IB Mathematics AA Definition of cos , sin and tan angles Study Notes
IB Mathematics AA Definition of cos , sin and tan angles Study Notes
IB Mathematics AA Definition of cos , sin and tan angles Study Notes Offer a clear explanation of Definition of cos , sin and tan angles , including various formula, rules, exam style questions as example to explain the topics. Worked Out examples and common problem types provided here will be sufficient to cover for topic Definition of cos , sin and tan angles.
Definition of \( \sin \theta \) and \( \cos \theta \)
Definition of \( \sin \theta \) and \( \cos \theta \) in terms of the Unit Circle
The unit circle is a circle with radius 1 centered at the origin (0, 0) in the coordinate plane.
Any point on the unit circle can be represented as: \( (x, y) = (\cos \theta, \sin \theta) \)
Here, \( \theta \) is the angle measured from the positive x-axis (in radians or degrees), going counterclockwise.
Key relationships:
- \( \sin^2 \theta + \cos^2 \theta = 1 \) (Pythagorean identity)
- The sign of \( \sin \theta \) and \( \cos \theta \) depends on the quadrant where \( \theta \) lies:
| Quadrant | sin \( \theta \) | cos \( \theta \) | tan \( \theta \) |
|---|---|---|---|
| I | + | + | + |
| II | + | − | – |
| III | − | − | + |
| IV | − | + | – |
Example
Find \( \sin 150^\circ \) and \( \cos 150^\circ \) using the unit circle.
▶️Answer/Explanation
150° lies in Quadrant II (where \( \sin \) is + and \( \cos \) is −).
Reference angle = 180° − 150° = 30°.
Values for 30°:
\( \sin 30^\circ = \frac{1}{2}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2} \)
Therefore:
\( \sin 150^\circ = +\frac{1}{2} \)
\( \cos 150^\circ = -\frac{\sqrt{3}}{2} \)
Definition of \( \tan \theta \)
Definition of \( \tan \theta \)
\( \tan \theta = \frac{\sin \theta}{\cos \theta} \) provided \( \cos \theta \ne 0 \).
\( \tan \theta \) represents the slope (gradient) of a line making angle \( \theta \) with the positive x-axis.
Equation of a line through the origin: \(y = x \tan \theta \) where \( \theta \) is the angle between the line and the positive x-axis.
Special cases:
\( \theta = 45^\circ \Rightarrow y = x \)
\( \theta = 0^\circ \Rightarrow y = 0 \) (x-axis)
\( \theta = 90^\circ \Rightarrow \) vertical line (undefined gradient)
Example
Find the equation of a line through the origin making an angle of 60° with the positive x-axis.
▶️Answer/Explanation
\( \theta = 60^\circ \).
\( \tan 60^\circ \): $\tan 60^\circ = \sqrt{3} \approx 1.732 $
The equation of the line is: $ y = x \tan 60^\circ = 1.732x$
Exact values of Trigonometric Ratios
Exact Values of Trigonometric Ratios
These are the exact values of sine, cosine, and tangent for common angles (in radians):
| Angle \( \theta \) | \( \sin \theta \) | \( \cos \theta \) | \( \tan \theta \) |
|---|---|---|---|
| \( 0 \) | 0 | 1 | 0 |
| \( \frac{\pi}{6} \) | \( \frac{1}{2} \) | \( \frac{\sqrt{3}}{2} \) | \( \frac{1}{\sqrt{3}} \) |
| \( \frac{\pi}{4} \) | \( \frac{\sqrt{2}}{2} \) | \( \frac{\sqrt{2}}{2} \) | 1 |
| \( \frac{\pi}{3} \) | \( \frac{\sqrt{3}}{2} \) | \( \frac{1}{2} \) | \( \sqrt{3} \) |
| \( \frac{\pi}{2} \) | 1 | 0 | undefined |
For multiples of these angles, apply quadrant rules (ASTC or sign chart).
Example
Find the exact value of \( \sin \frac{5 \pi}{6} \) and \( \cos \frac{5 \pi}{6} \).
▶️Answer/Explanation
\( \frac{5 \pi}{6} \) lies in Quadrant II (where \( \sin \) is + and \( \cos \) is −).
Reference angle = \( \pi – \frac{5 \pi}{6} = \frac{\pi}{6} \).
\( \sin \frac{5 \pi}{6} = +\frac{1}{2} \)
\( \cos \frac{5 \pi}{6} = -\frac{\sqrt{3}}{2} \)
Extension of the Sine Rule – Ambiguous Case
Extension of the Sine Rule – Ambiguous Case
The ambiguous case arises when using the sine rule with two sides and an angle not included between them (SSA).
Given: two sides \( a \), \( b \) and angle \( A \) opposite side \( a \).
The Sine Rule: \( \frac{a}{\sin A} = \frac{b}{\sin B} \)
The number of possible triangles depends on the value of \( a \sin B \) or equivalently \( \sin B = \frac{b \sin A}{a} \):
No solution: If \( \sin B > 1 \) → no triangle.
One solution: If \( \sin B = 1 \) or the triangle is right-angled.
Two solutions: If \( 0 < \sin B < 1 \) → two possible angles for \( B \).
Remember: When \( \sin B \) gives two solutions, \( B \) can be \( B_1 = \arcsin(\sin B) \) or \( B_2 = 180^\circ – B_1 \).
Example
Given \( A = 40^\circ \), \( a = 7 \), and \( b = 10 \). Determine how many possible triangles exist and find the possible values of angle \( B \).
▶️Answer/Explanation
Use the sine rule: $ \sin B = \frac{b \sin A}{a} = \frac{10 \sin 40^\circ}{7} $
Compute: $ \sin B = \frac{10 \times 0.6428}{7} = 0.918 $
Since \( 0 < \sin B < 1 \), two solutions exist:
$ B_1 = \arcsin(0.918) \approx 66.8^\circ $
$ B_2 = 180^\circ – 66.8^\circ = 113.2^\circ $
Check for valid triangles:
Triangle 1: \( C = 180^\circ – 40^\circ – 66.8^\circ = 73.2^\circ \)
Triangle 2: \( C = 180^\circ – 40^\circ – 113.2^\circ = 26.8^\circ \)