Trigonometric Identities & The Unit Circle

Deriving Properties and Proofs using Symmetry

How can you use the unit circle to find properties and trigonometric identities for the sine and cosine functions?

A circle has an infinite number of symmetries. Here, we are specifically interested in symmetries with respect to the x-axis, the y-axis, the origin, and the line \( y = x \). Using these geometric symmetries allows us to visually deduce and write several foundational identities in trigonometry.

Identities due to Symmetry on the Origin, X, and Y Axes

Four angles \( \theta \), \( \pi - \theta \), \( \pi + \theta \), and \( 2\pi - \theta \) are shown below in a unit circle. Each angle corresponds to a point (\( A, B, C \), or \( D \)) on the circle.

Symmetry in the unit circle showing a rectangle formed by 4 points

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 \( ABCD \). Given the coordinates \( (a, b) \) of point \( A \), we can find the coordinates of the others through reflection:

Deduced Identities

From Points A and B (Y-Axis Symmetry):

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

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

From Points A and C (Origin Symmetry):

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

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

From Points A and D (X-Axis Symmetry):

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

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

Negative Angles (X-Axis Symmetry)

Two angles \( \theta \) and \( -\theta \) are shown below, corresponding to points \( A \) and \( D \).

Symmetry in unit circle showing positive and negative angles

Points \( A \) and \( D \) are reflections of each other across the x-axis. As seen previously, their coordinates are \( A(a,b) \) and \( D(a,-b) \). This yields the "Even/Odd" trigonometric identities:

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

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

Identities due to Symmetry Across the line \( y = x \)

Points \( A \) and \( B \) shown below are reflections of each other across the diagonal line \( y = x \). Because of this symmetry, the corresponding angles are \( \theta \) and \( \dfrac{\pi}{2} - \theta \).

Symmetry in unit circle with respect to the line y = x

Reflecting across \( y = x \) swaps the x and y coordinates. If \( A \) is \( (a, b) \), then \( B \) is \( (b, a) \).

Cofunction Identities

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

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

Algebraic Verification of Identities

We can prove all the visual symmetries we just discovered algebraically using the standard angle addition and subtraction formulas:

1. Verify: \( \cos(\pi - \theta) = -\cos\theta \)

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

Using \( \cos\pi = -1 \) and \( \sin\pi = 0 \):

\[ = (-1)\cos\theta + (0)\sin\theta = \mathbf{-\cos\theta} \]

2. Verify: \( \sin(\pi - \theta) = \sin\theta \)

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

Using \( \sin\pi = 0 \) and \( \cos\pi = -1 \):

\[ = (0)\cos\theta - (-1)\sin\theta = \mathbf{\sin\theta} \]

3. Verify: \( \cos(\pi + \theta) = -\cos\theta \)

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

\[ = (-1)\cos\theta - (0)\sin\theta = \mathbf{-\cos\theta} \]

4. Verify: \( \sin(\pi + \theta) = -\sin\theta \)

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

\[ = (0)\cos\theta + (-1)\sin\theta = \mathbf{-\sin\theta} \]

5. Verify: \( \cos(2\pi - \theta) = \cos\theta \)

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

Using \( \cos 2\pi = 1 \) and \( \sin 2\pi = 0 \):

\[ = (1)\cos\theta + (0)\sin\theta = \mathbf{\cos\theta} \]

6. Verify: \( \sin(2\pi - \theta) = -\sin\theta \)

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

\[ = (0)\cos\theta - (1)\sin\theta = \mathbf{-\sin\theta} \]

7. Verify: \( \cos(-\theta) = \cos\theta \)

Rewrite as \( \cos(0 - \theta) \):

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

Using \( \cos 0 = 1 \) and \( \sin 0 = 0 \):

\[ = (1)\cos\theta + (0)\sin\theta = \mathbf{\cos\theta} \]

8. Verify: \( \sin(-\theta) = -\sin\theta \)

Rewrite as \( \sin(0 - \theta) \):

