Trigonometry

Trigonometry Solver with Step-by-Step Answers

Solve triangles, prove identities, and find every solution in the interval you were given — not just the one your calculator returns.

Solve a trigonometry problem now

First solution free

Trigonometry punishes two habits harder than any other topic. The first is trusting the inverse function: $sin^{- 1} (0.5)$ returns $3 0^{\circ}$ and nothing else, while the equation $sin x = 0.5$ has infinitely many answers. The second is treating identities as things to recall rather than things to derive. Every solution here states which identity was used and why that side of the equation was chosen as the starting point.

Give the interval with the equation — for 0 <= x < 2pi or between 0 and 360 degrees — and the full solution set comes back rather than a principal value. Triangle questions can be typed as a list of the parts you know.

What this solver handles

Right-triangle ratios, angles of elevation and depression, and bearings.
The law of sines and the law of cosines, with the ambiguous two-sides-and-an-angle case handled explicitly.
Exact values for the standard angles, derived from the unit circle rather than recalled from a table.
Trig equations on a stated interval, including ones that factor as quadratics in sine or cosine.
Identity proofs using the Pythagorean, double angle, and sum formulas.
Graphs: amplitude, period, phase shift, and vertical translation read straight from the equation.
Inverse trig functions and their restricted ranges, which is where most sign confusion starts.
Radian and degree conversion, arc length, and sector area.

Two details change the answer more than anything else in a trig question: the interval and the angle unit. Give both. For a triangle, say which letters label which parts — lower-case $a$ opposite upper-case $A$ is the usual convention, but not every textbook follows it, and a mislabelled diagram produces a confident answer to a different triangle. For an identity proof, name the side you are required to start from if your teacher insists on one.

Three problems, worked

Worked example

2 sin x - 1 = 0, 0 \leq x < 2 π

1
Isolate the trig function first. Treat sin x as a single object, the way you would treat any variable.
$sin x = \frac{1}{2}$
2
The reference angle is the acute angle whose sine is one half.
$x_{ref} = \frac{π}{6}$
3
Sine is positive in the first and second quadrants, so there are two solutions in one revolution. The second is pi minus the reference angle.
$x = \frac{π}{6} and x = π - \frac{π}{6} = \frac{5 π}{6}$
4
Both lie inside the stated interval, so both are kept. Outside that interval you would add 2 pi n to each.
$x = \frac{π}{6} + 2 π n, x = \frac{5 π}{6} + 2 π n$

Answer

x = \frac{π}{6}, \frac{5 π}{6}

Worked example

a = 7, b = 9, C = 4 0^{\circ} — solve the triangle

1
Two sides with the angle between them. The law of sines cannot start, because every ratio it offers contains two unknowns, so use the law of cosines.
$c^{2} = a^{2} + b^{2} - 2 ab cos C$
2
Substitute. The cosine of 40 degrees is 0.7660, so the correction term is 126 times that.
$c^{2} = 49 + 81 - 126 (0.7660) = 33.48$
3
Take the positive root — a side length cannot be negative.
$c = 5.79$
4
Now use the law of sines for a second angle. Choose A rather than B, because a is the shorter of the two known sides, so A must be acute and the inverse sine is safe.
$\frac{sin A}{7} = \frac{sin 4 0 ^{\circ}}{5.79} ⟹ sin A = 0.7777$
5
Take the inverse sine, then subtract from 180 to get the last angle.
$A = 51. 0^{\circ}, B = 18 0^{\circ} - 4 0^{\circ} - 51. 0^{\circ} = 89. 0^{\circ}$
6
Check against the largest-side rule: b is the longest side and B is the largest angle, which is consistent.

Answer

c \approx 5.79, A \approx 51. 0^{\circ}, B \approx 89. 0^{\circ}

Worked example

2 cos^{2} x + cos x - 1 = 0, 0 \leq x < 2 π

1
This is a quadratic in cos x. Substituting a temporary letter makes that visible if it helps.
$2 u^{2} + u - 1 = 0 where u = cos x$
2
Factor as you would any quadratic with leading coefficient 2.
$(2 u - 1) (u + 1) = 0$
3
Set each factor to zero and translate back to cosine.
$cos x = \frac{1}{2} or cos x = - 1$
4
Cosine is positive in the first and fourth quadrants, so the first case gives two angles symmetric about the x-axis.
$x = \frac{π}{3}, \frac{5 π}{3}$
5
Cosine equals −1 at exactly one point in the interval, at the far left of the unit circle.
$x = π$
6
Three solutions in total. Note that cos x = 3/2 would have been rejected here — cosine never leaves the interval from −1 to 1, and discarding impossible roots is part of the method.

