# IB Mathematics (HL)/Trigonometry

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Topic 3: Circular functions and trigonometry discusses functions based on the circle, how trigonometric functions are derived, and various trigonometric identities and rules.

## 3.1 The circle and radian measure

Section content: The circle: radian measure of angles; length of an arc; area of a sector
* 1 radian is defined as the angle that is subtended by an arc of a length equal to the radius of the circle. 180° = π radians.
* The angle θ radians subtended by an arc of length l in a circle of radius r is defined as: $\theta ={\frac {l}{r}}$.
* The arc length l can be calculated using: $l=\theta r$, where θ is in radians or $l={\frac {\theta }{360}}(2\pi r)$, where θ is in degrees. These are derived from the formula for the circumference of a circle. (The former equation is included in the formula booklet)
* The area of a sector is calculated using: $A={\frac {1}{2}}\theta r^{2}$, where θ is in radians or $A={\frac {\theta }{360}}(\pi r^{2})$, where θ is in degrees. (The former equation is included in the formula booklet)

## 3.2 Trigonometry as circular functions

Section content: Definition of cos θ, sin θ and tan θ in terms of the unit circle; exact values of sin, cos and tan of 0, π/6, π/4, π/3, π/2 and their multiples; definition of the reciprocal trigonometric ratios sec θ, csc θ and cot θ; Pythagorean identities
Consider a unit circle (circle of radius 1) on a coordinate plane with centre at origin O(0, 0) and a point P(x, y) on its circumference. The angle θ is the angle made by OP from the positive x-axis and the angle α is the angle made by OP from the x-axis (either positive or negative). OP = 1 unit.
* In the first quadrant (0 < θ < π/2): $\sin {\theta }={\frac {y}{OP}}=y$, $\cos {\theta }={\frac {x}{OP}}=x$, and $\tan {\theta }={\frac {y}{x}}$. All ratios are positive.
* In the second quadrant (π/2 < θ < π): $\sin {\theta }=\sin {(\pi -\alpha )}={\frac {y}{1}}=y=\sin {\alpha }$, $\cos {\theta }=\cos {(\pi -\alpha )}={\frac {-x}{1}}=-x=-\cos {\alpha }$, and $\tan {\theta }=\tan {(\pi -\alpha )}={\frac {y}{-x}}=-{\frac {y}{x}}=-\tan {\alpha }$. Only the sine ratio is positive.
* In the third quadrant (π < θ < 3π/2): $\sin {\theta }=\sin {(\pi +\alpha )}={\frac {-y}{1}}=-y=-\sin {\alpha }$, $\cos {\theta }=\cos {(\pi +\alpha )}={\frac {-x}{1}}=-x=-\cos {\alpha }$, and $\tan {\theta }=\tan {(\pi +\alpha )}={\frac {-y}{-x}}={\frac {y}{x}}=\tan {\alpha }$. Only the tangent ratio is positive.
* In the fourth quadrant (3π/2 < θ < 2π): $\sin {\theta }=\sin {(2\pi -\alpha )}={\frac {-y}{1}}=-y=-\sin {\alpha }$, $\cos {\theta }=\cos {(2\pi -\alpha )}={\frac {x}{1}}=x=\cos {\alpha }$, and $\tan {\theta }=\tan {(2\pi -\alpha )}={\frac {-y}{x}}=-{\frac {y}{x}}=-\tan {\alpha }$. Only the cosine ratio is positive.
Tip: These relationships are not found in the formula booklet. However, you can use ASTC (all, sine, tangent, cosine) to make a phrase to make this relationship with the quadrants easier to remember; for example: "Add sugar to coffee".
A straight line passing through the origin O(0, 0) has the equation $y=\tan {\theta }$.
Exact values for acute angles are summarised in the following table. Paper 1 uses this knowledge and you may also be asked to use this in Paper 2.
| $\theta$ | $0$ | ${\frac {\pi }{6}}$ | ${\frac {\pi }{4}}$ | ${\frac {\pi }{3}}$ | ${\frac {\pi }{2}}$ |
| $\sin {\theta }$ | $0$ | ${\frac {1}{2}}$ | ${\frac {1}{\sqrt {2}}}$ | ${\frac {\sqrt {3}}{2}}$ | $1$ |
| $\cos {\theta }$ | $1$ | ${\frac {\sqrt {3}}{2}}$ | ${\frac {1}{\sqrt {2}}}$ | ${\frac {1}{2}}$ | $0$ |
| $\tan {\theta }$ | $0$ | ${\frac {1}{\sqrt {3}}}$ | $1$ | ${\sqrt {3}}$ | $\infty$ |
Tip: The exact values of sine and cosine ratios are the reverse of each other, so you only have to remember one set of values. Also tan θ is just (sin θ)/(cos θ).
The exact values for angles greater than π/2 radians can be derived by using the quadrant relationships above.
The reciprocal trigonometric functions are:
* $\sec {\theta }={\frac {1}{\cos {\theta }}}$
* $\csc {\theta }={\frac {1}{\sin {\theta }}}$
* $\cot {\theta }={\frac {1}{\tan {\theta }}}$
The first two of these are included in the formula booklet.
The Pythagorean identities are:
* $\cos ^{2}{\theta }+\sin ^{2}{\theta }=1$
* $1+\tan ^{2}{\theta }=\sec ^{2}{\theta }$
* $1+\cot ^{2}{\theta }=\csc ^{2}{\theta }$
All three of these are included in the formula booklet.

