Unit 6, lesson 52

Trigonometric Identities from the Unit Circle

How the unit circle gives the Pythagorean, quotient and reciprocal identities, how to use them as replacements, and why every original denominator restriction survives cancellation.

  • Theory part 103
  • Practice part 104
  • Identities and Equations

Calculus often hands you a trigonometric expression that is correct but awkward, such as 1−cos⁡2xsin⁡xcos⁡x\frac{1 - \cos^2 x}{\sin x \cos x}, and the next step only works once it is rewritten. Every honest rewrite comes from one fact about the unit circle, that the point (cos⁡x,sin⁡x)(\cos x, \sin x) lies on it. The skill is to replace pieces on purpose and to keep track of the inputs where the original expression was never defined.

Key terms

Trigonometric identity

Definition A trigonometric identity is an equation between two trigonometric expressions that is true for every real xx in their common domain, the set of inputs where both sides are defined. To prove an identity you show by valid algebra that the two sides agree on that whole set. Checking a few angles can expose a false claim, because one input where the sides differ is enough to refute it, but agreement at a few angles never proves an identity.

In plain words An identity is an equation that is true everywhere both sides make sense, not just for a few lucky angles.

Example The equation sin⁡x+cos⁡x=1\sin x + \cos x = 1 is true at x=0x = 0 and at x=π2x = \frac{\pi}{2}. It is still not an identity, because at x=π4x = \frac{\pi}{4} the left side is

22+22=2≠1.\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \ne 1.

Pythagorean identity

Definition For every real number xx,

sin⁡2x+cos⁡2x=1.\sin^2 x + \cos^2 x = 1.

The reason is that (cos⁡x,sin⁡x)(\cos x, \sin x) is a point on the unit circle u2+v2=1u^2 + v^2 = 1. Dividing by cos⁡2x\cos^2 x, which is allowed only when cos⁡x≠0\cos x \ne 0, and dividing by sin⁡2x\sin^2 x, allowed only when sin⁡x≠0\sin x \ne 0, give

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

One more consequence matters often. For real xx, sin⁡x=±1\sin x = \pm 1 exactly when cos⁡x=0\cos x = 0, because sin⁡2x=1\sin^2 x = 1 is the same as cos⁡2x=0\cos^2 x = 0.

In plain words The two coordinates of a point on the unit circle, squared and added, always give 11. The other two forms are the same fact divided through, and each carries the condition that makes the division legal.

Example Suppose sin⁡x=35\sin x = \frac{3}{5} and π2<x<π\frac{\pi}{2} < x < \pi. Then cos⁡2x=1−925=1625\cos^2 x = 1 - \frac{9}{25} = \frac{16}{25}, and cosine is negative in that interval, so cos⁡x=−45\cos x = -\frac{4}{5}. That gives tan⁡x=−34\tan x = -\frac{3}{4} and sec⁡x=−54\sec x = -\frac{5}{4}, and the second form checks out,

1+(−34)2=2516=(−54)2.1 + \left(-\frac{3}{4}\right)^2 = \frac{25}{16} = \left(-\frac{5}{4}\right)^2.

Quotient identities

Definition For real xx,

tan⁡x=sin⁡xcos⁡xwhen cos⁡x≠0,cot⁡x=cos⁡xsin⁡xwhen sin⁡x≠0.\begin{gathered} \tan x = \frac{\sin x}{\cos x} \quad \text{when } \cos x \ne 0, \\[0.4em] \cot x = \frac{\cos x}{\sin x} \quad \text{when } \sin x \ne 0. \end{gathered}

These come straight from the definitions of tangent and cotangent as ratios of the coordinates of the unit circle point.

In plain words Tangent and cotangent can always be traded for sine and cosine, as long as you remember which denominator must not be zero.

