Trigonometric identities

An iden­tity is an equa­tion that is true for every value of the vari­able for which both sides are defined. Trigonom­e­try rests on three iden­ti­ties, all com­ing from one fact, $x^2 + y^2 = r^2$. With them we can turn any ratio into any other, sim­plify com­pli­cated expres­sions and prove new results. This les­son proves the three iden­ti­ties, rearranges them into the forms used most often, and works through prob­lems that depend on them.

The three fun­da­men­tal iden­ti­ties#

Let $P(x, y)$ be on the ter­mi­nal side of $\theta$ with $OP = r$, so $x^2 + y^2 = r^2$.

Iden­tity 1. Divide $x^2 + y^2 = r^2$ by $r^2$:

$$\displaystyle \left(\frac{x}{r}\right)^2 + \left(\frac{y}{r}\right)^2 = 1 \implies \sin^2\theta + \cos^2\theta = 1$$

Iden­tity 2. Divide by $x^2$ (when $x \neq 0$): $\displaystyle 1 + \left(\frac{y}{x}\right)^2 = \left(\frac{r}{x}\right)^2$, so

$$\sec^2\theta - \tan^2\theta = 1$$

Iden­tity 3. Divide by $y^2$ (when $y \neq 0$): $\displaystyle \left(\frac{x}{y}\right)^2 + 1 = \left(\frac{r}{y}\right)^2$, so

$$\csc^2\theta - \cot^2\theta = 1$$

(Here $\sin^2\theta$ means $(\sin\theta)^2$.)

Use­ful forms#

From Iden­tity 1:

$$\sin^2\theta = 1 - \cos^2\theta, \quad \cos^2\theta = 1 - \sin^2\theta, \quad \sin\theta = \pm\sqrt{1 - \cos^2\theta}, \quad \cos\theta = \pm\sqrt{1 - \sin^2\theta}$$

(The sign comes from the quad­rant, as we will see; recall that $x^2 = a$ gives $x = \pm\sqrt{a}$.)

From Iden­tity 2: $\sec^2\theta = 1 + \tan^2\theta$ and $\tan^2\theta = \sec^2\theta - 1$. Fac­toris­ing the dif­fer­ence of squares,

$$\displaystyle (\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = 1, \quad \text{so} \quad \sec\theta - \tan\theta = \frac{1}{\sec\theta + \tan\theta}$$

From Iden­tity 3 in the same way: $\csc^2\theta = 1 + \cot^2\theta$, $\cot^2\theta = \csc^2\theta - 1$, and

$$\displaystyle \csc\theta - \cot\theta = \frac{1}{\csc\theta + \cot\theta}$$

Exam­ple 1#

If $\sec\theta + \tan\theta = 4$, find $\sec\theta$, $\tan\theta$ and $\sin\theta$.

$\displaystyle \sec\theta - \tan\theta = \frac{1}{4}$. Adding the two equa­tions: $\displaystyle 2\sec\theta = 4 + \frac{1}{4} = \frac{17}{4}$, so $\displaystyle \sec\theta = \frac{17}{8}$. Sub­tract­ing: $\displaystyle 2\tan\theta = \frac{15}{4}$, so $\displaystyle \tan\theta = \frac{15}{8}$.

$$\displaystyle \sin\theta = \frac{\tan\theta}{\sec\theta} = \frac{15/8}{17/8} = \frac{15}{17}$$

Higher pow­ers of sine and cosine#

Writ­ing $s = \sin^2\theta$ and $c = \cos^2\theta$, so that $s + c = 1$, three more results fol­low.

(i) $\sin^2\theta + \cos^4\theta = \cos^2\theta + \sin^4\theta = 1 - \sin^2\theta\cos^2\theta$.

$s + c^2 = s + c(1 - s) = s + c - sc = 1 - sc$, and in the same way $c + s^2 = 1 - sc$.

(ii) $\sin^4\theta + \cos^4\theta = 1 - 2\sin^2\theta\cos^2\theta$.

$s^2 + c^2 = (s + c)^2 - 2sc = 1 - 2sc$.

(iii) $\sin^6\theta + \cos^6\theta = 1 - 3\sin^2\theta\cos^2\theta$.

$s^3 + c^3 = (s + c)^3 - 3sc(s + c) = 1 - 3sc$.

