International Baccalaureate · DP 1: Analysis and Approaches HL

Further Trigonometry

⌚ ~130 min ▢ 14 blocks ✓ 70 exercises ➣ Prerequisites : The Unit Circle and Radian Measure, Trigonometric Functions, Equations, Unit Circle and Trigonometric Identities

I Reciprocal Trigonometric Functions

I.1 The Secant, the Cosecant and the Cotangent

Definition — Reciprocal Trigonometric Functions

The cosecant is $\csc(x) = \frac{1}{\sin(x)}$, for $\sin(x) \neq 0$. It is also written $\operatorname{cosec}(x)$.
The secant is $\sec(x) = \frac{1}{\cos(x)}$, for $\cos(x) \neq 0$.
The cotangent is $\cot(x) = \frac{\cos(x)}{\sin(x)}$, for $\sin(x) \neq 0$. It equals $\frac{1}{\tan(x)}$ where $\tan(x)$ exists and is non-zero; at $x = \frac{\pi}{2}$, for instance, $\tan(x)$ does not exist and $\cot(x) = 0$.

Proposition — Graph of the Cosecant Function

The graph of $y = \csc(x) = \frac{1}{\sin(x)}$ is shown below.

Period: $2\pi$.
Vertical Asymptotes: At $x = k\pi$, $k \in \mathbb{Z}$ (where $\sin x = 0$).
Range: $({-\infty}, -1] \cup [1, \infty)$.

Proposition — Graph of the Secant Function

The graph of $y = \sec(x) = \frac{1}{\cos(x)}$ is shown below.

Period: $2\pi$.
Vertical Asymptotes: At $x = \frac{\pi}{2} + k\pi$, $k \in \mathbb{Z}$ (where $\cos x = 0$).
Range: $({-\infty}, -1] \cup [1, \infty)$.

Proposition — Graph of the Cotangent Function

The graph of $y = \cot(x) = \frac{\cos(x)}{\sin(x)}$ is shown below.

Period: $\pi$.
Vertical Asymptotes: At $x = k\pi$, $k \in \mathbb{Z}$ (where $\sin x = 0$).
Zeros: At $x = \frac{\pi}{2} + k\pi$, $k \in \mathbb{Z}$ (where $\cos x = 0$): there $\tan x$ does not exist, but $\cot x = 0$.
Range: $\mathbb{R}$.

Skills to practice

Recognising Asymptotes and Simplifying
Evaluating the Reciprocal Functions at Special Angles without Technology

I.2 Pythagorean Identities for the Secant and the Cosecant

Proposition — Pythagorean Identities for the Secant and the Cosecant

For every real number $x$ such that $\cos x \neq 0$: $$1 + \tan^2 x = \sec^2 x.$$ For every real number $x$ such that $\sin x \neq 0$: $$1 + \cot^2 x = \csc^2 x.$$ Together with $\cos^2 x + \sin^2 x = 1$, they are the three Pythagorean identities. The cosecant is also written $\operatorname{cosec} x$.

Proof

Let $x$ be a real number such that $\cos x \neq 0$. Starting from the left side: $$\begin{aligned}[t] 1 + \tan^2 x &= 1 + \frac{\sin^2 x}{\cos^2 x} && \text{(definition of the tangent)}\\ &= \frac{\cos^2 x + \sin^2 x}{\cos^2 x} && \text{(common denominator)}\\ &= \frac{1}{\cos^2 x} && \text{(Pythagorean identity)}\\ &= \sec^2 x, \end{aligned}$$ by the definition of the secant. In the same way, let $x$ be a real number such that $\sin x \neq 0$, so that $\cot x = \dfrac{\cos x}{\sin x}$: $$\begin{aligned}[t] 1 + \cot^2 x &= 1 + \frac{\cos^2 x}{\sin^2 x} && \text{(definition of the cotangent)}\\ &= \frac{\sin^2 x + \cos^2 x}{\sin^2 x} && \text{(common denominator)}\\ &= \frac{1}{\sin^2 x} && \text{(Pythagorean identity)}\\ &= \csc^2 x, \end{aligned}$$ by the definition of the cosecant. Hence both identities hold wherever their two sides are defined.