Example At x=5π6x = \frac{5\pi}{6} the point is (−32,12)\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right), so

tan⁡5π6=1/2−3/2=−33,cot⁡5π6=−3.\begin{gathered} \tan \frac{5\pi}{6} = \frac{1/2}{-\sqrt{3}/2} = -\frac{\sqrt{3}}{3}, \\[0.4em] \cot \frac{5\pi}{6} = -\sqrt{3}. \end{gathered}

Reciprocal identities

Definition For real xx,

sec⁡x=1cos⁡xwhen cos⁡x≠0,csc⁡x=1sin⁡xwhen sin⁡x≠0,\begin{gathered} \sec x = \frac{1}{\cos x} \quad \text{when } \cos x \ne 0, \\[0.4em] \csc x = \frac{1}{\sin x} \quad \text{when } \sin x \ne 0, \end{gathered}

and cot⁡x=1tan⁡x\cot x = \frac{1}{\tan x} when both sin⁡x≠0\sin x \ne 0 and cos⁡x≠0\cos x \ne 0. The last one needs both conditions, because at x=π2+kπx = \frac{\pi}{2} + k\pi the cotangent is 00 while tangent is undefined.

In plain words Secant, cosecant and cotangent are one over cosine, sine and tangent. Each holds only where its denominator is a real, nonzero number.

Example For x≠kπx \ne k\pi the product sin⁡xcsc⁡x\sin x \csc x equals sin⁡x⋅1sin⁡x=1\sin x \cdot \frac{1}{\sin x} = 1. At x=πx = \pi the product is undefined, not 11, since csc⁡π\csc \pi does not exist.

Common domain

Definition The common domain of an equation F(x)=G(x)F(x) = G(x) is the set of real xx at which both F(x)F(x) and G(x)G(x) are defined. A trigonometric identity is a claim about that set only. Before any algebra, list every denominator in both sides and every function, such as tangent or secant, that brings its own restriction.

In plain words First decide where the two sides even exist. The identity only promises equality there.

Example Consider

11−sin⁡x+11+sin⁡x=2sec⁡2x.\frac{1}{1 - \sin x} + \frac{1}{1 + \sin x} = 2\sec^2 x.

The left side needs sin⁡x≠1\sin x \ne 1 and sin⁡x≠−1\sin x \ne -1, and the right side needs cos⁡x≠0\cos x \ne 0. Since sin⁡x=±1\sin x = \pm 1 exactly when cos⁡x=0\cos x = 0, both sides need the same thing, so the common domain is x≠π2+kπx \ne \frac{\pi}{2} + k\pi. On that set, a common denominator and the Pythagorean identity give

(1+sin⁡x)+(1−sin⁡x)(1−sin⁡x)(1+sin⁡x)=21−sin⁡2x=2cos⁡2x=2sec⁡2x.\begin{aligned} &\frac{(1 + \sin x) + (1 - \sin x)}{(1 - \sin x)(1 + \sin x)} \\[0.4em] &= \frac{2}{1 - \sin^2 x} = \frac{2}{\cos^2 x} \\[0.4em] &= 2\sec^2 x. \end{aligned}

Replacement

Definition A replacement substitutes for part of an expression another expression that equals it at every input under consideration, using an identity whose conditions hold there. The result equals the original only on the original domain. A replacement can never make the expression defined at an input where the original was undefined.

In plain words Swap a piece for something equal to it, say which identity justifies the swap, and keep the old restrictions.

Example Simplify (sec⁡2x−1)cos⁡2x(\sec^2 x - 1)\cos^2 x. It is defined when cos⁡x≠0\cos x \ne 0, and there 1+tan⁡2x=sec⁡2x1 + \tan^2 x = \sec^2 x lets you replace sec⁡2x−1\sec^2 x - 1 by tan⁡2x\tan^2 x.

(sec⁡2x−1)cos⁡2x=tan⁡2xcos⁡2x=sin⁡2xcos⁡2x⋅cos⁡2x=sin⁡2x,cos⁡x≠0.\begin{aligned} (\sec^2 x - 1)\cos^2 x &= \tan^2 x \cos^2 x \\[0.4em] &= \frac{\sin^2 x}{\cos^2 x} \cdot \cos^2 x \\[0.4em] &= \sin^2 x, \\[0.4em] &\cos x \ne 0. \end{aligned}

The expression sin⁡2x\sin^2 x is defined everywhere, but the original is not, so the condition cos⁡x≠0\cos x \ne 0 stays.

