Trigonometric Equations

Section 1 of 6

The unit circle and ASTC

Picture a circle of radius $1$ centred at the origin. We call it the unit circle.

Pick any point $(x, y)$ on the circle. Drop a perpendicular down to the $x$-axis — you get a right triangle.

(i) Sin, Cos, Tan from the triangle

From SOH CAH TOA, with hypotenuse $= 1$:

$\cos A = \dfrac{\text{adj}}{\text{hyp}} = \dfrac{x}{1} = x$

$\sin A = \dfrac{\text{opp}}{\text{hyp}} = \dfrac{y}{1} = y$

$\tan A = \dfrac{\text{opp}}{\text{adj}} = \dfrac{y}{x} = \dfrac{\sin A}{\cos A}$

So on the unit circle: $\cos A$ is the $x$-coordinate and $\sin A$ is the $y$-coordinate.

(ii) The four quadrants

Going anticlockwise from the positive $x$-axis:

Q1: $0°$ to $90°$ (both $x$ and $y$ positive)

Q2: $90°$ to $180°$ ($x$ negative, $y$ positive)

Q3: $180°$ to $270°$ (both negative)

Q4: $270°$ to $360°$ ($x$ positive, $y$ negative)

Since $\cos A = x$ and $\sin A = y$, the signs of sin/cos/tan flip from quadrant to quadrant.

(iii) ASTC

We remember which is positive where with ASTC:

Q1: All positive

Q2: Sin positive (cos, tan negative)

Q3: Tan positive (sin, cos negative)

Q4: Cos positive (sin, tan negative)

Some students learn it as All Students Take Calculus.

(iv) The reference angle

When we solve a trig equation, we first find the reference angle — the angle in Q1 that gives the same positive ratio.

Then we use it to land in whichever quadrants the sign matches.

Foundations

1.On the unit circle: $\cos A = x$, $\sin A = y$, $\tan A = \dfrac{y}{x} = \dfrac{\sin A}{\cos A}$.

2.ASTC: Q1 All, Q2 Sin, Q3 Tan, Q4 Cos — positive.

3.The reference angle is the Q1 angle from the positive value.

4.Quadrant formulas: Q2 $= 180° - A$, Q3 $= 180° + A$, Q4 $= 360° - A$.

Section 2 of 6

Basic equations — degrees

Now we use ASTC to solve equations for $0° \le A \le 360°$.

(i) Worked example — $\cos A = -\dfrac{1}{2}$

Solve $\cos A = -\dfrac{1}{2}$ for $0° \le A \le 360°$.

Step 1 — ask yourself: is it sin/cos/tan, and is it positive or negative?

$\cos A$, and it's negative.

Step 2 — from ASTC: cos is positive in Q1 and Q4, so cos is negative in Q2 and Q3.

Step 3 — find the reference angle using the positive value:

$\cos A = \dfrac{1}{2} \quad \Rightarrow \quad A = 60°$

Step 4 — apply the quadrant formulas:

Q2: $180° - 60° = 120°$

Q3: $180° + 60° = 240°$

$A = 120°$ or $240°$

(ii) Worked example — $\tan A = -1$

Solve $\tan A = -1$ for $0° \le A \le 360°$.

$\tan A$ is negative. From ASTC: tan positive in Q1, Q3 — so negative in Q2 and Q4.

Reference angle: $\tan A = 1 \quad \Rightarrow \quad A = 45°$

Q2: $180° - 45° = 135°$

Q4: $360° - 45° = 315°$

$A = 135°$ or $315°$

Process — every basic equation

1.Sin, cos, or tan? And: positive or negative?

2.Use ASTC to pick the two quadrants.

3.Reference angle: use the positive value to get the Q1 angle.

4.Apply: Q2 $= 180-A$, Q3 $= 180+A$, Q4 $= 360-A$.

YOU TRY · 1

Solve $\cos A = \dfrac{\sqrt{3}}{2}$ for $0° \le A \le 360°$.

Cos positive — which two quadrants? Then find the reference angle.

Cos positive: Q1 and Q4.