## 3.3 Compound and double angle identities

Section content: Compound angle identities; double angle identities
The compound angle identities expresses trigonometric ratios of the sum of two angles in terms of ratios of separate angles. These are:
* $\sin {(A\pm B)}=\sin {A}\cos {B}\pm \cos {A}\sin {B}$
* $\cos {(A\pm B)}=\cos {A}\cos {B}\mp \sin {A}\sin {B}$ (note that the signs between the terms are opposite of that in the original function)
* $\tan {(A\pm B)}={\frac {\tan {A}\pm \tan {B}}{1\mp \tan {A}\tan {B}}}$
The proofs of these have been explicitly excluded from the syllabus.
The double angle identities can be derived from the compound angle identities.
| Identity | Proof |
| $\sin {2\theta }=2\sin {\theta }\cos {\theta }$ | $\sin {(A+B)}=\sin {A}\cos {B}+\cos {A}\sin {B}$ $\sin {(\theta +\theta )}=\sin {\theta }\cos {\theta }+\cos {\theta }\sin {\theta }$ $\therefore \sin {2\theta }=2\sin {\theta }\cos {\theta }$ |
| $\cos {2\theta }=\cos ^{2}{\theta }-\sin ^{2}{\theta }$ $\cos {2\theta }=2\cos ^{2}{\theta }-1$ $\cos {2\theta }=1-2\sin ^{2}{\theta }$ | $\cos {(A+B)}=\cos {A}\cos {B}-\sin {A}\sin {B}$ $\cos {(\theta +\theta )}=\cos {\theta }\cos {\theta }-\sin {\theta }\sin {\theta }$ $\therefore \cos {2\theta }=\cos ^{2}{\theta }-\sin ^{2}{\theta }$ If we apply $\cos ^{2}{\theta }+\sin ^{2}{\theta }=1$ we can get another two equivalent expressions: $\cos {2\theta }=\cos ^{2}{\theta }-(1-\cos ^{2}{\theta })$ $\therefore \cos {2\theta }=2\cos ^{2}{\theta }-1$ ...and... $\cos {2\theta }=(1-\sin ^{2}{\theta })-\sin ^{2}{\theta }$ $\therefore \cos {2\theta }=1-2\sin ^{2}{\theta }$ |
| $\therefore \tan {(2\theta )}={\frac {2\tan {\theta }}{1-\tan ^{2}{\theta }}}$ | $\tan {(A+B)}={\frac {\tan {A}+\tan {B}}{1-\tan {A}\tan {B}}}$ $\tan {(\theta +\theta )}={\frac {\tan {\theta }+\tan {\theta }}{1-\tan {\theta }\tan {\theta }}}$ $\therefore \tan {(2\theta )}={\frac {2\tan {\theta }}{1-\tan ^{2}{\theta }}}$ |
All identities in this section are included in the formula booklet.

## 3.4 Composite trigonometric functions

Section content: Composite functions of the form f(x) = a sin(b(x + c)) + d Rewind: 2.3 - Transformation of graphs: translations; stretches; reflections in the axes
Consider a composite trigonometric function f where $f(x)=a\sin {(b(x+c))}+d$
* |a| is the amplitude, half the vertical distance between the maximum and minimum height. When a is negative the graph is also reflected along y = d.
* b is the number of cycles of the wave in the normal period (π radians for tangent, 2π radians for sine and cosine). The new period is hence the normal period divided by b i.e. in this case period = 2π/b.
* c is the phase shift, or horizontal shift (to the left or right). When c is positive the graph moves to the left; when c is negative the graph moves to the right.
* d is the vertical shift (up or down). When d is positive the graph moves up; when d is negative the graph moves down.
Tip: The phase shift c is sometimes confused with bc due to the common perception that b = 1. Examination questions may trick you by expanding the expression in the trigonometric function. For example $f(x)=\sin {(2x+8)}$, c is not 8 but instead 8 divided by 2, which is 4.

