos(\theta) \)
B - \( \qquad \cos(2 \theta) = 1 - 2 \sin^2(\theta) = 2 \cos^2(\theta) - 1 \)
C - \( \qquad \tan(2 \theta) = \dfrac{2 \tan \theta}{1 - \tan^2 \theta} \)
Example 1
Given that \( \sin(\theta) = 0.4 \) and \( \theta \) is in quadrant 2 , find the exact and approximate values of
a) \( \quad \sin(2 \theta) \)
b) \( \quad \cos(2 \theta) \).
Solution to Example 1
a)
According to the above formulas in A, we have
\( \qquad \sin(2 \theta) = 2 \sin(\theta) \cos(\theta) \qquad (I) \)
\( \sin(\theta) \) is given, we therefore need to calculate \( \cos(\theta) \).
There are many ways to calculate \( \cos(\theta) \).
Let us use the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \) which gives
\( \qquad \cos^2(\theta) = 1 - \sin^2(\theta) \)
Solve for \( \cos(\theta) \) to obtain
\( \qquad \cos(\theta) = \pm \sqrt {1 - \sin^2(\theta) } \)
Angle \( \theta \) is in quadrant 2 (given) and therefore \( \cos(\theta) \) is negative. Hence
\( \qquad \cos(\theta) = - \sqrt {1 - 0.4^2 } = - \sqrt {0.84} \)
Substitute \( \sin(\theta) \) and \( \cos(\theta) \) by their values in \( (I) \) to obtain
\( \qquad \sin(2 \theta) = 2 \cdot 0.4 (- \sqrt {0.84}) = - 0.16 \sqrt {21} \approx -0.73321\)
b)
According to the above formulas in B, we have
\( \qquad \cos(2 \theta) = 1 - 2 \sin^2(\theta) \qquad (II) \)
Substitute \( \sin(\theta) \) by its value in \( (II) \) to obtain
\( \qquad \cos(2 \theta) = 1 - 2 \cdot 0.4^2 = 0.68\)
Example 2
Given that \( \quad \tan(\theta) = 5 \) and \( \theta \) is in quadrant 3 , find the exact and approximate values of \( \sin(2 \theta) \) .
Solution to Example 2
According to the formulas given above in A, we have
\( \qquad \sin(2 \theta) = 2 \sin(\theta) \cos(\theta) \qquad (I) \)
We are given \( \tan(\theta) \) and the quadrant of \( \theta \).
Let us use the definition of \( \tan(\theta) \) in a right triangle
\( \qquad \tan(\theta) = \dfrac{\text{Opposite Side}}{\text{Adjacent Side}} \)
Given that \( \qquad \tan(\theta) = 5 \), we can write
\( \qquad \tan(\theta) = \dfrac{5}{1} = \dfrac{\text{Opposite Side}}{\text{Adjacent Side}}\),
and say that
\( \text{Opposite Side} = 5 \) 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
\( \qquad \text{Hypotenuse} = \sqrt{5^2 + 1^2} = \sqrt {26} \)
We now calculate \( \sin(\theta) \) and \( \cos(\theta) \) taking into account the fact that both \( \sin(\theta) \) and \( \cos(\theta) \) are negative in quadrant 3.
\( \qquad \sin(\theta) = - \dfrac{\text{Opposite Side}}{\text{Hypotenuse}} = - \dfrac{5}{\sqrt {26}}\)
\( \qquad \cos(\theta) = - \dfrac{\text{Adjacent Side}}{\text{Hypotenuse}} = - \dfrac{1}{\sqrt {26}}\)
Substitute \( \sin(\theta) \) and \( \cos(\theta) \) by their values in \( (I) \), to obtain
\( \qquad \sin(2 \theta) = 2 \sin(\theta) \cos(\theta) = 2 (- \dfrac{5}{\sqrt {26}}) (- \dfrac{1}{\sqrt {26}}) \)
Simplify
\( \qquad \sin(2 \theta) = \dfrac{10}{26} = \dfrac{5}{13} \approx 0.38461\)
Example 3
Given that \( \cos(\theta) = 0.8 \) and \( \theta \) is in quadrant 4 , find the exact and approximate values of
a) \( \quad \sec(2 \theta) \)
b) \( \quad \csc(2 \theta) \)
c) \( \quad \cot(2 \theta) \).
Solution to Example 3
a)
\( \sec(2 \theta) \) is given by
\( \qquad \sec(2 \theta) = \dfrac{1}{\cos(2 \theta)} \)
Use the formula of \( \cos(2 \theta) \) given in B above, we obtain
\( \qquad \sec(2 \theta) = \dfrac{1}{2 \cos^2(\theta) - 1} \)
Substitute \( \cos(\theta) = 0.8 \) to obtain
\( \qquad \sec(2 \theta) = \dfrac{1}{2 \cdot 0.8^2 - 1} = \dfrac{1}{0.28} \approx 3.57142 \)
b)
\( \csc(2 \theta) \) is given by
\( \qquad \csc(2 \theta) = \dfrac{1}{\sin(2 \theta)} \)
Use the formula of \( \sin(2 \theta) \) given in A above, we obtain
\( \qquad \csc(2 \theta) = \dfrac{1}{2 \cos(\theta) \sin(\theta) } \qquad (I) \)
We need to find \( \sin(\theta) \) given \( \cos(\theta) = 0.8 \). The identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \) gives
\( \qquad \sin^2(\theta) = 1 - \cos^2(\theta) \)
Solve for \( \sin(\theta) \) to obtain
\( \qquad \sin(\theta) = \pm \sqrt {1 - \cos^2(\theta) } \)
Angle \( \theta \) is in quadrant 4 (given) and therefore \( \sin(\theta) \) is negative. Hence
\( \qquad \sin(\theta) = - \sqrt {1 - 0.8^2} = - 0.6 \)
Substitute \( \cos(\theta) \) and \( \sin(\theta) \) by their values in \( (I) \) above, we obtain
\( \qquad \csc(2 \theta) = \dfrac{1}{2 \cdot 0.8 \cdot (-0.6)} = - \dfrac{1}{0.96} \approx -1.04166\)
c)
We use the identity
\( \qquad \cot(2 \theta) = \dfrac{\cos(2 \theta)}{\sin(2 \theta)} = \dfrac{\csc(2 \theta)}{\sec(2 \theta)} \)
Substitute \( \csc(2 \theta) \) and \( \sec(2 \theta) \) by their exact values calculated in parts a) and b)
\( \qquad \cot(2 \theta) = \dfrac{- \dfrac{1}{0.96}}{\dfrac{1}{0.28} } = -\dfrac{7}{24} \approx -0.29166\)