Method — Proving a Trigonometric Identity

To prove that an equality between trigonometric expressions is an identity:

State where both sides are defined: no denominator is $0$, and every $\tan x$, $\sec x$, $\csc x$ or $\cot x$ that appears exists.
Start from one side, usually the more complicated one, and transform it through a chain of equalities until the other side appears. Do not transform both sides at once, unless every step is an equivalence.
Useful moves: write $\tan x$, $\cot x$, $\sec x$ and $\csc x$ in terms of $\sin x$ and $\cos x$; put fractions over a common denominator; replace $\cos^2 x + \sin^2 x$ by $1$, $1 + \tan^2 x$ by $\sec^2 x$ and $1 + \cot^2 x$ by $\csc^2 x$, or the reverse; factorise, for instance $\sec^2 x - 1 = (\sec x - 1)(\sec x + 1)$.
Name the rule used at each step, in the justification column.

Example

Given that $\tan x = -2$ and $\dfrac{\pi}{2} < x < \pi$, find the exact values of $\sec x$, $\cos x$, $\sin x$, $\csc x$ and $\cot x$.

Answer

Since $\dfrac{\pi}{2} < x < \pi$, the point of the unit circle associated with $x$ lies in the second quadrant: $\cos x < 0$ and $\sin x > 0$. $$\begin{aligned}[t] \sec^2 x &= 1 + \tan^2 x && \text{(Pythagorean identity for the secant)}\\ &= 1 + (-2)^2\\ &= 5, \end{aligned}$$ so $\sec x = \sqrt{5}$ or $\sec x = -\sqrt{5}$. Since $\cos x < 0$, its reciprocal $\sec x$ is negative: $\sec x = -\sqrt{5}$. Then $$\begin{aligned}[t] \cos x &= \frac{1}{\sec x} = -\frac{1}{\sqrt{5}} = -\frac{\sqrt{5}}{5}, && \text{(definition of the secant)}\\ \sin x &= \tan x \cos x && \text{(definition of the tangent)}\\ &= (-2) \times \left(-\frac{\sqrt{5}}{5}\right) = \frac{2\sqrt{5}}{5},\\ \csc x &= \frac{1}{\sin x} = \frac{5}{2\sqrt{5}} = \frac{\sqrt{5}}{2}, && \text{(definition of the cosecant)}\\ \cot x &= \frac{\cos x}{\sin x} = \frac{1}{\tan x} = -\frac{1}{2}. && \text{(definition of the cotangent; } \tan x \neq 0\text{)} \end{aligned}$$ Check: $\sin x > 0$, as expected in the second quadrant, and $\cos^2 x + \sin^2 x = \dfrac{5}{25} + \dfrac{20}{25} = 1$.

Example

Show that each equality is an identity, and state for which real numbers $x$ it holds.

$(\sec x - 1)(\sec x + 1) = \tan^2 x$;
$\tan x + \cot x = \sec x \csc x$.

Answer

Both sides are defined when $\cos x \neq 0$. Let $x$ be such a real number. Starting from the left side: $$\begin{aligned}[t] (\sec x - 1)(\sec x + 1) &= \sec^2 x - 1 && \text{(difference of two squares)}\\ &= (1 + \tan^2 x) - 1 && \text{(Pythagorean identity for the secant)}\\ &= \tan^2 x, \end{aligned}$$ which is the right side. The identity holds for every real $x$ such that $\cos x \neq 0$.
Both sides are defined when $\cos x \neq 0$ and $\sin x \neq 0$. Let $x$ be such a real number. Starting from the left side: $$\begin{aligned}[t] \tan x + \cot x &= \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} && \text{(definitions of the tangent and the cotangent)}\\ &= \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} && \text{(common denominator)}\\ &= \frac{1}{\sin x \cos x} && \text{(Pythagorean identity)}\\ &= \frac{1}{\cos x} \times \frac{1}{\sin x}\\ &= \sec x \csc x, \end{aligned}$$ which is the right side. The identity holds for every real $x$ such that $\cos x \neq 0$ and $\sin x \neq 0$.

