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Maths 9 โ€“ Trigonometric identities

๐Ÿ“˜ Notes

Trigonometric Identities

In these questions, you use the main trigonometric identities, the signs of trigonometric functions in different quadrants, and the exact values of special angles.

A typical question has two steps: first simplify using an identity, then calculate the final value.

1. Main identities

\[ \sin^2\alpha + \cos^2\alpha = 1 \] \[ \tan\alpha = \frac{\sin\alpha}{\cos\alpha} \] \[ \cot\alpha = \frac{\cos\alpha}{\sin\alpha} \] \[ \tan\alpha \cdot \cot\alpha = 1 \]

2. Signs in the quadrants

Quadrant Interval \(\sin\alpha\) \(\cos\alpha\) \(\tan\alpha\)
I \(0 < \alpha < \frac{\pi}{2}\) + + +
II \(\frac{\pi}{2} < \alpha < \pi\) + - -
III \(\pi < \alpha < \frac{3\pi}{2}\) - - +
IV \(-\frac{\pi}{2} < \alpha < 0\) - + -

3. Exact values you need

Angle \(\sin\) \(\cos\) \(\tan\) \(\cot\)
\(30^\circ\) \(\frac12\) \(\frac{\sqrt3}{2}\) \(\frac1{\sqrt3}\) \(\sqrt3\)
\(45^\circ\) \(\frac{\sqrt2}{2}\) \(\frac{\sqrt2}{2}\) \(1\) \(1\)
\(60^\circ\) \(\frac{\sqrt3}{2}\) \(\frac12\) \(\sqrt3\) \(\frac1{\sqrt3}\)
\(120^\circ\) \(\frac{\sqrt3}{2}\) \(-\frac12\) \(-\sqrt3\) \(-\frac1{\sqrt3}\)

4. How to find \(\sin\alpha\) if \(\cos\alpha\) is known

  1. Use \(\sin^2\alpha + \cos^2\alpha = 1\).
  2. Find \(\sin^2\alpha = 1 - \cos^2\alpha\).
  3. Take the square root.
  4. Choose the correct sign from the quadrant.

5. How to find \(\cos\alpha\) if \(\sin\alpha\) is known

  1. Use \(\sin^2\alpha + \cos^2\alpha = 1\).
  2. Find \(\cos^2\alpha = 1 - \sin^2\alpha\).
  3. Take the square root.
  4. Choose the correct sign from the quadrant.

6. Useful quick simplifications

\[ \sin^2 x + \cos^2 x = 1 \] \[ \tan x \cdot \cot x = 1 \] \[ \tan x \cdot \frac{\cos x}{\sin x} = 1 \] \[ \cot x \cdot \frac{\sin x}{\cos x} = 1 \]

7. Worked examples

Example 1

Find \(\sin\alpha\), if \(\cos\alpha = -\frac{\sqrt{21}}{5}\) and \(\frac{\pi}{2} < \alpha < \pi\).

Solution

\[ \sin^2\alpha = 1 - \cos^2\alpha \] \[ \sin^2\alpha = 1 - \frac{21}{25} = \frac{4}{25} \] \[ \sin\alpha = \pm \frac{2}{5} \]

Since \(\alpha\) is in quadrant II, \(\sin\alpha > 0\), so \[ \sin\alpha = \frac{2}{5} \]

Example 2

Find the value of \(\tan 10^\circ \cdot \cot 10^\circ + 4\sin 30^\circ\).

Solution

\[ \tan 10^\circ \cdot \cot 10^\circ = 1 \] \[ \sin 30^\circ = \frac12 \] \[ 4\sin 30^\circ = 4 \cdot \frac12 = 2 \]
\[ 1 + 2 = 3 \]

Example 3

Find \(\cos\alpha\), if \(\sin\alpha = -\frac{\sqrt{15}}{4}\) and \(-\frac{\pi}{2} < \alpha < 0\).

Solution

\[ \cos^2\alpha = 1 - \sin^2\alpha \] \[ \cos^2\alpha = 1 - \frac{15}{16} = \frac{1}{16} \] \[ \cos\alpha = \pm \frac14 \]

Since \(\alpha\) is in quadrant IV, \(\cos\alpha > 0\), so \[ \cos\alpha = \frac14 \]

Example 4

Find the value of \(\sqrt3\tan 60^\circ - \sin^2 20^\circ - \cos^2 20^\circ\).

Solution

\[ \tan 60^\circ = \sqrt3 \] \[ \sqrt3 \cdot \sqrt3 = 3 \] \[ \sin^2 20^\circ + \cos^2 20^\circ = 1 \]
\[ 3 - 1 = 2 \]

Example 5

Find \(\sin\alpha\), if \(\cos\alpha = -\frac35\) and \(\pi < \alpha < \frac{3\pi}{2}\).

Solution

\[ \sin^2\alpha = 1 - \cos^2\alpha = 1 - \frac{9}{25} = \frac{16}{25} \] \[ \sin\alpha = \pm \frac45 \]

Since \(\alpha\) is in quadrant III, \(\sin\alpha < 0\), so \[ \sin\alpha = -\frac45 \]

Example 6

Find the value of \(\sin^2 110^\circ + \cos^2 110^\circ + 5\sqrt2\sin 45^\circ\).

Solution

\[ \sin^2 110^\circ + \cos^2 110^\circ = 1 \] \[ \sin 45^\circ = \frac{\sqrt2}{2} \] \[ 5\sqrt2 \cdot \frac{\sqrt2}{2} = 5 \]
\[ 1 + 5 = 6 \]

8. Common mistakes

  • After taking a square root, forgetting to choose the correct sign.
  • Ignoring the quadrant.
  • Forgetting that \(\sin^2 x + \cos^2 x = 1\).
  • Using the wrong exact values for \(30^\circ\), \(45^\circ\), \(60^\circ\), or \(120^\circ\).
  • Not simplifying \(\tan x \cdot \cot x\) to \(1\).

9. Summary

\[ \sin^2\alpha + \cos^2\alpha = 1 \] \[ \tan\alpha = \frac{\sin\alpha}{\cos\alpha} \] \[ \cot\alpha = \frac{\cos\alpha}{\sin\alpha} \] \[ \tan\alpha \cdot \cot\alpha = 1 \]

First use the identity, then choose the correct sign using the quadrant.