Trigonometric Identities and the Unit Circle
Questions With Solutions

How to use the unit circle to find properties and trigonometric identities of the sine and cosine functions? Grade 11 trigonometry questions are presented along with detailed solutions and explanations.
A circle has an infinite number of symmetries with respects to lines through the center and a symmetry with respect to its center. We are interested here on the symmetries with respect to its center, the x-axis, the y-axis an the line y = x. It will be shown how the use of these symmetries allows us to write several identities in trigonometry.

Identities due to Symmetry of the Unit Circle on the origin, x and y axes


Four angles (θ, π - θ, π + θ and 2π - θ) are shown below in a unit circle. To each angle corresponds a point (A, B, C or D) on the unit circle.

symmetry in unit circle.


The four angles have the same reference angle equal to \( \theta \). Due to the symmetry of the unit circle, the four points form a rectangle \( ABCD \), as shown in the diagram.

Points \( A \) and \( B \) are reflections of each other across the y-axis. Points \( A \) and \( C \) are reflections of each other across the origin. Points \( A \) and \( D \) are reflections of each other across the x-axis.

Given the coordinates \( a \) and \( b \) of point \( A \), and using the symmetries of the unit circle, the coordinates of the points \( A \), \( B \), \( C \), and \( D \) are:

\[ A: (a , b), \quad B: (-a , b), \quad C: (-a , -b), \quad D: (a , -b) \]

We now express the coordinates of each point in terms of the sine and cosine of the corresponding angle:

\[ A: (a , b) = (\cos \theta , \sin \theta) \] \[ B: (-a , b) = (\cos(\pi - \theta) , \sin(\pi - \theta)) \] \[ C: (-a , -b) = (\cos(\pi + \theta) , \sin(\pi + \theta)) \] \[ D: (a , -b) = (\cos(2\pi - \theta) , \sin(2\pi - \theta)) \]

Examples of Identities

Comparing the x and y-coordinates of points A and B, we can write:

\[ \cos(\pi - \theta) = -\cos \theta \] \[ \sin(\pi - \theta) = \sin \theta \]

Comparing the x and y-coordinates of points A and C, we can write:

\[ \cos(\pi + \theta) = -\cos \theta \] \[ \sin(\pi + \theta) = -\sin \theta \]

Comparing the x and y-coordinates of points A and D, we can write:

\[ \cos(2\pi - \theta) = \cos \theta \] \[ \sin(2\pi - \theta) = -\sin \theta \]

More Identities due to Symmetry of the Unit Circle on the x axis (Negative angles)

Two angles \(\theta\) and \(-\theta\) are shown below in a unit circle, corresponding to points \(A\) and \(D\) on the unit circle.

Symmetry in unit circle and negative angles


Points \(A\) and \(D\) are reflections of each other across the x-axis. Given the coordinates \(a\) and \(b\) of point \(A\), the coordinates of \(D\) are:
\[ D = (a, -b) \] We can express the coordinates of points \(A\) and \(D\) in terms of the sine and cosine of the corresponding angles as follows:
\[ A = (a, b) = (\cos \theta, \sin \theta) \] \[ D = (a, -b) = (\cos(-\theta), \sin(-\theta)) \]
Examples of identities that can be deduced:
\[ \cos(-\theta) = \cos \theta \] \[ \sin(-\theta) = -\sin \theta \]

Identities due to Symmetry of the Unit Circle on the line y = x

Points A and B shown in the unit circle below are reflection of each other on the line y = x. Because of the symmetry of the unit circle with respect to the line y = x, the corresponding angles to these points are θ and π/2 - θ as shown below.

symmetry in unit circle with respect to the line y = x.
Points \( A \) and \( B \) are reflections of each other across the line \( y = x \). Given the coordinates \( a \) and \( b \) of point \( A \), the coordinates of point \( B \) are given by: \[ B: (b, a) \] We now express the coordinates of points \( A \) and \( B \) in terms of the sine and cosine of the corresponding angle as follows: \[ A: (a, b) = (\cos \theta, \sin \theta) \] \[ B: (b, a) = \left( \cos \left(\dfrac{\pi}{2} - \theta \right), \sin \left(\dfrac{\pi}{2} - \theta \right) \right) \] Examples of Identities that may be deduced \[ \cos\left(\dfrac{\pi}{2} - \theta \right) = \sin \theta \] \[ \sin\left(\dfrac{\pi}{2} - \theta \right) = \cos \theta \]

Use the following standard trigonometric identities:

Now verify the following trigonometric identities:

  1. \( \cos(\pi - \theta) = -\cos \theta \)
  2. \( \sin(\pi - \theta) = \sin \theta \)
  3. \( \cos(\pi + \theta) = -\cos \theta \)
  4. \( \sin(\pi + \theta) = -\sin \theta \)
  5. \( \cos(2\pi - \theta) = \cos \theta \)
  6. \( \sin(2\pi - \theta) = -\sin \theta \)
  7. \( \cos(-\theta) = \cos \theta \)
  8. \( \sin(-\theta) = -\sin \theta \)
  9. \( \cos\left(\dfrac{\pi}{2} - \theta\right) = \sin \theta \)
  10. \( \sin\left(\dfrac{\pi}{2} - \theta\right) = \cos \theta \)