Skills to practice

Finding the Other Ratios from One Ratio
Proving Identities with the Secant, the Cosecant or the Cotangent

II Inverse Trigonometric Functions

II.1 The Arcsine, Arccosine and Arctangent Functions

The inverse trigonometric functions, also known as arc functions, give the angle from the value of a trigonometric ratio. They are the inverses of the sine, cosine and tangent functions, each restricted to an interval on which it is one-to-one.

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Definition — Inverse Sine Function

The inverse sine function, denoted $\arcsin$ or $\sin^{-1}$, is defined on $[-1, 1]$: for every $x$ in $[-1, 1]$, $\arcsin x$ is the number $y$ in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ such that $\sin y = x$.

Domain: $[-1, 1]$
Range: $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

The $-1$ in $\sin^{-1} x$ is not an exponent: $\sin^{-1} x$ is $\arcsin x$, not $\frac{1}{\sin x}$, which is $\csc x$.

Proposition — Graph of the Inverse Sine Function

The graph of $\textcolor{colordef}{y = \arcsin x}$ is the reflection in the line $\textcolor{olive}{y = x}$ of the graph of $\textcolor{colorprop}{y = \sin x}$ restricted to $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. It stops at its two endpoints, $\left(-1, -\frac{\pi}{2}\right)$ and $\left(1, \frac{\pi}{2}\right)$.

Definition — Inverse Cosine Function

The inverse cosine function, denoted $\arccos$ or $\cos^{-1}$, is defined on $[-1, 1]$: for every $x$ in $[-1, 1]$, $\arccos x$ is the number $y$ in $[0, \pi]$ such that $\cos y = x$.

Domain: $[-1, 1]$
Range: $[0, \pi]$

Proposition — Graph of the Inverse Cosine Function

The graph of $\textcolor{colordef}{y = \arccos x}$ is the reflection in the line $\textcolor{olive}{y = x}$ of the graph of $\textcolor{colorprop}{y = \cos x}$ restricted to $[0, \pi]$. It stops at its two endpoints, $(-1, \pi)$ and $(1, 0)$.

Definition — Inverse Tangent Function

The inverse tangent function, denoted $\arctan$ or $\tan^{-1}$, is defined on $\mathbb{R}$: for every real number $x$, $\arctan x$ is the number $y$ in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ such that $\tan y = x$.

Domain: $\mathbb{R}$
Range: $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

Proposition — Graph of the Inverse Tangent Function

The graph of $y = \arctan x$ is the reflection in the line $y = x$ of the graph of $y = \tan x$ restricted to $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. The lines $y = -\frac{\pi}{2}$ and $y = \frac{\pi}{2}$ are horizontal asymptotes.

Proposition — Properties of Inverse Trigonometric Functions

$\arcsin(\sin x) = x$ for $x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, and $\sin(\arcsin x) = x$ for $x \in [-1, 1]$
$\arccos(\cos x) = x$ for $x \in [0, \pi]$, and $\cos(\arccos x) = x$ for $x \in [-1, 1]$
$\arctan(\tan x) = x$ for $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, and $\tan(\arctan x) = x$ for every real number $x$

Example

Evaluate $\arcsin\left(\frac{\sqrt{3}}{2}\right)$.

Answer

We look for the number $y$ in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ such that $\sin y = \frac{\sqrt{3}}{2}$. Since $\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$ and $\frac{\pi}{3} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, $$\arcsin\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}.$$ The number $\frac{2\pi}{3}$ also has sine $\frac{\sqrt{3}}{2}$, but it is not in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Example

Find the exact value of:

$\sin\left(\arccos\dfrac{1}{3}\right)$;
$\tan\left(\arcsin\left(-\dfrac{2}{3}\right)\right)$.