Cancellation

Definition If B(x)≠0B(x) \ne 0 and C(x)≠0C(x) \ne 0, then

A(x) C(x)B(x) C(x)=A(x)B(x).\frac{A(x)\,C(x)}{B(x)\,C(x)} = \frac{A(x)}{B(x)}.

Cancelling the common factor C(x)C(x) is valid only at inputs where it is nonzero and where everything in the original is defined. The simplified expression may be defined at more inputs than the original, but the equality holds only on the original domain. Cancelled factors do not restore excluded inputs.

In plain words You may cross out a factor that appears on top and bottom, but the places where that factor was zero remain forbidden.

Example Simplify the quotient from the opening. The original needs sin⁡x≠0\sin x \ne 0 and cos⁡x≠0\cos x \ne 0, that is, x≠kπ2x \ne \frac{k\pi}{2}. Replace 1−cos⁡2x1 - \cos^2 x by sin⁡2x\sin^2 x and cancel one factor of sin⁡x\sin x.

1−cos⁡2xsin⁡xcos⁡x=sin⁡2xsin⁡xcos⁡x=sin⁡xcos⁡x=tan⁡x,x≠kπ2.\begin{aligned} \frac{1 - \cos^2 x}{\sin x \cos x} &= \frac{\sin^2 x}{\sin x \cos x} \\[0.4em] &= \frac{\sin x}{\cos x} = \tan x, \\[0.4em] &x \ne \frac{k\pi}{2}. \end{aligned}

At x=πx = \pi the tangent is 00, yet the original is undefined there because its denominator sin⁡πcos⁡π\sin \pi \cos \pi is 00. So the simplified form agrees with the original only for x≠kπ2x \ne \frac{k\pi}{2}.

Check yourself

  1. Simplify tan⁡xsec⁡x\frac{\tan x}{\sec x} and state where the result equals the original.
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The original needs cos⁡x≠0\cos x \ne 0, and there

tan⁡xsec⁡x=sin⁡xcos⁡x⋅cos⁡x=sin⁡x.\frac{\tan x}{\sec x} = \frac{\sin x}{\cos x} \cdot \cos x = \sin x.

The equality holds for x≠π2+kπx \ne \frac{\pi}{2} + k\pi. At x=π2x = \frac{\pi}{2} the sine is 11 but the original is undefined.

  1. Verify the identity below and give its common domain.

    11+cos⁡x+11−cos⁡x=2csc⁡2x\frac{1}{1 + \cos x} + \frac{1}{1 - \cos x} = 2\csc^2 x
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The left side needs cos⁡x≠±1\cos x \ne \pm 1, and the right side needs sin⁡x≠0\sin x \ne 0. These are the same condition, because cos⁡x=±1\cos x = \pm 1 exactly when sin⁡x=0\sin x = 0, so the common domain is x≠kπx \ne k\pi. On that set

11+cos⁡x+11−cos⁡x=(1−cos⁡x)+(1+cos⁡x)(1+cos⁡x)(1−cos⁡x)=21−cos⁡2x=2sin⁡2x=2csc⁡2x.\begin{aligned} \frac{1}{1 + \cos x} + \frac{1}{1 - \cos x} &= \frac{(1 - \cos x) + (1 + \cos x)}{(1 + \cos x)(1 - \cos x)} \\[0.4em] &= \frac{2}{1 - \cos^2 x} = \frac{2}{\sin^2 x} = 2\csc^2 x. \end{aligned}

The step from 1−cos⁡2x1 - \cos^2 x to sin⁡2x\sin^2 x is the Pythagorean identity.

  1. Show with one angle that cos⁡x=1−sin⁡x\cos x = 1 - \sin x is not an identity.
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Take x=πx = \pi. The left side is cos⁡π=−1\cos \pi = -1 and the right side is 1−sin⁡π=11 - \sin \pi = 1. The two sides differ, so the equation is not an identity, even though it happens to hold at x=0x = 0 and x=π2x = \frac{\pi}{2}.

How this lesson connects