Double Angle Trigonometry Problems with Solutions

This page explains how to find the exact and approximate values of trigonometric functions involving double angles using the double angle formulas. Fully worked examples and exercises with solutions are included.

Review of Double Angle Formulas

The double angle identities are:

• \(\sin(2\theta) = 2\sin\theta\cos\theta\)
• \(\cos(2\theta) = 1 - 2\sin^2\theta = 2\cos^2\theta - 1\)
• \(\tan(2\theta) = \dfrac{2\tan\theta}{1 - \tan^2\theta}\)

Examples with Step-by-Step Solutions

Example 1

Given \(\sin\theta = 0.4\) and \(\theta\) is in Quadrant II, find the exact and approximate values of:

1. \(\sin(2\theta)\)
2. \(\cos(2\theta)\)

Solution

a) Using \(\sin(2\theta) = 2\sin\theta\cos\theta\), we must first find \(\cos\theta\).

From the identity \(\sin^2\theta + \cos^2\theta = 1\),

Since \(\theta\) is in Quadrant II, \(\cos\theta\) is negative:

Substitute:

b) Using \(\cos(2\theta) = 1 - 2\sin^2\theta\):

Example 2

Given \(\tan\theta = 5\) and \(\theta\) is in Quadrant III, find \(\sin(2\theta)\).

Solution

Interpret \(\tan\theta = \dfrac{5}{1}\) as a right triangle with:

• Opposite side = 5
• Adjacent side = 1

The hypotenuse is:

In Quadrant III, both sine and cosine are negative:

Now apply the double angle formula:

Example 3

Given \(\cos\theta = 0.8\) and \(\theta\) is in Quadrant IV, find:

1. \(\sec(2\theta)\)
2. \(\csc(2\theta)\)
3. \(\cot(2\theta)\)

Solution

a)

b)

First find \(\sin\theta\):

c)

Exercises

1. Given \(\cos\theta = -0.2\) in Quadrant II, find \(\tan(2\theta)\).
2. Given \(\cot\theta = 4\) in Quadrant III, find \(\sin(2\theta)\).
3. Given \(\sec\theta = -4\) in Quadrant III, find \(\sec(2\theta)\).

Solutions to Exercises

1. 
   Formula given in C above is written as \[ \qquad \tan(2 \theta) = \dfrac{2 \tan \theta}{1 - \tan^2 \theta} \quad (I) \] \[ \sin \theta = \sqrt{1-(-0.2)^2} = \sqrt{ 0.96} \] . \[ \tan \theta = \dfrac{\sin \theta}{\cos \theta} = \dfrac{\sqrt{0.96}}{-0.2} = - \sqrt{24} \] Substitute \( \tan \theta = - \sqrt{24} \) in \( (I) \) \[ \qquad \tan(2 \theta) = \dfrac{2 (- \sqrt{24})}{-23} = \dfrac{4\sqrt{6}}{23} \approx 0.42599 \]
2. 
   Let us use the definition of \( \cot(\theta) \) in a right triangle \[ \qquad \cot(\theta) = \dfrac{\text{Adjacent Side}}{\text{ Opposite Side}} \] Given that \( \qquad \cot(\theta) = 4 \), we can write \[ \qquad \cot(\theta) = \dfrac{4}{1} = \dfrac{\text{Adjacent Side}}{\text{Opposite Side}}\] and write that \[ \text{Adjacent Side} = 4 \quad \text{and} \quad \text{Opposite Side} = 1 \] Calculate the Hypotenuse of the right triangle using the Pythagorean theorem \[ \qquad \text{Hypotenuse} = \sqrt { \text{Adjacent Side }^2 + \text{Opposite Side}^2 } = \sqrt{4^2+1^2} = \sqrt{17} \] In quadrant \( 3 \), \( \sin (\theta) \) and \( \cos (\theta) \) are negative and using the definition of \( \sin (\theta) \) and \( \cos(\theta) \), we have \[ \sin (\theta) = - \dfrac{\text{Opposite Side}}{\text{Hypotenuse}} = - \dfrac{1}{\sqrt{17}} \] \[ \sin (\theta) = - \dfrac{\text{Adjacent Side}}{\text{Hypotenuse}} = - \dfrac{4}{\sqrt{17}} \] Substitute in the formula \( \qquad \sin(2 \theta) = 2 \sin(\theta) \cos(\theta) \) in A above, to obtain \[ \qquad \sin(2 \theta) = 2 (- \dfrac{1}{\sqrt{17}}) (- \dfrac{4}{\sqrt{17}}) = \dfrac{8}{17} \approx 0.47058 \]
3. 
   Given \( \sec \theta = -4 \) and \( \sec \theta = \dfrac{1}{\cos \theta}\), we have \[ \dfrac{1}{\cos \theta} = - 4\] Solve for \( \cos \theta \) \[ \cos \theta = - \dfrac{1}{4}\] We are asked to calculate \( \sec (2 \theta) \) which is given by \[ \sec (2 \theta) = \dfrac{1}{\cos (2 \theta)} \] Use formula B, given above, to write \[ \sec (2 \theta) = \dfrac{1}{ 2 \cos^2(\theta) - 1 } \] Substitute \( \cos \theta \) by its value found above, \[ \sec (2 \theta) = \dfrac{1}{ 2 \left(- \dfrac{1}{4} \right)^2 - 1 } = - \dfrac{8}{7} \approx -1.14285 \]

More References

• Double Angle Formulas
• Trigonometric Tables
• More Trigonometry Problems