Answer

Let $\theta = \arccos\dfrac{1}{3}$. By the definition of $\arccos$, $\cos\theta = \dfrac{1}{3}$ and $0 \le \theta \le \pi$, so $\sin\theta \ge 0$. $$\begin{aligned}[t] \sin^2\theta &= 1 - \cos^2\theta && \text{(Pythagorean identity)}\\ &= 1 - \frac{1}{9}\\ &= \frac{8}{9}. \end{aligned}$$ Since $\sin\theta \ge 0$, $\sin\theta = \sqrt{\dfrac{8}{9}} = \dfrac{2\sqrt{2}}{3}$. Hence $\sin\left(\arccos\dfrac{1}{3}\right) = \dfrac{2\sqrt{2}}{3}$.
Let $\theta = \arcsin\left(-\dfrac{2}{3}\right)$. By the definition of $\arcsin$, $\sin\theta = -\dfrac{2}{3}$ and $-\dfrac{\pi}{2} \le \theta \le \dfrac{\pi}{2}$, so $\cos\theta \ge 0$. $$\begin{aligned}[t] \cos^2\theta &= 1 - \sin^2\theta && \text{(Pythagorean identity)}\\ &= 1 - \frac{4}{9}\\ &= \frac{5}{9}, \end{aligned}$$ so $\cos\theta = \dfrac{\sqrt{5}}{3}$, which is not $0$. Then $$\begin{aligned}[t] \tan\theta &= \frac{\sin\theta}{\cos\theta} && \text{(definition of the tangent)}\\ &= -\frac{2}{3} \times \frac{3}{\sqrt{5}}\\ &= -\frac{2\sqrt{5}}{5}. \end{aligned}$$ Hence $\tan\left(\arcsin\left(-\dfrac{2}{3}\right)\right) = -\dfrac{2\sqrt{5}}{5}$.

In both cases the range of the inverse function gives the sign of the square root: this is where the restricted domain of the definition is used.

Skills to practice

Evaluating Inverse Trigonometric Functions at Special Values without Technology
Simplifying an Inverse Function of a Trigonometric Function
Finding Exact Values of Compositions

III Double Angle Identities

III.1 Compound Angle and Double Angle Formulas

Recappreviously studied — click to reread

Proposition — Cosine of Difference

$$\cos (A - B) = \cos A \cos B + \sin A \sin B.$$

Proof

Consider the two unit vectors whose tips are the points of the unit circle associated with $A$ and $B$: $$\vec{u} = \begin{pmatrix}\cos A \\ \sin A \end{pmatrix} \quad\text{and}\quad \vec{v} = \begin{pmatrix}\cos B \\ \sin B \end{pmatrix}.$$ The geometric angle $\theta \in [0, \pi]$ between $\vec{u}$ and $\vec{v}$ is equal to $A - B$ or to $B - A$, up to a multiple of $2\pi$. Since cosine is even and $2\pi$-periodic, $\cos\theta = \cos(A - B)$.
The scalar product $\vec{u} \cdot \vec{v}$ can be computed in two ways:

From the angle between the vectors: $$\vec{u} \cdot \vec{v} = \|\vec{u}\| \,\|\vec{v}\| \cos\theta = 1 \cdot 1 \cdot \cos(A - B) = \cos(A - B).$$
From the components: $$\vec{u} \cdot \vec{v} = \begin{pmatrix}\cos A \\ \sin A \end{pmatrix} \cdot \begin{pmatrix}\cos B \\ \sin B \end{pmatrix} = (\cos A)(\cos B) + (\sin A)(\sin B).$$

Equating the two expressions gives the identity $$\cos(A - B) = \cos A \cos B + \sin A \sin B.$$

