Unit 5, lesson 58

Fundamental Trigonometric Identities

Why sine squared plus cosine squared is always 1, how the reciprocal, quotient and even and odd identities follow from the unit circle, and why each identity holds only where both sides are defined.

  • Theory part 115
  • Practice part 116
  • Fundamental Identities

Knowing one trigonometric value often pins down all the others. If you know that sin⁡θ=35\sin \theta = \frac{3}{5} and which quadrant θ\theta lies in, the other five values follow without a calculator. The tools for this are a handful of identities, and every one of them comes from the point (cos⁡θ,sin⁡θ)(\cos \theta, \sin \theta) on the unit circle and the symmetry of that circle.

Key terms

Unit-circle derivation

Definition For a real number θ\theta, let P(θ)=(x,y)P(\theta) = (x, y) be the point where the terminal side of the angle of θ\theta radians in standard position meets the unit circle, as in the unit circle lesson. The six trigonometric functions are defined from its coordinates,

cos⁡θ=x,sin⁡θ=y,tan⁡θ=yx,cot⁡θ=xy,sec⁡θ=1x,csc⁡θ=1y,\begin{gathered} \cos \theta = x, \quad \sin \theta = y, \\[0.4em] \tan \theta = \frac{y}{x}, \quad \cot \theta = \frac{x}{y}, \\[0.8em] \sec \theta = \frac{1}{x}, \quad \csc \theta = \frac{1}{y}, \end{gathered}

where tan⁡θ\tan \theta and sec⁡θ\sec \theta require x≠0x \ne 0, and cot⁡θ\cot \theta and csc⁡θ\csc \theta require y≠0y \ne 0. The fundamental identities are derived from these definitions, from the circle equation x2+y2=1x^2 + y^2 = 1, and from the fact that P(−θ)P(-\theta) is the mirror image of P(θ)P(\theta) across the xx-axis.

In plain words Every identity on this page is a fact about one point on the circle, written with trigonometric names. If you forget one, you can rebuild it from the picture.

Example Suppose P(θ)=(−35,45)P(\theta) = \left(-\frac{3}{5}, \frac{4}{5}\right). This point is on the unit circle, since 925+1625=1\frac{9}{25} + \frac{16}{25} = 1. Reading off the coordinates and dividing gives all six values.

cos⁡θ\cos \thetasin⁡θ\sin \thetatan⁡θ\tan \thetacot⁡θ\cot \thetasec⁡θ\sec \thetacsc⁡θ\csc \theta
−35-\frac{3}{5}45\frac{4}{5}−43-\frac{4}{3}−34-\frac{3}{4}−53-\frac{5}{3}54\frac{5}{4}

Identities and their common domain

Definition The common domain of an equation between two expressions in θ\theta is the set of all real θ\theta for which both sides are defined. The equation is an identity if it is true for every θ\theta in its common domain. An identity makes no claim about values outside the common domain. An equation that is true for some values in its common domain but not for all, such as sin⁡θ=12\sin \theta = \frac{1}{2}, is a conditional equation, not an identity.

In plain words Before you call two expressions equal, find where both of them make sense. The identity is a promise only for those inputs.

Example In the equation tan⁡θcos⁡θ=sin⁡θ\tan \theta \cos \theta = \sin \theta the left side is defined only when cos⁡θ≠0\cos \theta \ne 0, and the right side is defined for every θ\theta. So the common domain is all real θ\theta except θ=π2+kπ\theta = \frac{\pi}{2} + k\pi for integers kk. On that set

tan⁡θcos⁡θ=sin⁡θcos⁡θ⋅cos⁡θ=sin⁡θ,\tan \theta \cos \theta = \frac{\sin \theta}{\cos \theta} \cdot \cos \theta = \sin \theta,

so the equation is an identity. At θ=π2\theta = \frac{\pi}{2} the right side equals 11 but the left side is undefined, and the identity says nothing there.

Reciprocal identities

Definition The reciprocal identities are

csc⁡θ=1sin⁡θ,sec⁡θ=1cos⁡θ,cot⁡θ=1tan⁡θ.\begin{gathered} \csc \theta = \frac{1}{\sin \theta}, \\[0.8em] \sec \theta = \frac{1}{\cos \theta}, \\[0.8em] \cot \theta = \frac{1}{\tan \theta}. \end{gathered}

The first holds when sin⁡θ≠0\sin \theta \ne 0, the second when cos⁡θ≠0\cos \theta \ne 0, and the third when both sin⁡θ≠0\sin \theta \ne 0 and cos⁡θ≠0\cos \theta \ne 0, that is when θ\theta is not an integer multiple of π2\frac{\pi}{2}. On the same sets they can be read the other way, as sin⁡θ=1csc⁡θ\sin \theta = \frac{1}{\csc \theta}, cos⁡θ=1sec⁡θ\cos \theta = \frac{1}{\sec \theta} and tan⁡θ=1cot⁡θ\tan \theta = \frac{1}{\cot \theta}. They follow from the definitions, for example cot⁡θ=xy=1÷yx\cot \theta = \frac{x}{y} = 1 \div \frac{y}{x} when x≠0x \ne 0 and y≠0y \ne 0.

