Radians and Trigonometric Identities

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$2π radii, so a full turn is $2\pi$2π radians, a half turn is $\pi$π radians, and a right angle is $\frac{\pi }{2}$π/2. One radian is about 57.3°. See radians.

The unit matters because the derivative of sin x is cos x only when x is in 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?

Measured in radians, the arc of a sector has length $s=r\theta$s = rθ and the sector has area ½$r^2\theta$r²θ. Both come from taking one fraction of the whole circle.

The arc is that fraction of the circumference $2\pi r$2πr, and the $2\pi$2π cancels to leave $r\theta$rθ. The area follows the same way.

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?

An identity is an equation true for every angle, not one to solve for particular angles. Two of them carry the rest: $\tan \theta =\sin \theta /\cos \theta$tan θ = sin θ / cos θ, and ${\sin }^2\theta +{\cos }^2\theta =1$sin²θ + cos²θ = 1.

The second is Pythagoras written in trigonometric notation. Draw a circle of radius 1 and take any point on it.

The everyday use is substitution. An equation containing both ${\sin }^{2}x$sin²x and cos x cannot be solved as it stands, but replacing ${\sin }^{2}x$sin²x with $1-{\cos }^{2}x$1 − cos²x turns it into a quadratic in cos x. See trigonometric identities.

What are secant, cosecant and cotangent?

They are the reciprocals of cosine, sine and tangent: $\sec \theta =\frac{1}{\cos }\theta$sec θ = 1/cos θ, $\mathrm{cosec}\theta =\frac{1}{\sin }\theta$cosec θ = 1/sin θ, and $\cot \theta =\frac{1}{\tan }\theta$cot θ = 1/tan θ.

Any value you know gives its reciprocal at once, so cos 60° = ½ makes sec 60° = 2. 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?

They give the sine and cosine of a sum of two angles. Begin with the natural guess and test it on a pair of angles.

The correct expansions are these.

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?

They turn an expression in 2A into one in A, so that a single-angle method applies. Setting B = A in the compound formulas produces them.

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. Use the form containing whichever function the rest of the problem uses.

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?

Write a sin x + b cos x as one shifted sine wave, $R\sin \left(x+\alpha \right)$R sin(x + α). Expand $R\sin \left(x+\alpha \right)$R sin(x + α) with the compound formula and match coefficients: a must be $R\cos \alpha$R cos α and b must be $R\sin \alpha$R sin α.

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.

One wave has a greatest value of R and a least value of −R. 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 repeats its answers every full turn, so the general solution uses an integer n to list all of them at once.

Sine has two answers in each turn, at $\theta$θ and at $180^{\circ }-\theta$180° − θ, and both repeat every 360°; the 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 $\frac{1}{\cos }\theta$1/cos θ and cosecant is $\frac{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.

Learn this properly in the app

Every lesson linked above is an illustrated screen in Math Challenge with a diagram, a worked example and questions to try. The triangle groundwork is in triangle trigonometry and bearings, and these identities are used again in the rules of differentiation and techniques of integration.