Another proof, with distances. Let $P$ and $Q$ be the points of the unit circle associated with $A$ and $B$: $P(\cos A, \sin A)$ and $Q(\cos B, \sin B)$. Then $$\begin{aligned}[t] PQ^2 &= (\cos A - \cos B)^2 + (\sin A - \sin B)^2 && \text{(distance between two points)}\\ &= \cos^2 A + \sin^2 A + \cos^2 B + \sin^2 B\\ &\quad - 2(\cos A\cos B + \sin A\sin B) && \text{(expanding)}\\ &= 2 - 2(\cos A\cos B + \sin A\sin B). && (\cos^2 + \sin^2 = 1) \end{aligned}$$ The rotation about $O$ through the angle $-B$ keeps distances. It sends $Q$ to the point $Q'(1, 0)$, of angle $0$, and $P$ to the point $P'\big(\cos(A - B), \sin(A - B)\big)$, of angle $A - B$. So $$\begin{aligned}[t] PQ^2 = P'Q'^2 &= \big(\cos(A - B) - 1\big)^2 + \sin^2(A - B) && \text{(distance between two points)}\\ &= \cos^2(A - B) + \sin^2(A - B) - 2\cos(A - B) + 1 && \text{(expanding)}\\ &= 2 - 2\cos(A - B). && (\cos^2 + \sin^2 = 1) \end{aligned}$$ The two expressions of $PQ^2$ are equal: $2 - 2\cos(A - B) = 2 - 2(\cos A\cos B + \sin A\sin B)$, which gives the identity again, without the scalar product.

Proposition — Cosine of Sum

$$\cos (A + B) = \cos A \cos B - \sin A \sin B.$$

Proof

Substitute $B$ by $-B$ in the cosine difference formula: $$\begin{aligned} \cos(A + B) &= \cos\bigl(A - (-B)\bigr)\\ &= \cos A \cos(-B) + \sin A \sin(-B)\\ &= \cos A \cos B + \sin A (-\sin B)\quad(\text{since } \cos(-\theta) = \cos \theta\text{ and }\sin(-\theta) = -\sin \theta)\\ &= \cos A \cos B - \sin A \sin B. \end{aligned}$$

Proposition — Sine of Sum and Difference

$$\sin (A + B) = \sin A \cos B + \cos A \sin B,$$ $$\sin (A - B) = \sin A \cos B - \cos A \sin B.$$

Proof

Using the complementary angle identity $\sin \theta = \cos\left(\dfrac{\pi}{2} - \theta\right)$: $$\begin{aligned} \sin(A + B) &= \cos\left(\frac{\pi}{2} - (A + B)\right)\\ &= \cos\left( \left(\frac{\pi}{2} - A\right) - B \right)\\ &= \cos\left(\frac{\pi}{2} - A\right) \cos B + \sin\left(\frac{\pi}{2} - A\right) \sin B\\ &= \sin A \cos B + \cos A \sin B. \end{aligned}$$ For the difference, write $$\begin{aligned} \sin(A - B) &= \sin\bigl(A + (-B)\bigr)\\ &= \sin A \cos(-B) + \cos A \sin(-B)\\ &= \sin A \cos B - \cos A \sin B. \end{aligned}$$

Proposition — Tangent of Sum and Difference

For angles $A$ and $B$ such that all expressions below are defined, $$\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B},$$ $$\tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}.$$

Proof

Assume $\cos A \cos B \neq 0$ and $\cos(A + B) \neq 0$ so that all ratios below are defined: $$\begin{aligned} \tan(A + B) &= \frac{\sin(A + B)}{\cos(A + B)} \\ &= \frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B - \sin A \sin B} \\ &= \frac{\dfrac{\sin A \cos B + \cos A \sin B}{\cos A \cos B}}{\dfrac{\cos A \cos B - \sin A \sin B}{\cos A \cos B}} \quad \text{(divide numerator and denominator by } \cos A \cos B\text{)} \\ &= \frac{\dfrac{\sin A}{\cos A} + \dfrac{\sin B}{\cos B}}{1 - \dfrac{\sin A \sin B}{\cos A \cos B}}\\ &= \frac{\tan A + \tan B}{1 - \tan A \tan B}. \end{aligned}$$ The formula for $\tan(A-B)$ is obtained by replacing $B$ with $-B$ and using $\tan(-B) = -\tan B$.

