Maths › Trigonometry › Trigonometric graphs and equations
Trigonometric graphs and equations
Past 90 degrees the triangle definitions give out, and the unit circle takes over: sine and cosine become coordinates, their graphs become waves, and every trig equation becomes a picture. The calculator hands you one answer; the graph tells you how many you actually owe.
Builds on Triangles and the sine and cosine rules.
IN THIS TOPIC
- Sketch and read the sine, cosine and tangent graphs, with exact values at the standard angles.
- Use tan θ = sin θ/cos θ and sin²θ + cos²θ = 1 to reshape and solve equations.
- Find every solution in a stated interval, including for multiple angles and hidden quadratics.
WHAT YOU PROBABLY THINK
To solve sin x cos x = cos x, divide both sides by cos x.
The graphs and the exact values
For angles beyond a triangle's reach, sine and cosine are defined as coordinates: an angle x measured from the positive horizontal axis marks a point on a circle of radius 1, and that point is (cos x, sin x). The definitions agree with the triangle versions below 90° and then keep going, around the circle and around again, which is why both graphs repeat every 360°. Tangent, being sin/cos, repeats every 180° and blows up wherever cos x = 0.
A small stock of exact values covers most non-calculator questions, and all of them come from two triangles, the half-square and the half-equilateral.
| x | sin x | cos x | tan x |
|---|---|---|---|
| 0° | 0 | 1 | 0 |
| 30° | 1/2 | √3/2 | 1/√3 |
| 45° | √2/2 | √2/2 | 1 |
| 60° | √3/2 | 1/2 | √3 |
| 90° | 1 | 0 | — |
Two identities
Two identities are on the must-learn list and do most of the algebraic work in this topic,
the first for converting a sin-and-cos equation into a tan equation, the second, which is the unit circle's Pythagoras, for trading sin² for cos² and back.
WORKED EXAMPLE
The equation you must not divide
Solve sin x cos x = cos x for 0 ≤ x < 360°.
Resist the opening lie. Gather and factorise instead: cos x (sin x − 1) = 0.
cos x = 0 gives x = 90° and 270°; sin x = 1 gives x = 90° again.
Solutions: x = 90° and 270°.
Dividing both sides by cos x would have thrown 270° away, because at 270° you would have divided by zero. Factorising never loses a solution; dividing by an expression that can vanish always might.
Every solution in the interval
A calculator's inverse button returns one principal value; the graph then supplies the rest. For sine, a second solution sits at 180° minus the principal value, and adding or subtracting full periods generates the others. The reliable routine for a transformed angle is to substitute θ for the whole bracket, solve for θ across the widened interval, and translate back at the end.
WORKED EXAMPLE
A shifted angle
Solve sin (x + 70°) = 0.5 for 0 < x < 360°.
Let θ = x + 70°. As x runs over its interval, θ runs from 70° to 430°.
sin θ = 0.5 at θ = 30°, 150°, 390°, … and inside (70°, 430°) that means θ = 150° or 390°.
Translate back: x = 80° or 320°.
The 30° root fell outside the widened interval and was discarded; the 390° root, one full wave later, survived and became 320°. Widening the interval first is what keeps both decisions honest.
YOUR TURN
A doubled angle
Solve 3 + 5 cos 2x = 1 for −180° < x < 180°, before opening the working.
Show the working
Rearranging gives cos 2x = −2/5. Let θ = 2x, so θ runs over (−360°, 360°).
The principal value is cos−1(−0.4) = 113.6°, and cosine's symmetry gives θ = ±113.6° and ±246.4°.
Halving: x = ±56.8° and ±123.2°, four solutions.
Doubling the angle doubled the solution count, because 2x sweeps two full waves while x sweeps one. Losing half the answers by solving in the x-interval is the standard casualty here.
TRY IT UNSEEN
A quadratic in disguise, trig edition
Solve 6 cos2 x + sin x − 5 = 0 for 0 ≤ x < 360°.
Show the working
Trade cos² for sin²: 6(1 − sin2 x) + sin x − 5 = 0, so 6 sin2 x − sin x − 1 = 0.
Factorise with s = sin x: (3s + 1)(2s − 1) = 0, so sin x = 1/2 or −1/3.
sin x = 1/2 gives x = 30°, 150°. sin x = −1/3 gives a principal value of −19.5°, landing in range as x = 199.5° and 340.5°.
Four solutions: 30°, 150°, 199.5°, 340.5°. The identity turned two trig ratios into one, and the algebra lesson's substitution habit did the rest.
THE EXAM BIT
- Sketch the relevant graph for the widened interval before collecting solutions; count the crossings first, then find them.
- For sin, the partner solution is 180° − θ; for cos it is −θ (or 360° − θ); for tan, add 180°. Quote the rule you are using.
- Substitute θ for the whole bracket, widen the interval to match, solve, translate back. Marks are lost at the widening step more than anywhere else.
- Never divide by cos x, sin x or any expression that can be zero; move everything to one side and factorise.
- Exact-value questions expect the table quoted, not decimals: sin 60° = √3/2, not 0.866.
CHECK YOURSELF
Solve tan x = √3 for 0 ≤ x < 360°, giving exact answers.
Show a hint
One exact value, then tangent's period.
Show the answer
The exact-value table gives the principal value x = 60°.
Tangent repeats every 180°, so the other solution in range is 60° + 180° = 240°.
Solutions: x = 60° and 240°, and the graph shows exactly two branches crossing √3 in the interval.
The unit circle defines sin and cos for every angle; the graphs repeat and the table of exact values covers the rest.
Widen the interval with the substituted angle, harvest every crossing, then translate back.
CHECK YOUR PROGRESS
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- Sketch and read the sine, cosine and tangent graphs, with exact values at the standard angles.
- Use tan θ = sin θ/cos θ and sin²θ + cos²θ = 1 to reshape and solve equations.
- Find every solution in a stated interval, including for multiple angles and hidden quadratics.
No animated video for this topic yet; these notes stand alone.