Radians and Trigonometric Identities

Updated 7 October 2026

Past triangles, trigonometry changes in two ways: angles are measured in radians instead of degrees, and identities let one trigonometric expression be rewritten as another. The second is what makes trigonometric equations solvable and many integrals possible.

What is a radian, and why bother?

A radian is the angle at the center of a circle that cuts off an arc as long as the radius. The circumference is $2\pi r$2πr, so a full turn is $2\pi$2π radians, a half turn is $\pi$π radians, and one radian is about 57°. To change degrees to radians, multiply by $\frac{\pi }{180}$π/180. For 135°, the fraction $\frac{135}{180}$135/180 simplifies to $\frac{3}{4}$3/4, so 135° is $\frac{3\pi }{4}$3π/4 radians; the fraction simplifies because 135 ÷ 45 = 3 and 180 ÷ 45 = 4. Radians are worth the change because they make an angle a pure ratio, arc length over radius, with no unit chosen by convention, and calculus depends on that: the derivative of sin x is cos x only when x is in radians. Going the other way reverses the factor. Multiply radians by $\frac{180}{\pi }$180/π, so $\frac{\pi }{6}$π/6 becomes $\frac{180}{6}=30$180/6 = 30 degrees. Check a conversion against an angle you know: $\frac{\pi }{2}$π/2 should give a right angle, and 180 ÷ 2 = 90.

A right angle is $\frac{\pi }{2}$π/2 radians, and to one decimal place one radian is 57.3°. See radians.

In degrees a factor of $\frac{\pi }{180}$π/180 appears at every differentiation, so calculus works in radians throughout.

The familiar angles keep their exact ratios under the new names: 30° is $\frac{\pi }{6}$π/6, 45° is $\frac{\pi }{4}$π/4, and 60° is $\frac{\pi }{3}$π/3. See exact trigonometric values in radians.

How do you find arc length and sector area?

In radians, a sector with radius r and angle $\theta$θ has arc length $s=r\theta$s = rθ and area $A=\frac{1}{2}{r}^2\theta$A = ½r²θ. Take r = 4 and $\theta =\frac{3}{2}$θ = 3/2: the arc is $4\times \frac{3}{2}=6$4 × 3/2 = 6 units long, and the area is $\frac{1}{2}\times 16\times \frac{3}{2}=12$½ × 16 × 3/2 = 12 square units. Both formulas take a fraction of the whole circle. The angle is $\frac{\theta }{\left(2\pi \right)}$θ/(2π) of a full turn, so the arc is that fraction of the circumference $2\pi r$2πr and the sector is that fraction of the area $\pi r^2$πr²; the $2\pi$2π cancels to leave $r\theta$rθ and $\frac{1}{2}{r}^2\theta$½r²θ. Given the arc and the radius, divide to find the angle: an arc of 10 on a radius of 5 makes 10 ÷ 5 = 2 radians. A segment, between the arc and its chord, needs one more step: subtract the triangle, leaving $\frac{1}{2}{r}^2\theta -\frac{1}{2}{r}^2\sin \theta$½r²θ − ½r² sin θ. Check on a half circle, where $\theta =\pi$θ = π: the arc is $\pi r$πr, half of $2\pi r$2πr.

A full turn, $\theta =2\pi$θ = 2π, gives back the whole circumference and the whole area.

A sector of radius 6 with angle $\frac{\pi }{3}$π/3 has arc length $2\pi$2π and area $6\pi$6π.

In degrees both results carry a factor of $\frac{\theta }{360}$θ/360 through every line. See arc length and sector area.

What is a trigonometric identity?

A trigonometric identity is an equation true for every angle, unlike $\sin x=\frac{1}{2}$sin x = 1/2, which holds only for particular angles. Two identities carry most of the others: $\tan \theta =\sin \theta /\cos \theta$tan θ = sin θ / cos θ, and ${\sin }^2\theta +{\cos }^2\theta =1$sin²θ + cos²θ = 1. At 30°, $\sin 30^{\circ }=\frac{1}{2}$sin 30° = 1/2 and $\cos 30^{\circ }=\frac{\sqrt{3}}{2}$cos 30° = √3/2, so the squares are $\frac{1}{4}$1/4 and $\frac{3}{4}$3/4, and $\frac{1}{4}+\frac{3}{4}=1$1/4 + 3/4 = 1. The second identity is Pythagoras on the unit circle: the point at angle $\theta$θ is $\left(\cos \theta ,\sin \theta \right)$(cos θ, sin θ), at distance 1 from the center, and squares are never negative, so it holds in every quadrant. To prove a new identity, start from one side and rewrite it until it becomes the other. When a line holds ${\sin }^2\theta$sin²θ but everything else is in $\cos \theta$cos θ, replace ${\sin }^2\theta$sin²θ with $1-{\cos }^2\theta$1 − cos²θ so that one function remains. Check with an angle: tan 45° = 1, and sin 45° ÷ cos 45° is $\left(\frac{\sqrt{2}}{2}\right)÷\left(\frac{\sqrt{2}}{2}\right)$(√2/2) ÷ (√2/2), also 1.