Proposition — Double Angle Formulas

For every real number $A$: $$\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A,$$ $$\sin 2A = 2\sin A\cos A.$$ They are obtained by taking $B = A$ in the formulas for $\cos(A + B)$ and $\sin(A + B)$, which gives $\sin 2A$ and $\cos^2 A - \sin^2 A$, then using $\cos^2 A + \sin^2 A = 1$ for the other two forms of $\cos 2A$.

Example

Compute the exact value of $\cos\dfrac{\pi}{12}$ in two ways:

with an addition formula, writing $\dfrac{\pi}{12} = \dfrac{\pi}{3} - \dfrac{\pi}{4}$;
with a double angle formula, since $2 \times \dfrac{\pi}{12} = \dfrac{\pi}{6}$.

Answer

With the cosine of a difference: $$\begin{aligned}[t] \cos\frac{\pi}{12} &= \cos\left(\frac{\pi}{3} - \frac{\pi}{4}\right) &&\\ &= \cos\frac{\pi}{3}\cos\frac{\pi}{4} + \sin\frac{\pi}{3}\sin\frac{\pi}{4} && \text{(cosine of a difference)}\\ &= \frac{1}{2} \times \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2} && \text{(special values)}\\ &= \frac{\sqrt{2} + \sqrt{6}}{4}. \end{aligned}$$
With $A = \dfrac{\pi}{12}$, the formula $\cos 2A = 2\cos^2 A - 1$ gives $\cos\dfrac{\pi}{6} = 2\cos^2\dfrac{\pi}{12} - 1$, so $$\begin{aligned}[t] \cos^2\frac{\pi}{12} &= \frac{1}{2}\left(1 + \cos\frac{\pi}{6}\right) &&\\ &= \frac{1}{2}\left(1 + \frac{\sqrt{3}}{2}\right) = \frac{2 + \sqrt{3}}{4}. \end{aligned}$$ Since $\dfrac{\pi}{12} \in \left[0, \dfrac{\pi}{2}\right]$, $\cos\dfrac{\pi}{12} \ge 0$, hence $\cos\dfrac{\pi}{12} = \dfrac{\sqrt{2 + \sqrt{3}}}{2}$.

The two results are the same number: $\left(\dfrac{\sqrt{2} + \sqrt{6}}{4}\right)^2 = \dfrac{2 + 2\sqrt{12} + 6}{16} = \dfrac{8 + 4\sqrt{3}}{16} = \dfrac{2 + \sqrt{3}}{4}$, and both are positive.

Skills to practice

Calculating Exact Values with the Compound and Double Angle Formulas

III.2 The Double Angle Formula for the Tangent

Proposition — Double Angle Formula for the Tangent

For every real number $A$ such that $\cos A \neq 0$ and $\cos 2A \neq 0$: $$\tan 2A = \frac{2\tan A}{1 - \tan^2 A}.$$ Under these conditions, $1 - \tan^2 A \neq 0$.

Proof

Let $A$ be a real number such that $\cos A \neq 0$ and $\cos 2A \neq 0$. Take $B = A$ in the formula for the tangent of a sum: its conditions $\cos A \cos B = \cos^2 A \neq 0$ and $\cos(A + B) = \cos 2A \neq 0$ hold, so $$\begin{aligned}[t] \tan 2A &= \tan(A + A)\\ &= \frac{\tan A + \tan A}{1 - \tan A \tan A} && \text{(tangent of a sum)}\\ &= \frac{2\tan A}{1 - \tan^2 A}. \end{aligned}$$ The denominator is not $0$: $$\begin{aligned}[t] 1 - \tan^2 A &= 1 - \frac{\sin^2 A}{\cos^2 A} && \text{(definition of the tangent)}\\ &= \frac{\cos^2 A - \sin^2 A}{\cos^2 A} && \text{(common denominator)}\\ &= \frac{\cos 2A}{\cos^2 A}, && \text{(double angle formula)} \end{aligned}$$ which is not $0$ since $\cos 2A \neq 0$.

