Trigonometry Beyond Right Triangles · Lesson 3

Identities you can use

Three identities carry most questions. The Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 comes from the unit circle: the point (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta) satisfies x2+y2=1x^2 + y^2 = 1. The quotient identity is tan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta}. The cofunction identities are sin⁡(90∘−θ)=cos⁡θ\sin(90^\circ - \theta) = \cos\theta and cos⁡(90∘−θ)=sin⁡θ\cos(90^\circ - \theta) = \sin\theta: the two acute angles of a right triangle swap opposite and adjacent sides.

The reciprocal functions are csc⁡θ=1sin⁡θ\csc\theta = \frac{1}{\sin\theta}, sec⁡θ=1cos⁡θ\sec\theta = \frac{1}{\cos\theta}, and cot⁡θ=1tan⁡θ=cos⁡θsin⁡θ\cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta}. The names cross over: cosecant goes with sine, and secant goes with cosine.

Identities and the forms you will use
IdentityUseful rearrangements
sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 11−cos⁡2θ=sin⁡2θ1 - \cos^2\theta = \sin^2\theta and 1−sin⁡2θ=cos⁡2θ1 - \sin^2\theta = \cos^2\theta
tan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta}sin⁡θ=tan⁡θcos⁡θ\sin\theta = \tan\theta \cos\theta
sin⁡(90∘−θ)=cos⁡θ\sin(90^\circ - \theta) = \cos\thetasin⁡70∘=cos⁡20∘\sin 70^\circ = \cos 20^\circ

To simplify, rewrite everything in sines and cosines, replace 1−cos⁡2x1 - \cos^2 x or 1−sin⁡2x1 - \sin^2 x wherever you see them, and cancel only common factors. A quick test catches mistakes: put in x=30∘x = 30^\circ and check that the expression and your answer give the same number.

Worked example Easy

For all values of xx where sin⁡x≠0\sin x \neq 0, which of the following is equivalent to 1−cos⁡2xsin⁡x\frac{1 - \cos^2 x}{\sin x}?

  1. −sin⁡x-\sin x
  2. sin⁡x\sin xAnswer
  3. csc⁡x−cos⁡2x\csc x - \cos^2 x
  4. csc⁡x−cos⁡x\csc x - \cos x

How to solve it

  1. Rearrange the Pythagorean identity: 1−cos⁡2x=sin⁡2x1 - \cos^2 x = \sin^2 x.
  2. So the expression is sin⁡2xsin⁡x\frac{\sin^2 x}{\sin x}.
  3. Cancel one factor of sin⁡x\sin x: the result is sin⁡x\sin x.
  4. Check at x=30∘x = 30^\circ: 1−3412=12=sin⁡30∘\frac{1 - \frac{3}{4}}{\frac{1}{2}} = \frac{1}{2} = \sin 30^\circ.

Why each choice is right or wrong

  • A. Incorrect. −sin⁡x-\sin x comes from rearranging the identity with the wrong sign, as 1−cos⁡2x=−sin⁡2x1 - \cos^2 x = -\sin^2 x. Since sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1, 1−cos⁡2x=sin⁡2x1 - \cos^2 x = \sin^2 x.
  • B. Correct. 1−cos⁡2x=sin⁡2x1 - \cos^2 x = \sin^2 x, and sin⁡2xsin⁡x=sin⁡x\frac{\sin^2 x}{\sin x} = \sin x.
  • C. Incorrect. csc⁡x−cos⁡2x\csc x - \cos^2 x divides only the 1 by sin⁡x\sin x. Every term of the numerator must be divided, and the identity 1−cos⁡2x=sin⁡2x1 - \cos^2 x = \sin^2 x makes that quick: sin⁡2xsin⁡x=sin⁡x\frac{\sin^2 x}{\sin x} = \sin x.
  • D. Incorrect. csc⁡x−cos⁡x\csc x - \cos x splits the fraction into 1sin⁡x−cos⁡2xsin⁡x\frac{1}{\sin x} - \frac{\cos^2 x}{\sin x} and then cancels a cos⁡x\cos x against sin⁡x\sin x. Only common factors cancel; use the identity on the numerator instead.

Worked example Medium

If sin⁡θ=2029\sin\theta = \frac{20}{29} and 0∘<θ<90∘0^\circ < \theta < 90^\circ, what is the value of sec⁡θ\sec\theta?

  1. 929\frac{9}{29}
  2. 2129\frac{21}{29}
  3. 2921\frac{29}{21}Answer
  4. 2920\frac{29}{20}

How to solve it

  1. cos⁡2θ=1−sin⁡2θ=1−400841=441841\cos^2\theta = 1 - \sin^2\theta = 1 - \frac{400}{841} = \frac{441}{841}.
  2. θ\theta is acute, so cos⁡θ=2129\cos\theta = \frac{21}{29}.
  3. sec⁡θ=1cos⁡θ=2921\sec\theta = \frac{1}{\cos\theta} = \frac{29}{21}.

Why each choice is right or wrong

  • A. Incorrect. 929\frac{9}{29} is 1−20291 - \frac{20}{29}, which leaves out the squares. The identity is sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1, which gives cos⁡θ=2129\cos\theta = \frac{21}{29}.
  • B. Incorrect. 2129\frac{21}{29} is cos⁡θ\cos\theta. That is one step short: the secant is its reciprocal, 2921\frac{29}{21}.
  • C. Correct. cos⁡θ=1−400841=2129\cos\theta = \sqrt{1 - \frac{400}{841}} = \frac{21}{29}, and sec⁡θ=1cos⁡θ=2921\sec\theta = \frac{1}{\cos\theta} = \frac{29}{21}.
  • D. Incorrect. 2920\frac{29}{20} is 1sin⁡θ\frac{1}{\sin\theta}, the cosecant. The secant is the reciprocal of the cosine, not the sine.

Trap Secant is not one over sine

sec⁡θ=1cos⁡θ\sec\theta = \frac{1}{\cos\theta} and csc⁡θ=1sin⁡θ\csc\theta = \frac{1}{\sin\theta}. The names cross over, so expect a choice built on 1sin⁡θ\frac{1}{\sin\theta} when the question asks for the secant.

Trap $\sin^2 x$ is not $\sin(x^2)$

sin⁡2x\sin^2 x means (sin⁡x)2(\sin x)^2: find the sine, then square it. sin⁡(x2)\sin(x^2) squares the angle first. For x=30∘x = 30^\circ, sin⁡2x=14\sin^2 x = \frac{1}{4}, while sin⁡(900∘)=0\sin(900^\circ) = 0.

Calculator When a calculator isn’t faster

Rewriting in sines and cosines by hand is usually the quickest route. A calculator is a good safety net: evaluate the expression and each choice at one angle that is not a special angle, such as x=0.7x = 0.7 in radian mode, and keep the choice that matches.

In the built-in Desmos calculator if you test online, you can also graph both sides, such as y=(1-cos(x)^2)/sin(x) and y=sin(x). If the two graphs lie on top of each other, the expressions are equal wherever both are defined.