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

\[ = (0)\cos\theta - (1)\sin\theta = \mathbf{-\sin\theta} \]

9. Verify: \( \cos\left(\dfrac{\pi}{2} - \theta\right) = \sin\theta \)

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

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

\[ = (0)\cos\theta + (1)\sin\theta = \mathbf{\sin\theta} \]

10. Verify: \( \sin\left(\dfrac{\pi}{2} - \theta\right) = \cos\theta \)

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

\[ = (1)\cos\theta - (0)\sin\theta = \mathbf{\cos\theta} \]

Challenge Questions

Apply your understanding of symmetry, identities, and angle addition formulas to solve the following challenge problems.

Challenge 1 (Tangent Symmetry): Using the sine and cosine symmetries derived above, prove that the tangent function has a period of \( \pi \). Meaning, prove that \( \tan(\pi + \theta) = \tan\theta \).

Show Proof

Step 1: Use the quotient identity to rewrite tangent in terms of sine and cosine.

\[ \tan(\pi + \theta) = \dfrac{\sin(\pi + \theta)}{\cos(\pi + \theta)} \]

Step 2: Substitute the symmetries for origin reflection (Quadrant III).

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

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

Step 3: Simplify the fraction.

\[ \dfrac{-\sin\theta}{-\cos\theta} = \dfrac{\sin\theta}{\cos\theta} = \mathbf{\tan\theta} \]

Challenge 2 (Applying Symmetries): Evaluate the exact value of the following expression without a calculator: \[ \sin\left(\dfrac{13\pi}{4}\right) \cos\left(-\dfrac{5\pi}{6}\right) \]

Show Solution

Step 1: Simplify the Sine Term.
Find a coterminal angle for \( \dfrac{13\pi}{4} \) by subtracting \( 2\pi \) (\( \dfrac{8\pi}{4} \)).
\( \sin\left(\dfrac{13\pi}{4}\right) = \sin\left(\dfrac{5\pi}{4}\right) \).
This is in Quadrant III, related to \( \dfrac{\pi}{4} \). Therefore, \( \sin\left(\dfrac{5\pi}{4}\right) = -\dfrac{\sqrt{2}}{2} \).

Step 2: Simplify the Cosine Term.
Use the even function identity: \( \cos(-\theta) = \cos(\theta) \).
\( \cos\left(-\dfrac{5\pi}{6}\right) = \cos\left(\dfrac{5\pi}{6}\right) \).
This is in Quadrant II, related to \( \dfrac{\pi}{6} \). Therefore, \( \cos\left(\dfrac{5\pi}{6}\right) = -\dfrac{\sqrt{3}}{2} \).

Step 3: Multiply the values.

\[ \left(-\dfrac{\sqrt{2}}{2}\right) \cdot \left(-\dfrac{\sqrt{3}}{2}\right) = \mathbf{\dfrac{\sqrt{6}}{4}} \]

Challenge 3 (Advanced Cofunctions): Use the angle addition formulas to verify the identity for reflecting across the negative y-axis (Quadrant III/IV boundary): \[ \cos\left(\dfrac{3\pi}{2} - \theta\right) = -\sin\theta \]

Show Proof

Step 1: Expand using the cosine difference formula: \( \cos(A - B) = \cos A \cos B + \sin A \sin B \).

\[ \cos\left(\dfrac{3\pi}{2} - \theta\right) = \cos\left(\dfrac{3\pi}{2}\right) \cos\theta + \sin\left(\dfrac{3\pi}{2}\right) \sin\theta \]

Step 2: Evaluate the trigonometric values at \( \dfrac{3\pi}{2} \) (270 degrees).

We know that at the bottom of the unit circle, the coordinates are \( (0, -1) \). So, \( \cos\left(\dfrac{3\pi}{2}\right) = 0 \) and \( \sin\left(\dfrac{3\pi}{2}\right) = -1 \).

Step 3: Substitute and simplify.

\[ = (0)\cos\theta + (-1)\sin\theta = \mathbf{-\sin\theta} \]

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