Example

Given that $\cos\theta = -\dfrac{1}{3}$ and $\pi < \theta < \dfrac{3\pi}{2}$, find the exact values of $\sin 2\theta$, $\cos 2\theta$ and $\tan 2\theta$, without finding $\theta$.

Answer

Since $\pi < \theta < \dfrac{3\pi}{2}$ (third quadrant), $\sin\theta < 0$. $$\begin{aligned}[t] \sin^2\theta &= 1 - \cos^2\theta && \text{(Pythagorean identity)}\\ &= 1 - \frac{1}{9} = \frac{8}{9}, \end{aligned}$$ so $\sin\theta = -\dfrac{2\sqrt{2}}{3}$. Then $$\begin{aligned}[t] \sin 2\theta &= 2\sin\theta\cos\theta && \text{(double angle formula)}\\ &= 2 \times \left(-\frac{2\sqrt{2}}{3}\right) \times \left(-\frac{1}{3}\right) = \frac{4\sqrt{2}}{9},\\ \cos 2\theta &= 2\cos^2\theta - 1 && \text{(double angle formula)}\\ &= \frac{2}{9} - 1 = -\frac{7}{9},\\ \tan 2\theta &= \frac{\sin 2\theta}{\cos 2\theta} && \text{(definition of the tangent)}\\ &= \frac{4\sqrt{2}}{9} \times \left(-\frac{9}{7}\right) = -\frac{4\sqrt{2}}{7}. \end{aligned}$$

Example

Given that $\cos\theta = -\dfrac{1}{3}$ and $\pi < \theta < \dfrac{3\pi}{2}$, find the exact value of $\tan\theta$, then use the double angle formula for the tangent to find $\tan 2\theta$, without finding $\theta$.

Answer

Since $\pi < \theta < \dfrac{3\pi}{2}$ (third quadrant), $\sin\theta < 0$, and $\sin^2\theta = 1 - \cos^2\theta = \dfrac{8}{9}$ (Pythagorean identity), so $\sin\theta = -\dfrac{2\sqrt{2}}{3}$. Then $$\begin{aligned}[t] \tan\theta &= \frac{\sin\theta}{\cos\theta} && \text{(definition of the tangent)}\\ &= \left(-\frac{2\sqrt{2}}{3}\right) \times (-3) = 2\sqrt{2}. \end{aligned}$$ $1 - \tan^2\theta = 1 - 8 = -7 \neq 0$, so the double angle formula for the tangent applies: $$\begin{aligned}[t] \tan 2\theta &= \frac{2\tan\theta}{1 - \tan^2\theta} && \text{(double angle formula for the tangent)}\\ &= \frac{4\sqrt{2}}{-7} = -\frac{4\sqrt{2}}{7}. \end{aligned}$$ Check: the double angle formulas for the sine and the cosine give $\sin 2\theta = 2\sin\theta\cos\theta = \dfrac{4\sqrt{2}}{9}$ and $\cos 2\theta = 2\cos^2\theta - 1 = -\dfrac{7}{9}$, and $\dfrac{\sin 2\theta}{\cos 2\theta} = -\dfrac{4\sqrt{2}}{7}$ as well.

Skills to practice

Finding the Ratios of the Double Angle from One Ratio
Proving Identities with the Double Angle Formulas

IV Equations and Identities

IV.1 Solving Equations with an Identity

Example

Solve each equation for $0 \le x \le 2\pi$, giving exact values. First use a double angle formula to write the equation with a single trigonometric function of $x$, or as a product, then solve it.

$\cos 2x + 3\sin x = 2$;
$\sin 2x = \sqrt{3}\cos x$.