Reference: $\cos A = \dfrac{\sqrt{3}}{2} \Rightarrow A = 30°$

Q1: $30°$, Q4: $360 - 30 = 330°$

$A = 30°$ or $330°$

YOU TRY · 2

Solve $\cos \theta = -\dfrac{1}{\sqrt{2}}$ for $0° \le \theta \le 360°$.

Cos negative → Q2 and Q3.

Cos negative: Q2 and Q3.

Reference: $\cos \theta = \dfrac{1}{\sqrt{2}} \Rightarrow \theta = 45°$

Q2: $180 - 45 = 135°$, Q3: $180 + 45 = 225°$

$\theta = 135°$ or $225°$

Section 3 of 6

Basic equations — radians

Same process — but now $A$ is in radians.

$\pi = 180° \qquad 2\pi = 360°$

The quadrant formulas in radians:

(i) Worked example — $\tan A = \sqrt{3}$

Solve $\tan A = \sqrt{3}$ for $0 \le A \le 2\pi$.

Tan positive — from ASTC, that's Q1 and Q3.

Reference angle: $\tan A = \sqrt{3} \quad \Rightarrow \quad A = \dfrac{\pi}{3}$

Q1: $\dfrac{\pi}{3}$

Q3: $\pi + \dfrac{\pi}{3} = \dfrac{3\pi}{3} + \dfrac{\pi}{3} = \dfrac{4\pi}{3}$

$A = \dfrac{\pi}{3}$ or $\dfrac{4\pi}{3}$

Radians — quadrant formulas

1.Q1: $A$, Q2: $\pi - A$, Q3: $\pi + A$, Q4: $2\pi - A$.

2.$\pi = 180°$, one full circle $= 2\pi$.

3.Common reference angles: $30° = \dfrac{\pi}{6}$, $45° = \dfrac{\pi}{4}$, $60° = \dfrac{\pi}{3}$.

YOU TRY · 3

Solve $\tan A = \dfrac{1}{\sqrt{3}}$ for $0 \le A \le 2\pi$.

Tan positive — Q1 and Q3. Reference angle from $\tan A = \dfrac{1}{\sqrt{3}}$.

Tan positive: Q1 and Q3.

Reference: $\tan A = \dfrac{1}{\sqrt{3}} \Rightarrow A = \dfrac{\pi}{6}$

Q1: $\dfrac{\pi}{6}$, Q3: $\pi + \dfrac{\pi}{6} = \dfrac{7\pi}{6}$

$A = \dfrac{\pi}{6}$ or $\dfrac{7\pi}{6}$

Section 4 of 6

General solutions

What if there's no range — what if we want every solution?

After we find the basic Q solutions, we keep adding full circles:

$+\, 360°n$ in degrees

$+\, 2n\pi$ in radians

where $n \in \mathbb{N}$ — natural numbers. Each $n$ is one more lap around the circle.

(i) Worked example — $\sin A = -0.35$ (nearest degree)

Find all values of $A$ to nearest degree such that $\sin A = -0.35$.

Sin negative — from ASTC, Q3 and Q4.

Reference angle: $\sin A = 0.35 \quad \Rightarrow \quad A = \sin^{-1}(0.35) = 20°$

Q3: $180° + 20° = 200°$

Q4: $360° - 20° = 340°$

Now add a full circle as many times as we like:

$A = 200° + 360°n$

$A = 340° + 360°n$

$n = 0$ gives the basic two solutions; $n = 1$ adds one full revolution; $n = 2$ adds two; and so on.

General solutions

1.Find the basic Q solutions in $[0°, 360°)$ or $[0, 2\pi)$ using ASTC.

2.Add $360°n$ (degrees) or $2n\pi$ (radians) to each.

3.$n \in \mathbb{N}$ — each $n$ is one more full revolution.

YOU TRY · 4

Find the general solution of $\tan A = -1$, with $A$ in radians.

Tan negative — Q2 and Q4. Reference $\tan A = 1$.

Tan negative: Q2 and Q4.