In plain words Cosecant, secant and cotangent are one over sine, cosine and tangent. A function and its reciprocal always have the same sign.

Example If sin⁡θ=−27\sin \theta = -\frac{2}{7}, then csc⁡θ=−72\csc \theta = -\frac{7}{2}. Now watch the third identity at θ=π2\theta = \frac{\pi}{2}, where P(θ)=(0,1)P(\theta) = (0, 1). There cot⁡π2=01=0\cot \frac{\pi}{2} = \frac{0}{1} = 0, but tan⁡π2\tan \frac{\pi}{2} is undefined, so 1tan⁡θ\frac{1}{\tan \theta} gives nothing. That is why the identity cot⁡θ=1tan⁡θ\cot \theta = \frac{1}{\tan \theta} excludes π2\frac{\pi}{2} even though cotangent itself is defined there.

Quotient identities

Definition The quotient identities are

tan⁡θ=sin⁡θcos⁡θfor cos⁡θ≠0,cot⁡θ=cos⁡θsin⁡θfor sin⁡θ≠0.\begin{gathered} \tan \theta = \frac{\sin \theta}{\cos \theta} \quad \text{for } \cos \theta \ne 0, \\[0.8em] \cot \theta = \frac{\cos \theta}{\sin \theta} \quad \text{for } \sin \theta \ne 0. \end{gathered}

They follow from tan⁡θ=yx\tan \theta = \frac{y}{x} and cot⁡θ=xy\cot \theta = \frac{x}{y} with x=cos⁡θx = \cos \theta and y=sin⁡θy = \sin \theta.

In plain words Tangent is sine over cosine, and cotangent is cosine over sine. With these and the reciprocal identities, any trigonometric expression can be written with sine and cosine alone, which is often the first step of a simplification.

Example At θ=7π6\theta = \frac{7\pi}{6} the sine is −12-\frac{1}{2} and the cosine is −32-\frac{\sqrt{3}}{2}, so

tan⁡7π6=−1/2−3/2=13=33,cot⁡7π6=−3/2−1/2=3.\begin{gathered} \tan \frac{7\pi}{6} = \frac{-1/2}{-\sqrt{3}/2} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}, \\[0.8em] \cot \frac{7\pi}{6} = \frac{-\sqrt{3}/2}{-1/2} = \sqrt{3}. \end{gathered}

Both are positive, as they should be in Quadrant III, where sine and cosine are both negative.

Pythagorean identities

Definition For every real θ\theta

sin⁡2θ+cos⁡2θ=1,\sin^2 \theta + \cos^2 \theta = 1,

where sin⁡2θ\sin^2 \theta means (sin⁡θ)2(\sin \theta)^2. Two more identities follow,

1+tan⁡2θ=sec⁡2θfor cos⁡θ≠0,1+cot⁡2θ=csc⁡2θfor sin⁡θ≠0.\begin{gathered} 1 + \tan^2 \theta = \sec^2 \theta \quad \text{for } \cos \theta \ne 0, \\[0.4em] 1 + \cot^2 \theta = \csc^2 \theta \quad \text{for } \sin \theta \ne 0. \end{gathered}

The first identity is the circle equation x2+y2=1x^2 + y^2 = 1 at the point P(θ)P(\theta). Dividing it by x2=cos⁡2θx^2 = \cos^2 \theta when x≠0x \ne 0 gives the second, and dividing it by y2=sin⁡2θy^2 = \sin^2 \theta when y≠0y \ne 0 gives the third.

In plain words Drop a perpendicular from P(θ)P(\theta) to the xx-axis. The right triangle you get has legs ∣cos⁡θ∣\lvert \cos \theta \rvert and ∣sin⁡θ∣\lvert \sin \theta \rvert and hypotenuse 11, so the first identity is the Pythagorean theorem, and it stays true when the triangle flattens onto an axis. Knowing one of sine and cosine fixes the size of the other, and the quadrant fixes its sign.

Example Suppose sin⁡θ=35\sin \theta = \frac{3}{5} and θ\theta lies in Quadrant II. Then cos⁡2θ=1−925=1625\cos^2 \theta = 1 - \frac{9}{25} = \frac{16}{25}, so cos⁡θ=±45\cos \theta = \pm\frac{4}{5}. Cosine is negative in Quadrant II, so cos⁡θ=−45\cos \theta = -\frac{4}{5}. The quotient and reciprocal identities then give

tan⁡θ=3/5−4/5=−34,cot⁡θ=−43,sec⁡θ=−54,csc⁡θ=53.\begin{aligned} \tan \theta &= \frac{3/5}{-4/5} = -\frac{3}{4}, \\[0.8em] \cot \theta &= -\frac{4}{3}, \\[0.8em] \sec \theta &= -\frac{5}{4}, \\[0.8em] \csc \theta &= \frac{5}{3}. \end{aligned}

As a check, 1+tan⁡2θ=1+916=25161 + \tan^2 \theta = 1 + \frac{9}{16} = \frac{25}{16}, which is sec⁡2θ\sec^2 \theta.