Solutions

  1. Use the general identity \( \cos(A - B) = \cos A \cos B + \sin A \sin B \) to expand \( \cos(\pi - \theta) \):

    \[ \cos(\pi - \theta) = \cos \pi \cos \theta + \sin \pi \sin \theta \]

    Using \( \cos \pi = -1 \) and \( \sin \pi = 0 \), we simplify to:

    \[ \cos(\pi - \theta) = -\cos \theta \]

  2. Use the general identity \( \sin(A - B) = \sin A \cos B - \cos A \sin B \) to expand \( \sin(\pi - \theta) \):

    \[ \sin(\pi - \theta) = \sin \pi \cos \theta - \cos \pi \sin \theta \]

    Using \( \sin \pi = 0 \) and \( \cos \pi = -1 \), we simplify to:

    \[ \sin(\pi - \theta) = \sin \theta \]

  3. Use the identity \( \cos(A + B) = \cos A \cos B - \sin A \sin B \) to expand \( \cos(\pi + \theta) \):

    \[ \cos(\pi + \theta) = \cos \pi \cos \theta - \sin \pi \sin \theta \]

    Using \( \cos \pi = -1 \) and \( \sin \pi = 0 \), we simplify to:

    \[ \cos(\pi + \theta) = -\cos \theta \]

  4. Use the identity \( \sin(A + B) = \sin A \cos B + \cos A \sin B \) to expand \( \sin(\pi + \theta) \):

    \[ \sin(\pi + \theta) = \sin \pi \cos \theta + \cos \pi \sin \theta \]

    Using \( \sin \pi = 0 \) and \( \cos \pi = -1 \), we simplify to:

    \[ \sin(\pi + \theta) = -\sin \theta \]

  5. Use the identity \( \cos(A - B) = \cos A \cos B + \sin A \sin B \) to expand \( \cos(2\pi - \theta) \):

    \[ \cos(2\pi - \theta) = \cos 2\pi \cos \theta + \sin 2\pi \sin \theta \]

    Using \( \cos 2\pi = 1 \) and \( \sin 2\pi = 0 \), we simplify to:

    \[ \cos(2\pi - \theta) = \cos \theta \]

  6. Use the identity \( \sin(A - B) = \sin A \cos B - \cos A \sin B \) to expand \( \sin(2\pi - \theta) \):

    \[ \sin(2\pi - \theta) = \sin 2\pi \cos \theta - \cos 2\pi \sin \theta \]

    Using \( \sin 2\pi = 0 \) and \( \cos 2\pi = 1 \), we simplify to:

    \[ \sin(2\pi - \theta) = -\sin \theta \]

  7. To verify \( \cos(-\theta) = \cos \theta \), rewrite it as \( \cos(0 - \theta) \):

    \[ \cos(-\theta) = \cos(0 - \theta) = \cos 0 \cos \theta + \sin 0 \sin \theta \]

    Using \( \cos 0 = 1 \) and \( \sin 0 = 0 \), we simplify to:

    \[ \cos(-\theta) = \cos \theta \]

  8. To verify \( \sin(-\theta) = -\sin \theta \), rewrite it as \( \sin(0 - \theta) \):

    \[ \sin(-\theta) = \sin(0 - \theta) = \sin 0 \cos \theta - \cos 0 \sin \theta \]

    Using \( \sin 0 = 0 \) and \( \cos 0 = 1 \), we simplify to:

    \[ \sin(-\theta) = -\sin \theta \]

  9. Use the identity \( \cos(A - B) = \cos A \cos B + \sin A \sin B \) to expand \( \cos\left(\dfrac{\pi}{2} - \theta\right) \):

    \[ \cos\left(\dfrac{\pi}{2} - \theta\right) = \cos\dfrac{\pi}{2} \cos \theta + \sin\dfrac{\pi}{2} \sin \theta \]

    Using \( \cos\dfrac{\pi}{2} = 0 \) and \( \sin\dfrac{\pi}{2} = 1 \), we simplify to:

    \[ \cos\left(\dfrac{\pi}{2} - \theta\right) = \sin \theta \]

  10. Use the identity \( \sin(A - B) = \sin A \cos B - \cos A \sin B \) to expand \( \sin\left(\dfrac{\pi}{2} - \theta\right) \):

    \[ \sin\left(\dfrac{\pi}{2} - \theta\right) = \sin\dfrac{\pi}{2} \cos \theta - \cos\dfrac{\pi}{2} \sin \theta \]

    Using \( \sin\dfrac{\pi}{2} = 1 \) and \( \cos\dfrac{\pi}{2} = 0 \), we simplify to:

    \[ \sin\left(\dfrac{\pi}{2} - \theta\right) = \cos \theta \]

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