Answer

The form $\cos 2x = 1 - 2\sin^2 x$ of the double angle formula leaves only $\sin x$: $$\begin{aligned}[t] \cos 2x + 3\sin x = 2 &\iff 1 - 2\sin^2 x + 3\sin x = 2 && \text{(double angle formula)}\\ &\iff 2\sin^2 x - 3\sin x + 1 = 0 && \text{(every term on one side)}\\ &\iff (2\sin x - 1)(\sin x - 1) = 0 && \text{(factorise)}\\ &\iff \sin x = \frac{1}{2} \ \text{ or } \ \sin x = 1. && \text{(null factor law)} \end{aligned}$$ On the unit circle, $\sin x = \dfrac{1}{2}$ for $x = \dfrac{\pi}{6}$ and $x = \dfrac{5\pi}{6}$, and $\sin x = 1$ for $x = \dfrac{\pi}{2}$. The solutions are $\dfrac{\pi}{6}$, $\dfrac{\pi}{2}$ and $\dfrac{5\pi}{6}$.
The form $\sin 2x = 2\sin x\cos x$ of the double angle formula leaves a common factor $\cos x$. The equation is not divided by $\cos x$, which can be $0$: $$\begin{aligned}[t] \sin 2x = \sqrt{3}\cos x &\iff 2\sin x\cos x - \sqrt{3}\cos x = 0 && \text{(double angle formula)}\\ &\iff \cos x\,(2\sin x - \sqrt{3}) = 0 && \text{(common factor)}\\ &\iff \cos x = 0 \ \text{ or } \ \sin x = \frac{\sqrt{3}}{2}. && \text{(null factor law)} \end{aligned}$$ $\cos x = 0$ for $x = \dfrac{\pi}{2}$ and $x = \dfrac{3\pi}{2}$; $\sin x = \dfrac{\sqrt{3}}{2}$ for $x = \dfrac{\pi}{3}$ and $x = \dfrac{2\pi}{3}$. The solutions, in increasing order, are $\dfrac{\pi}{3}$, $\dfrac{\pi}{2}$, $\dfrac{2\pi}{3}$ and $\dfrac{3\pi}{2}$. Dividing by $\cos x$ would have lost $\dfrac{\pi}{2}$ and $\dfrac{3\pi}{2}$.

Example

Solve $\sec^2 x = 3\tan x - 1$ for $0 \le x \le 2\pi$, giving exact values. First use the Pythagorean identity for the secant to write the equation with $\tan x$ only, then solve it as an equation of quadratic type.

Answer

The equation is defined when $\cos x \neq 0$. The Pythagorean identity for the secant leaves only $\tan x$: $$\begin{aligned}[t] \sec^2 x = 3\tan x - 1 &\iff 1 + \tan^2 x = 3\tan x - 1 && \text{(Pythagorean identity for the secant)}\\ &\iff \tan^2 x - 3\tan x + 2 = 0 && \text{(every term on one side)}\\ &\iff (\tan x - 1)(\tan x - 2) = 0 && \text{(factorise)}\\ &\iff \tan x = 1 \ \text{ or } \ \tan x = 2. && \text{(null factor law)} \end{aligned}$$ $\tan x = 1$ for $x = \dfrac{\pi}{4}$ and $x = \dfrac{5\pi}{4}$. The value $2$ is not the tangent of a special angle: the solution of $\tan x = 2$ between $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$ is $\arctan 2$, which lies between $\dfrac{\pi}{4}$ and $\dfrac{\pi}{2}$ since $1 < 2$; the tangent has period $\pi$, so the other solution on the interval is $\pi + \arctan 2$. None of these values makes $\cos x = 0$. The solutions, in increasing order, are $\dfrac{\pi}{4}$, $\arctan 2$, $\dfrac{5\pi}{4}$ and $\pi + \arctan 2$.

Skills to practice

Solving Equations with the Secant, the Cosecant or the Cotangent with Technology
Solving Equations with the Secant, the Cosecant or the Cotangent
Solving Equations with a Double Angle

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Skills to practice

Reasoning and Solving Problems
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