Any point on the circle of radius 1 shows it.

Once ${\sin }^{2}x$sin²x is replaced, an equation in ${\sin }^{2}x$sin²x and cos x is a quadratic in cos x. See trigonometric identities.

What are secant, cosecant and cotangent?

Secant, cosecant and cotangent are the reciprocals of cosine, sine and tangent: $\sec \theta =1/\cos \theta$sec θ = 1 / cos θ, $\mathrm{cosec}\theta =1/\sin \theta$cosec θ = 1 / sin θ, and $\cot \theta =1/\tan \theta$cot θ = 1 / tan θ, which is $\cos \theta /\sin \theta$cos θ / sin θ. Pair them by the third letter: sec goes with cos, and cosec with sin. To evaluate one, find the ordinary ratio and turn it upside down. Since $\sin 30^{\circ }=\frac{1}{2}$sin 30° = 1/2, cosec 30° = 2, and since $\tan 30^{\circ }=\frac{1}{\sqrt{3}}$tan 30° = 1/√3, $\cot 30^{\circ }=\sqrt{3}$cot 30° = √3. The names earn their place because many expressions divide by a ratio, and one symbol for that quotient shortens identities and derivatives. Each reciprocal is undefined where its ratio is zero, so sec 90° does not exist, because cos 90° = 0, and neither does cot 0°. Check a value by multiplying the pair: sec 60° × cos 60° should be 1, and $2\times \frac{1}{2}=1$2 × 1/2 = 1.

Divide ${\sin }^2\theta +{\cos }^2\theta =1$sin²θ + cos²θ = 1 through by ${\cos }^2\theta$cos²θ and a second identity appears in one step.

Dividing by ${\sin }^2\theta$sin²θ instead gives $1+{\cot }^2\theta ={\mathrm{cosec}}^2\theta$1 + cot²θ = cosec²θ. Derive them rather than memorizing them. The pair matters in calculus, where the derivative of tan x is ${\sec }^{2}x$sec²x, so an integral containing $1+{\tan }^{2}x$1 + tan²x is done once it is rewritten as ${\sec }^{2}x$sec²x. See secant, cosecant and cotangent and the identity $1+{\tan }^2\theta ={\sec }^2\theta$1 + tan²θ = sec²θ.

Where do the compound angle formulas come from?

The compound angle formulas expand the sine and cosine of a sum: sin(A + B) = sin A cos B + cos A sin B, and cos(A + B) = cos A cos B − sin A sin B. The guess sin A + sin B fails: with A = B = 30°, $\sin 60^{\circ }=\frac{\sqrt{3}}{2}$sin 60° = √3/2, but the guess gives $\frac{1}{2}+\frac{1}{2}=1$1/2 + 1/2 = 1. The true formula gives $\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}$√3/4 + √3/4, which is $\frac{\sqrt{3}}{2}$√3/2. They come from stacking a right triangle with angle B on one with angle A: the far corner's height splits into sin A cos B and cos A sin B, and the horizontal distance gives the cosine formula. For a difference, replace B with −B, which flips the middle signs, and 15° = 45° − 30° then gives exact values. Check the cosine formula at A = B = 45°: it gives $\frac{1}{2}-\frac{1}{2}$1/2 − 1/2, which is 0, and cos 90° = 0.

The three sum forms are below; writing −B for B gives each difference form.

Sine keeps the plus sign and cosine takes a minus.

That single difference accounts for most errors here. See the compound angle formulas.

What are the double angle formulas for?

The double angle formulas rewrite a function of 2A in terms of A: sin 2A = 2 sin A cos A, and $\cos 2A={\cos }^{2}A-{\sin }^{2}A$cos 2A = cos²A − sin²A. They are the compound formulas with B = A. Their use is to put every term of an equation in the same angle. To solve sin 2x = sin x for $0^{\circ }\le x\le 360^{\circ }$0° ≤ x ≤ 360°, write 2 sin x cos x = sin x, move everything to one side and factor: sin x (2 cos x − 1) = 0. So sin x = 0, giving 0°, 180° and 360°, or $\cos x=\frac{1}{2}$cos x = 1/2, giving 60° and 300°. Factor rather than divide, because dividing by sin x would lose the first three solutions. With cos 2x, pick the cosine form that matches the other terms. Check x = 60°: sin 120° and sin 60° are both $\frac{\sqrt{3}}{2}$√3/2, since 180 − 60 = 120 makes the angles supplementary.