Exam­ple 2#

If $u_n = \cos^n\theta + \sin^n\theta$, find $2u_6 - 3u_4$.

Using (ii) and (iii):

$$2u_6 - 3u_4 = 2(1 - 3\sin^2\theta\cos^2\theta) - 3(1 - 2\sin^2\theta\cos^2\theta) = 2 - 3 = -1$$

The terms in $\sin^2\theta\cos^2\theta$ can­cel, so the answer is $-1$ for every $\theta$.

Exam­ple 3#

If $\displaystyle \sin\theta + \cos\theta = \frac{7}{5}$, find $\sin\theta\cos\theta$ and $\sin^4\theta + \cos^4\theta$.

Square: $\displaystyle \sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta = \frac{49}{25}$, so $\displaystyle 1 + 2\sin\theta\cos\theta = \frac{49}{25}$ and $\displaystyle \sin\theta\cos\theta = \frac{12}{25}$.

$$\displaystyle \sin^4\theta + \cos^4\theta = 1 - 2\left(\frac{12}{25}\right)^2 = 1 - \frac{288}{625} = \frac{337}{625}$$

Exam­ple 4#

If $\displaystyle \tan\theta = p - \frac{1}{4p}$, find $\sec\theta - \tan\theta$.

$$\displaystyle \sec^2\theta = 1 + \left(p - \frac{1}{4p}\right)^2 = 1 + p^2 - \frac{1}{2} + \frac{1}{16p^2} = p^2 + \frac{1}{2} + \frac{1}{16p^2} = \left(p + \frac{1}{4p}\right)^2$$

So $\displaystyle \sec\theta = \pm\left(p + \frac{1}{4p}\right)$, and

$$\displaystyle \sec\theta - \tan\theta = \left(p + \frac{1}{4p}\right) - \left(p - \frac{1}{4p}\right) = \frac{1}{2p} \quad \text{or} \quad -\left(p + \frac{1}{4p}\right) - \left(p - \frac{1}{4p}\right) = -2p$$

Exam­ple 5#

If $\tan^2\theta = 1 - k^2$, find $\sec\theta + \tan^3\theta\csc\theta$.

Since $\displaystyle \tan\theta\csc\theta = \frac{\sin\theta}{\cos\theta} \times \frac{1}{\sin\theta} = \sec\theta$,

$$\sec\theta + \tan^2\theta \cdot \tan\theta\csc\theta = \sec\theta + \tan^2\theta\sec\theta = \sec\theta(1 + \tan^2\theta) = \sec^3\theta$$

and $\sec^2\theta = 1 + \tan^2\theta = 2 - k^2$, so the value is $(2 - k^2)^{3/2}$.

Prov­ing iden­ti­ties#

Start from the more com­pli­cated side and sim­plify it to the other side, using the iden­ti­ties and the rec­i­p­ro­cal and quo­tient rela­tions. Do not "cross over" terms from one side to the other as if solv­ing an equa­tion.

Exam­ple 6#

Prove that $(1 - \cos^2\theta)(1 + \cot^2\theta) = 1$.

$$\displaystyle \text{LHS} = \sin^2\theta \times \csc^2\theta = \sin^2\theta \times \frac{1}{\sin^2\theta} = 1 = \text{RHS}$$

Prac­tice#

1. Prove that $(1 + \tan^2\theta)\cos^2\theta = 1$.
2. Prove that $\sec^4\theta - \sec^2\theta = \tan^4\theta + \tan^2\theta$.
3. If $\sec\theta + \tan\theta = 5$, find $\sec\theta - \tan\theta$, $\sec\theta$ and $\tan\theta$.
4. If $\csc\theta + \cot\theta = 3$, find $\cos\theta$.
5. If $\displaystyle \sin\theta\cos\theta = \frac{1}{3}$, find $\sin^6\theta + \cos^6\theta$.
6. If $\displaystyle \sin\theta + \cos\theta = \frac{7}{5}$, find $\sin^2\theta + \cos^4\theta$.
7. Prove that $(\sin\theta + \csc\theta)^2 + (\cos\theta + \sec\theta)^2 = 7 + \tan^2\theta + \cot^2\theta$.
8. Prove that $\displaystyle \frac{1}{\sec\theta - \tan\theta} - \frac{1}{\cos\theta} = \tan\theta$.

Answers#