Parity identities

Definition For every real θ\theta

cos⁡(−θ)=cos⁡θ,sin⁡(−θ)=−sin⁡θ,\begin{gathered} \cos(-\theta) = \cos \theta, \\[0.4em] \sin(-\theta) = -\sin \theta, \end{gathered}

and for every θ\theta where the functions are defined

tan⁡(−θ)=−tan⁡θ,cot⁡(−θ)=−cot⁡θ,sec⁡(−θ)=sec⁡θ,csc⁡(−θ)=−csc⁡θ.\begin{gathered} \tan(-\theta) = -\tan \theta, \\[0.4em] \cot(-\theta) = -\cot \theta, \\[0.4em] \sec(-\theta) = \sec \theta, \\[0.4em] \csc(-\theta) = -\csc \theta. \end{gathered}

Each of these four functions is defined at −θ-\theta exactly when it is defined at θ\theta, so the identities say that cosine and secant are even functions, and sine, tangent, cotangent and cosecant are odd functions. The reason is that the angle −θ-\theta is a turn of the same size in the opposite direction, so P(−θ)P(-\theta) is the reflection of P(θ)=(x,y)P(\theta) = (x, y) across the xx-axis, the point (x,−y)(x, -y).

In plain words Changing the sign of the angle flips the point to the other side of the xx-axis. The first coordinate stays the same and the second changes sign, so cosine and secant do not change, and every function built with the second coordinate changes sign.

Example With θ=π3\theta = \frac{\pi}{3}

sin⁡(−π3)=−32,cos⁡(−π3)=12,tan⁡(−π3)=−3,sec⁡(−π3)=2.\begin{gathered} \sin\left(-\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}, \\[0.8em] \cos\left(-\frac{\pi}{3}\right) = \frac{1}{2}, \\[0.8em] \tan\left(-\frac{\pi}{3}\right) = -\sqrt{3}, \\[0.8em] \sec\left(-\frac{\pi}{3}\right) = 2. \end{gathered}

Parity also simplifies expressions. For every θ\theta with cos⁡θ≠0\cos \theta \ne 0

sin⁡(−θ)cos⁡(−θ)=−sin⁡θcos⁡θ=−tan⁡θ.\frac{\sin(-\theta)}{\cos(-\theta)} = \frac{-\sin \theta}{\cos \theta} = -\tan \theta.

Check yourself

  1. Suppose cos⁡θ=−513\cos \theta = -\frac{5}{13} and θ\theta lies in Quadrant III. Find the other five trigonometric values.
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The Pythagorean identity gives sin⁡2θ=1−25169=144169\sin^2 \theta = 1 - \frac{25}{169} = \frac{144}{169}. Sine is negative in Quadrant III, so sin⁡θ=−1213\sin \theta = -\frac{12}{13}. Then

tan⁡θ=125,cot⁡θ=512,sec⁡θ=−135,csc⁡θ=−1312.\begin{gathered} \tan \theta = \frac{12}{5}, \qquad \cot \theta = \frac{5}{12}, \\[0.8em] \sec \theta = -\frac{13}{5}, \qquad \csc \theta = -\frac{13}{12}. \end{gathered}
  1. Show that (sec⁡2x−1)cot⁡x=tan⁡x(\sec^2 x - 1)\cot x = \tan x on its common domain, and give that domain.
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The left side needs cos⁡x≠0\cos x \ne 0 for the secant and sin⁡x≠0\sin x \ne 0 for the cotangent. The right side needs cos⁡x≠0\cos x \ne 0. So the common domain is all real xx that are not integer multiples of π2\frac{\pi}{2}. On that set sec⁡2x−1=tan⁡2x\sec^2 x - 1 = \tan^2 x and tan⁡xcot⁡x=1\tan x \cot x = 1, so

(sec⁡2x−1)cot⁡x=tan⁡2xcot⁡x=tan⁡x.\begin{aligned} (\sec^2 x - 1)\cot x &= \tan^2 x \cot x \\[0.4em] &= \tan x. \end{aligned}

At x=0x = 0 the right side is 00 while the left side is undefined, which is why 00 is left out.

  1. Simplify cos⁡(−θ)tan⁡(−θ)\cos(-\theta)\tan(-\theta) and say for which θ\theta the result holds.
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By the parity and quotient identities

cos⁡(−θ)tan⁡(−θ)=cos⁡θ⋅(−tan⁡θ)=−cos⁡θ⋅sin⁡θcos⁡θ=−sin⁡θ,\begin{aligned} \cos(-\theta)\tan(-\theta) &= \cos \theta \cdot (-\tan \theta) \\[0.4em] &= -\cos \theta \cdot \frac{\sin \theta}{\cos \theta} \\[0.8em] &= -\sin \theta, \end{aligned}

valid whenever cos⁡θ≠0\cos \theta \ne 0, that is θ≠π2+kπ\theta \ne \frac{\pi}{2} + k\pi for integers kk. At θ=π2\theta = \frac{\pi}{2} the original expression is undefined even though −sin⁡θ=−1-\sin \theta = -1.

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