The cosine version has three forms, because ${\sin }^{2}A+{\cos }^{2}A=1$sin²A + cos²A = 1 removes either squared term: $\cos 2A={\cos }^{2}A-{\sin }^{2}A=2{\cos }^{2}A-1=1-2{\sin }^{2}A$cos 2A = cos²A − sin²A = 2cos²A − 1 = 1 − 2sin²A.

Rearranging the last form isolates ${\sin }^{2}A$sin²A, and the same move on $2{\cos }^{2}A-1$2cos²A − 1 isolates ${\cos }^{2}A$cos²A.

Those results, ${\cos }^{2}A=$cos²A = ½(1 + cos 2A) and ${\sin }^{2}A=$sin²A = ½(1 − cos 2A), turn a squared trigonometric function into a plain one, which is how ${\sin }^{2}x$sin²x and ${\cos }^{2}x$cos²x are integrated. See the double angle formulas.

How do you combine a sine and a cosine into one wave?

Any a sin x + b cos x can be written as one wave, $R\sin \left(x+\alpha \right)$R sin(x + α), with R > 0. Expand it with the compound formula and match coefficients: $R\cos \alpha =a$R cos α = a and $R\sin \alpha =b$R sin α = b. Squaring and adding gives $R=\sqrt{a^2+b^2}$R = √(a² + b²), and dividing gives $\tan \alpha =\frac{b}{a}$tan α = b/a. For $\sqrt{3}\sin x+\cos x$√3 sin x + cos x, $R=\sqrt{3+1}=2$R = √(3 + 1) = 2 and $\tan \alpha =\frac{1}{\sqrt{3}}$tan α = 1/√3, giving 2 sin(x + 30°). It works because $R\cos \alpha$R cos α and $R\sin \alpha$R sin α are the legs of a right triangle with hypotenuse R. When a or b is negative, find the quadrant of $\alpha$α from the signs of $\cos \alpha$cos α and $\sin \alpha$sin α, not from the inverse tangent alone. The greatest value is then R and the least is −R. Check at x = 0°: the original gives cos 0° = 1, and 2 sin 30° is $2\times \frac{1}{2}=1$2 × 1/2 = 1.

3 sin x + 4 cos x = 5 sin(x + 53.1°), because $R=\sqrt{9+16}=5$R = √(9 + 16) = 5 and $\tan \alpha =\frac{4}{3}$tan α = 4/3.

To solve an equation, divide by R, solve $\sin \left(x+\alpha \right)=\frac{c}{R}$sin(x + α) = c/R for the bracket, then subtract $\alpha$α from each answer. The bracket carries the range with it, so x + 53.1° runs from 53.1° when x runs from 0°. See the harmonic form $R\sin \left(x+\alpha \right)$R sin(x + α) and solving a sin x + b cos x = c with the R form.

How do you write every solution, not just the ones in range?

A trigonometric equation has infinitely many solutions, because each function repeats: sine and cosine every 360°, or $2\pi$2π radians, and tangent every 180°. The general solution lists them with a whole number n. For $\sin x=\frac{1}{2}$sin x = 1/2, the calculator gives 30°, the symmetry of the sine curve gives 180° − 30°, which is 150°, and the full list is x = 30° + 360°n or x = 150° + 360°n. Adding a whole turn returns a point on the unit circle to the same place, so its sine cannot change. Tangent needs one family, $x=\theta +180^{\circ }n$x = θ + 180°n, because it repeats after half a turn; for tan x = 1 that is 45° + 180°n. In radians the sine lists read $\frac{\pi }{6}+2n\pi$π/6 + 2nπ and $\frac{5\pi }{6}+2n\pi$5π/6 + 2nπ. Test a member: n = 1 in the first family gives 30 + 360 = 390, a full turn past 30°, where the sine is again $\frac{1}{2}$1/2.

The sine pair is often written together as $180^{\circ }n+{\left(-1\right)}^n\theta$180°n + (−1)ⁿθ. Cosine's answers are $\theta$θ and $-\theta$−θ, so its list is $\pm \theta +360^{\circ }n$±θ + 360°n.

An interval question is a filter on that list. Substitute n = −1, 0, 1, 2 in turn and keep the answers that land in range. See every solution of a trigonometric equation.

The mistakes worth naming

• Leaving the calculator in degrees. Every formula above assumes radians, so the answers come out wrong.
• Expanding sin(A + B) as sin A + sin B. No trigonometric function distributes over addition.
• Losing the minus sign in cos(A + B). Sine keeps the plus and cosine takes the minus.
• Pairing sec with sin. Secant is $1/\cos \theta$1 / cos θ and cosecant is $1/\sin \theta$1 / sin θ.
• Stopping at the first solution. The question states which solutions to report.
• Rearranging both sides of an identity at once. A proof starts on one side and walks it to the other.

See proving a trigonometric identity.

The triangle groundwork is in triangle trigonometry and bearings, and these identities are used again in the rules of differentiation and techniques of integration.