Reference: $\tan A = 1 \Rightarrow A = \dfrac{\pi}{4}$

Q2: $\pi - \dfrac{\pi}{4} = \dfrac{3\pi}{4}$, Q4: $2\pi - \dfrac{\pi}{4} = \dfrac{7\pi}{4}$

Add $2n\pi$:

$A = \dfrac{3\pi}{4} + 2n\pi$ or $A = \dfrac{7\pi}{4} + 2n\pi$

Section 5 of 6

Multiple angle — degrees

When the equation has $2A$ or $3A$ instead of $A$, we change the range.

If $0° \le A \le 180°$, then $0° \le 3A \le 540°$ — we need every $3A$ solution in that bigger range.

(i) Worked example — $\sin 3A = \dfrac{1}{2}$

Solve $\sin 3A = \dfrac{1}{2}$ for $0° \le A \le 180°$.

Range for $3A$: $\quad 0° \le 3A \le 540°$ (1.5 full circles)

Sin positive — Q1 and Q2.

Reference angle: $\sin A = \dfrac{1}{2} \quad \Rightarrow \quad A = 30°$

Find all $3A$ values in $0°$ to $540°$:

Q1: $30°$, and $30° + 360° = 390°$

Q2: $180° - 30° = 150°$, and $150° + 360° = 510°$

$3A = 30°, \; 150°, \; 390°, \; 510°$

Divide each by $3$:

$A = 10°, \; 50°, \; 130°, \; 170°$

(ii) Worked example — $\cos 2A = -\dfrac{\sqrt{3}}{2}$

Solve $\cos 2A = -\dfrac{\sqrt{3}}{2}$ for $0° \le A \le 360°$.

Range for $2A$: $\quad 0° \le 2A \le 720°$ (two full circles)

Cos negative — Q2 and Q3.

Reference angle: $\cos A = \dfrac{\sqrt{3}}{2} \quad \Rightarrow \quad A = 30°$

Q2: $180° - 30° = 150°$, $+ 360° = 510°$

Q3: $180° + 30° = 210°$, $+ 360° = 570°$

$2A = 150°, \; 210°, \; 510°, \; 570°$

Divide each by $2$:

$A = 75°, \; 105°, \; 255°, \; 285°$

Multiple angle process

1.Multiply the range by the coefficient ($2$ for $2A$, $3$ for $3A$).

2.Find all solutions in that bigger range — keep adding $360°$ until you exceed it.

3.Divide each solution by the coefficient at the end.

YOU TRY · 5

Solve $\tan 3A = -\dfrac{1}{\sqrt{3}}$ for $0° \le A \le 180°$.

Range for $3A$: $0$ to $540°$. Tan negative — Q2 and Q4.

$0° \le 3A \le 540°$. Tan negative: Q2 and Q4.

Reference: $\tan A = \dfrac{1}{\sqrt{3}} \Rightarrow A = 30°$

Q2: $180 - 30 = 150°$, $+360 = 510°$

Q4: $360 - 30 = 330°$ (also within range)

$3A = 150°, \; 330°, \; 510°$

$A = 50°, \; 110°, \; 170°$

YOU TRY · 6

Solve $\cos 3A = -\dfrac{1}{2}$ for $0° \le A \le 360°$.

Range for $3A$: $0$ to $1080°$. Cos negative — Q2 and Q3.

$0° \le 3A \le 1080°$. Cos negative: Q2, Q3.

Reference: $\cos A = \dfrac{1}{2} \Rightarrow A = 60°$

Q2: $120°, \; 480°, \; 840°$

Q3: $240°, \; 600°, \; 960°$

$3A = 120°, 240°, 480°, 600°, 840°, 960°$

$A = 40°, \; 80°, \; 160°, \; 200°, \; 280°, \; 320°$

Section 6 of 6

Multiple angle — radians

Same idea — multiply the range by the coefficient. Now in radians.

(i) Worked example — $\tan 2A = \sqrt{3}$

Solve $\tan 2A = \sqrt{3}$ for $0 \le A \le 2\pi$.

Range for $2A$: $\quad 0 \le 2A \le 4\pi$ (two full circles)

Tan positive — Q1 and Q3.

