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Year 10 Maths - 10A Trigonometry

The Unit Circle

20 practice questions 2 video lessons Theory + worked examples
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Theory

Any angle, even one over \(360^\circ\) or a negative one, has a sine, cosine and tangent. On the unit circle, the point at angle \(\theta\) is \((\cos\theta, \sin\theta)\). Find the quadrant and the related acute angle, then take the sign from the ASTC rule. This Year 10A maths page covers exact values in all four quadrants, obtuse and negative angles, and finding the other ratios from one ratio or a point.

On the unit circle (radius \(1\), centre the origin), the point at angle \(\theta\), measured anticlockwise from the positive \(x\)-axis, is \(P(\cos\theta, \sin\theta)\). This defines sine and cosine for every angle, and \(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\).

The axes split the plane into four quadrants. The signs of \(x\) and \(y\) give the ASTC rule: All positive in quadrant 1, Sine in 2, Tangent in 3, Cosine in 4.

The related acute angle is the angle between the arm \(OP\) and the nearest part of the \(x\)-axis. A ratio of \(\theta\) equals the same ratio of the related angle, with the sign from ASTC.

A negative angle is measured clockwise. Adding or subtracting \(360^\circ\) does not change the direction, so it does not change any ratio: \(\sin 400^\circ = \sin 40^\circ\). For any point \((x, y)\) on the arm, at distance \(r\) from the origin: \(\sin\theta = \dfrac{y}{r}\), \(\cos\theta = \dfrac{x}{r}\), \(\tan\theta = \dfrac{y}{x}\).

ASTC: which ratios are positive in each quadrantThe number plane split into four quadrants. Quadrant 1, 0 to 90 degrees: all ratios positive. Quadrant 2, 90 to 180 degrees: sine positive. Quadrant 3, 180 to 270 degrees: tangent positive. Quadrant 4, 270 to 360 degrees: cosine positive. x y A all positive S sin positive T tan positive C cos positive Q1 Q2 Q3 Q4 0°, 360° 90° 180° 270°
All Stations To Central: the ratios that are positive, quadrant by quadrant, going anticlockwise from the first.
Point on the unit circle in the third quadrantA circle of radius 1 centred at the origin O. The arm OP makes an angle theta of about 215 degrees with the positive x-axis, measured anticlockwise, so P is in the third quadrant. P has coordinates cos theta, sin theta, both negative. The related acute angle alpha is between OP and the negative x-axis. θ α P(cos θ, sin θ) O 1
On the unit circle, \(P = (\cos\theta, \sin\theta)\). The related acute angle \(\alpha\) is measured to the nearest part of the \(x\)-axis and sets the size of each ratio; the quadrant sets the sign.
QuadrantAngle \(\theta\)Related anglePositive
1\(0^\circ\)–\(90^\circ\)\(\theta\)all
2\(90^\circ\)–\(180^\circ\)\(180^\circ - \theta\)\(\sin\)
3\(180^\circ\)–\(270^\circ\)\(\theta - 180^\circ\)\(\tan\)
4\(270^\circ\)–\(360^\circ\)\(360^\circ - \theta\)\(\cos\)
\[\sin(-\theta) = -\sin\theta \qquad \cos(-\theta) = \cos\theta \qquad \tan(-\theta) = -\tan\theta\]
sin(−θ)=−sinθ,cos(−θ)=cosθ
\[\sin(\theta \pm 360^\circ) = \sin\theta \qquad \cos(\theta \pm 360^\circ) = \cos\theta\]
sin(θ+360°)=sinθ
Obtuse angles. For \(90^\circ < \theta < 180^\circ\): \(\sin(180^\circ - \theta) = \sin\theta\), \(\cos(180^\circ - \theta) = -\cos\theta\) and \(\tan(180^\circ - \theta) = -\tan\theta\).
Boundary angles. At \(0^\circ\), \(90^\circ\), \(180^\circ\), \(270^\circ\) the point is on an axis: \((1, 0)\), \((0, 1)\), \((-1, 0)\), \((0, -1)\). So \(\sin 270^\circ = -1\), \(\cos 180^\circ = -1\), and \(\tan 90^\circ\) and \(\tan 270^\circ\) are undefined.

How to find a ratio of any angle

  1. Bring the angle into \(0^\circ\)–\(360^\circ\) by adding or subtracting \(360^\circ\).
  2. Find the quadrant and the related acute angle (table above).
  3. Write the ratio of the related angle, with the sign from ASTC.