Answer

x = \frac{π}{3}, π, \frac{5 π}{3}

Before checking any individual answer, count how many there should be. Across one revolution $sin x = k$ has two solutions for any $k$ strictly between −1 and 1, exactly one at either extreme, and none beyond them. Cosine behaves the same way. The third problem therefore had to yield three: two from $cos x = \frac{1}{2}$ and one from $cos x = - 1$ , which sits at the single leftmost point of the circle.

The count changes when the argument does. $sin 2 x = \frac{1}{2}$ on the same interval has four solutions, because as $x$ runs from 0 to $2 π$ the angle $2 x$ travels twice around the circle. Multiply the interval by the coefficient before you start, solve for $2 x$ there, and divide at the end. Getting the expected number first turns the substitution check into a formality rather than a discovery.

Read the values off the circle

Almost every exact value in a first trigonometry course comes from two triangles: half an equilateral triangle, which gives the 30 and 60 degree values, and half a square, which gives 45. Everything else is one of those values with a sign attached by the quadrant. Click through the angles below and watch the coordinates repeat with different signs.

Exact values, one click each

Select an angle to see its exact coordinates.

30°

π/6 rad

cos θ: √3/2
sin θ: 1/2

The coordinate pair is $(cos θ, sin θ)$ , in that order. Reading the pair the wrong way round is common enough that it is worth saying out loud the first few times: cosine is the horizontal one.

Choosing the right rule for a triangle

Which rule solves this triangle?

Does the triangle contain a right angle?

Where students go wrong

Stopping at the principal value

The inverse sine key returns one angle from a restricted range, because a function may only return one value. The equation has more. For a positive sine, add the second-quadrant partner $18 0^{\circ} - θ$ ; for a positive cosine, add the fourth-quadrant partner $36 0^{\circ} - θ$ ; for tangent, add $18 0^{\circ}$ .

Working in the wrong angle mode

An answer of $0.0175$ where you expected roughly 1 means the calculator was in radians and the question was in degrees. The two agree only at zero. Set the mode from the question — a question containing $π$ is in radians, a question containing a degree symbol is not — and check it again after anyone else borrows your calculator.

Cancelling a trig function off both sides

Dividing $2 sin x cos x = sin x$ by $sin x$ throws away every solution where the sine is zero. Factor instead. The same rule that applies to $x$ in algebra applies here: you may only divide by something you have shown is non-zero.

Assuming the law of sines gives one triangle

With two sides and a non-included angle, the inverse sine hands you the acute option. Its supplement has the same sine. If the supplement plus the known angle is still under 180 degrees, a genuine second triangle exists and a complete answer names both.

Which line loses two solutions?

One of these lines is wrong. Click it.

Three of those four end in the same place: an answer that is correct and incomplete. That is what makes them dangerous, because nothing on the page looks wrong and rechecking the arithmetic finds nothing. The defence is the graph. A horizontal line drawn across a sine curve visibly crosses it twice in every period, which is the picture behind the guide to function transformations and the reason a sketch belongs in the margin of any trig equation.

Formulas worth knowing cold

Trigonometry reference

Tap any formula with a derivation to see where it comes from.

Identities

Solving triangles

Exact values

Practice

Four to try

Answers are checked here — nothing is sent anywhere.

1
$Solve sin θ = \frac{3}{2}, 0^{\circ} \leq θ < 36 0^{\circ}$
2
$Right triangle: opposite 5, hypotenuse 13. sin θ = ?$
3
$Exact value of cos \frac{2 π}{3}$
4
$Sides 5 and 8 with 6 0^{\circ} between them. Third side = ?$

Frequently asked questions

Is the trigonometry solver free?

The first solution costs nothing and needs no account. A free account then allows three solutions a day. Gauth Plus is $11.99 a month with a three-day free trial if you need more during an exam period.

Does it show the steps or just the answer?

Solutions name the identity or rule at each line — Pythagorean identity applied, law of sines used because two angles were known, second solution added from the symmetry of the sine curve.

Will it give every solution in the interval, not just one?

Yes, provided you state the interval. Say zero to two pi, or zero to 360 degrees, and the answer lists every solution in that range plus the general form with the period attached.

Does it work in degrees or radians?

Both. If the question uses a degree symbol the answer stays in degrees; if it uses pi the answer stays in radians. State the mode explicitly when the question is ambiguous.

Can it prove identities rather than solve equations?

Yes. Identity proofs start from the messier side, state which substitution is being made at each line, and stop when both sides match. Ask for a specific starting side if your teacher requires one.

Does it handle the ambiguous case for the law of sines?

It does. When two sides and a non-included angle are given, the solution checks whether a second triangle exists and reports both, or explains why the second is impossible.

Keep going

The unit circle you actually needDerive the exact values instead of memorising sixteen of them.The trig identities worth memorisingWhich to know cold, which to rebuild on the spot during a proof.Geometry solverFor the diagram problems that turn into triangles.Domain and range calculatorWhy the inverse trig functions have the ranges they do.