Reference angle: $\tan A = \sqrt{3} \quad \Rightarrow \quad A = \dfrac{\pi}{3}$

Q1: $\dfrac{\pi}{3}$, $+ 2\pi = \dfrac{\pi}{3} + \dfrac{6\pi}{3} = \dfrac{7\pi}{3}$

Q3: $\pi + \dfrac{\pi}{3} = \dfrac{4\pi}{3}$, $+ 2\pi = \dfrac{10\pi}{3}$

$2A = \dfrac{\pi}{3}, \; \dfrac{4\pi}{3}, \; \dfrac{7\pi}{3}, \; \dfrac{10\pi}{3}$

Divide each by $2$:

$A = \dfrac{\pi}{6}, \; \dfrac{2\pi}{3}, \; \dfrac{7\pi}{6}, \; \dfrac{5\pi}{3}$

(ii) Worked example — $\cos 3A = -\dfrac{1}{2}$ (general)

Find the general solution of $\cos 3A = -\dfrac{1}{2}$ in radians.

Cos negative — Q2 and Q3.

Reference angle: $\cos A = \dfrac{1}{2} \quad \Rightarrow \quad A = \dfrac{\pi}{3}$

Q2: $\pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}$

Q3: $\pi + \dfrac{\pi}{3} = \dfrac{4\pi}{3}$

General solution for $3A$:

$3A = \dfrac{2\pi}{3} + 2n\pi$ or $3A = \dfrac{4\pi}{3} + 2n\pi$

Divide everything by $3$ (so $2n\pi$ becomes $\dfrac{2n\pi}{3}$):

$A = \dfrac{2\pi}{9} + \dfrac{2n\pi}{3}$

$A = \dfrac{4\pi}{9} + \dfrac{2n\pi}{3}$

Multiple angle — radians

1.Multiply the range by the coefficient ($2\pi$ becomes $4\pi$ for $2A$, $6\pi$ for $3A$).

2.For general solutions, divide $+2n\pi$ by the coefficient too — becomes $+\dfrac{2n\pi}{3}$ for $3A$.

3.Always check: do you have the right number of solutions for the range?

YOU TRY · 7

Solve $\sin 3A = -\dfrac{1}{2}$ for $0 \le A \le 2\pi$.

Range for $3A$: $0$ to $6\pi$. Sin negative — Q3 and Q4.

$0 \le 3A \le 6\pi$. Sin negative: Q3, Q4.

Reference: $\sin A = \dfrac{1}{2} \Rightarrow A = \dfrac{\pi}{6}$

Q3: $\pi + \dfrac{\pi}{6} = \dfrac{7\pi}{6}$, $+2\pi = \dfrac{19\pi}{6}$, $+4\pi = \dfrac{31\pi}{6}$

Q4: $2\pi - \dfrac{\pi}{6} = \dfrac{11\pi}{6}$, $+2\pi = \dfrac{23\pi}{6}$, $+4\pi = \dfrac{35\pi}{6}$

$3A = \dfrac{7\pi}{6}, \; \dfrac{11\pi}{6}, \; \dfrac{19\pi}{6}, \; \dfrac{23\pi}{6}, \; \dfrac{31\pi}{6}, \; \dfrac{35\pi}{6}$

$A = \dfrac{7\pi}{18}, \; \dfrac{11\pi}{18}, \; \dfrac{19\pi}{18}, \; \dfrac{23\pi}{18}, \; \dfrac{31\pi}{18}, \; \dfrac{35\pi}{18}$

SUM

The lot in one box

Trig equations toolkit

1.Unit circle: $\cos A = x$, $\sin A = y$, $\tan A = \dfrac{\sin A}{\cos A}$.

2.ASTC: Q1 All, Q2 Sin, Q3 Tan, Q4 Cos — positive.

3.Ask yourself: sin/cos/tan? And positive or negative?

4.Reference angle = Q1 angle from the positive value.

5.Degrees: Q2 $= 180-A$, Q3 $= 180+A$, Q4 $= 360-A$.

6.Radians: Q2 $= \pi - A$, Q3 $= \pi + A$, Q4 $= 2\pi - A$.

7.General solution: add $360°n$ (deg) or $2n\pi$ (rad) to each basic solution.

8.Multiple angle: multiply the range by the coefficient; divide the final answers (and the $+2n\pi$) by the coefficient.

End of lesson

Trigonometric Equations — HL · Mathslive.ie