Coordinates and expressions

  1. A point on the unit circle at angle \(\theta\) is \((\cos\theta, \sin\theta)\): find each ratio with the steps above.
  2. A point \((x, y)\) on the arm but not on the unit circle: find \(r = \sqrt{x^2 + y^2}\) first.
  3. Expressions such as \(\cos 405^\circ + \sin 30^\circ\): replace each ratio by its exact value, then simplify the surds and fractions.

Two conditions

  1. If you are told, for example, \(\sin\theta > 0\) and \(\tan\theta < 0\), list the quadrants for each condition and take the one they share (here quadrant 2).
  2. Then draw the related-angle triangle, find the third side, and attach the signs for that quadrant (Example 3).
Example 1 — Exact values in any quadrant
Find the exact value of:
(i) \(\sin 240^\circ\)
(ii) \(\cos 315^\circ\)
(iii) \(\tan(-150^\circ)\)
(iv) \(\sin 480^\circ\)
Solution

For each angle: find the quadrant, the related acute angle, then the sign from ASTC.

(i) Quadrant 3, related angle \(240^\circ - 180^\circ = 60^\circ\); sine is negative there.

\(\sin 240^\circ\)\(=\)\(-\sin 60^\circ = \mathbf{-\dfrac{\sqrt{3}}{2}}\)

(ii) Quadrant 4, related angle \(360^\circ - 315^\circ = 45^\circ\); cosine is positive there.

\(\cos 315^\circ\)\(=\)\(\cos 45^\circ = \mathbf{\dfrac{1}{\sqrt{2}}}\)

(iii) Add \(360^\circ\): \(-150^\circ\) points the same way as \(210^\circ\). That is quadrant 3, where tangent is positive.

\(\tan(-150^\circ)\)\(=\)\(\tan 210^\circ\)
\(\)\(=\)\(\tan 30^\circ = \mathbf{\dfrac{1}{\sqrt{3}}}\)

(iv) Subtract \(360^\circ\) to get \(120^\circ\), in quadrant 2, where sine is positive.

\(\sin 480^\circ\)\(=\)\(\sin 120^\circ = \sin 60^\circ = \mathbf{\dfrac{\sqrt{3}}{2}}\)
sin240°=−32
Example 2 — Points on the unit circle
Points \(P\) and \(Q\) lie on the unit circle at angles of \(135^\circ\) and \(300^\circ\) from the positive \(x\)-axis. Find
(i) the exact coordinates of \(P\)
(ii) the exact coordinates of \(Q\)
(iii) the exact value of \(\cos 300^\circ - \sin 135^\circ \cos 135^\circ\).
Solution

(i) \(135^\circ\) is in quadrant 2, related angle \(45^\circ\); only sine is positive there.

\(P\)\(=\)\((\cos 135^\circ, \sin 135^\circ)\)
\(\)\(=\)\(\left(\mathbf{-\dfrac{1}{\sqrt{2}}}, \mathbf{\dfrac{1}{\sqrt{2}}}\right)\)

(ii) \(300^\circ\) is in quadrant 4, related angle \(60^\circ\); only cosine is positive there.

\(Q\)\(=\)\((\cos 300^\circ, \sin 300^\circ)\)
\(\)\(=\)\(\left(\mathbf{\dfrac{1}{2}}, \mathbf{-\dfrac{\sqrt{3}}{2}}\right)\)

(iii) Substitute the exact values:

\(\cos 300^\circ\)
\(- \sin 135^\circ \cos 135^\circ\)
\(=\)\(\dfrac{1}{2} - \dfrac{1}{\sqrt{2}} \times \left(-\dfrac{1}{\sqrt{2}}\right)\)
\(\)\(=\)\(\dfrac{1}{2} + \dfrac{1}{2} = \mathbf{1}\)
P=(−12,12)

Check: each point is \(1\) from the origin, since \(\cos^2\theta + \sin^2\theta = 1\).

Example 3 — One ratio and the quadrant
Given that \(\cos\theta = \dfrac{7}{25}\) and \(270^\circ < \theta < 360^\circ\), find the exact values of \(\sin\theta\) and \(\tan\theta\).
Solution

Draw the triangle for the related angle \(\alpha\) with adjacent \(7\) and hypotenuse \(25\).

Triangle for the related angle alphaA right-angled triangle for the related angle alpha: adjacent 7, hypotenuse 25 and the third side 24, found by Pythagoras' theorem. α 7 24 25
\(\text{opposite}\)\(=\)\(\sqrt{25^2 - 7^2} = \sqrt{576} = 24\)

\(\theta\) is in quadrant 4, where only cosine is positive.

\(\sin\theta\)\(=\)\(\mathbf{-\dfrac{24}{25}}\)
\(\tan\theta\)\(=\)\(\dfrac{\sin\theta}{\cos\theta} = \mathbf{-\dfrac{24}{7}}\)
sinθ=−2425,tanθ=−247
Example 4 — A point on the terminal arm
The point \((-5, 12)\) lies on the terminal arm of an angle \(\theta\), where \(0^\circ \le \theta \le 360^\circ\). Find
(i) the exact values of \(\sin\theta\), \(\cos\theta\) and \(\tan\theta\)
(ii) \(\theta\), correct to the nearest degree.
Solution

(i) Find \(r\), the distance from the origin, then use \(\sin\theta = \dfrac{y}{r}\), \(\cos\theta = \dfrac{x}{r}\), \(\tan\theta = \dfrac{y}{x}\). The signs come out automatically.

Point (-5, 12) on the terminal arm of thetaThe arm from the origin to the point negative 5, 12 in the second quadrant. Its length r is the hypotenuse of a right-angled triangle with sides 5 and 12. Theta is measured anticlockwise from the positive x-axis; alpha is the related angle to the negative x-axis. θ α (−5, 12) 12 5 r
\(r\)\(=\)\(\sqrt{(-5)^2 + 12^2} = 13\)
\(\sin\theta\)\(=\)\(\mathbf{\dfrac{12}{13}}\)
\(\cos\theta\)\(=\)\(\mathbf{-\dfrac{5}{13}}\)
\(\tan\theta\)\(=\)\(\dfrac{12}{-5} = \mathbf{-\dfrac{12}{5}}\)

(ii) Quadrant 2. The related angle has \(\tan\alpha = \dfrac{12}{5}\).

\(\alpha\)\(=\)\(67.380\ldots^\circ\)
\(\theta\)\(=\)\(180^\circ - \alpha = 112.619\ldots^\circ \approx \mathbf{113^\circ}\)
r=(−5)2+122=13

A calculator's \(\tan^{-1}\left(-\dfrac{12}{5}\right)\) gives \(-67.4^\circ\), which is the wrong quadrant. Always work from the related angle.

Common pitfalls

Use the x-axis, not the y-axis. The related angle of \(240^\circ\) is \(60^\circ\) (to the negative \(x\)-axis), not \(30^\circ\).
Find the related angle from the positive value. For a point such as \((-5, 12)\), use \(\tan\alpha = \dfrac{12}{5}\), then place \(\alpha\) in the right quadrant. The calculator's \(\tan^{-1}\left(-\dfrac{12}{5}\right)\) gives an angle in the wrong quadrant.
Coordinates are (cos, sin), in that order. The \(x\)-coordinate is the cosine. Swapping them is a common slip with \(30^\circ\) and \(60^\circ\) values.
Cosine ignores the sign of the angle. \(\cos(-\theta) = \cos\theta\), but \(\sin(-\theta) = -\sin\theta\) and \(\tan(-\theta) = -\tan\theta\).
Attach the sign last. Find the size from the related angle, then the sign from the quadrant. A triangle drawn for the related angle has positive sides only.

Frequently asked questions

What does ASTC mean in trigonometry?

ASTC tells you which ratios are positive in each quadrant, going anticlockwise from the first: All in quadrant 1, Sine in quadrant 2, Tangent in quadrant 3 and Cosine in quadrant 4. A common way to remember it is All Stations To Central.

How do you find the related angle?

It is the acute angle between the arm of the angle and the x-axis. In quadrant 2 it is 180 minus theta, in quadrant 3 it is theta minus 180, and in quadrant 4 it is 360 minus theta.

How do you find sin of a negative angle?

A negative angle is measured clockwise. Add \(360^\circ\) to get the same direction as a positive angle, or use \(\sin(-\theta) = -\sin\theta\).

What is the unit circle definition of sine and cosine?

On a circle of radius 1 centred at the origin, the point at angle theta from the positive x-axis has coordinates cos theta, sin theta. Cosine is the x-coordinate and sine is the y-coordinate.

Why is the sine of an obtuse angle positive but its cosine negative?

An obtuse angle puts the point in quadrant 2, where \(y\) is positive and \(x\) is negative. Sine is the \(y\)-coordinate and cosine is the \(x\)-coordinate, so \(\sin 150^\circ = \dfrac{1}{2}\) but \(\cos 150^\circ = -\dfrac{\sqrt{3}}{2}\).