## 3.5 Inverse trigonometric functions

Section content: The inverse functions arcsin x, arccos x, arctan x; their domains and ranges; their graphs Rewind: 2.1 - Inverse function, including domain restriction; 2.3 - The graph of the inverse function as a reflection in y = x
* The inverse trigonometric functions (arcsin x, arccos x and arctan x) finds the angle given the trigonometric ratio (sin θ, cos θ and tan θ respectively); it effectively is the reverse of the standard trigonometric functions.
* From topic 2.1 you would have known that inverse functions will only exist for one-to-one functions and not for many-to-one functions, hence in technicality the inverse of a trigonometric function is not a function. However the domain of f(x) can be restricted so that it becomes a one-to-one function and hence has an inverse.
* The domain of $f(x)$ is the range of $f^{-1}(x)$, and the range of $f(x)$ is the domain of $f^{-1}(x)$. Hence the domain of arcsin x and arccos x is -1 ≤ x ≤ 1, while arctan x has the domain x: x exist for all real numbers.
* Alternative notation (especially on calculators; will also be used throughout the guide): $\arcsin {x}=\sin ^{-1}{x};\arccos {x}=\cos ^{-1}{x};\arctan {x}=\tan ^{-1}{x}$
* The graphs of the inverse functions can be obtained by first restricting the domains to -π/2 ≤ x ≤ π/2, then reflecting the graph in y = x.
* You should know the identity for sum of arctan functions
* And that arcsin(x) + arccos(x) = π/2

## 3.6 Solutions to trigonometric equations

Section content: Algebraic and graphical methods of solving trigonometric equations in a finite interval, including the use of trigonometric identities and factorisation Rewind: 2.6 - Solving polynomial equations both graphically and algebraically; use of technology to solve a variety of equations, including those where there is no appropriate analytic approach
Trigonometric equations can be solved through algebraic and graphical means. To solve such an equation algebraically:
* The key idea is to have all trigonometric ratios reduced to the same basic one: either sine or cosine (or tangent if algebra forces it to sin θ/cos θ). Use the trigonometric identities to help you do so. If this is not possible it may be easier to use the graphical method of solutions.
* Usually this will reduce to a single linear equation. However it can also be reduced to a polynomial whereby x is replaced with a trigonometric function. In this case solve as you would with a standard polynomial equation e.g. by factorisation, quadratic formula or the binomial theorem. You could also use the graphing function of your GDC to solve the equation when it is too difficult to solve manually.
* After reducing to the form sin θ = x, check the domain - what do they want your answers to be in between? Use the quadrant relationships in 3.2 to help you solve the equation for that domain. For example, if the domain is π/2 ≤ θ ≤ 3π/2, then your solutions would be $\theta =\pi -\sin ^{-1}{x},\pi +\sin ^{-1}{x}$.
To solve graphically, plot the left hand side of the equation as one graph and right hand side of the equation as another graph on the same pair of axes. Make sure that your GDC is on the correct angle mode (degrees or radians) - assume radians unless stated by the exam paper. The intersection points will be the solutions to the equation.

## 3.7 Sine and cosine rules

Section content: The cosine rule; the sine rule including the ambiguous case; area of a triangle as 1/2 ab sin C
The sine and cosine rules relate the sides and angles in any triangle.
Consider a triangle ABC where a = BC, b = AC and c = AB.
* If you are given the lengths of any two sides a and b and the angle C between those two sides (the included angle), then you can use the cosine rule to calculate the length of the remaining side c: $c^{2}=a^{2}+b^{2}-2ab\cos {C}$. A specific case of this is the Pythagoras' theorem for right-angled triangles, where C = π/2 radians or 90 degrees, then $c^{2}=a^{2}+b^{2}$.
* If you are given the lengths of all three sides a, b and c, then you can use the cosine rule to calculate any angle within the triangle; for angle C this is: $\cos {C}={\frac {a^{2}+b^{2}-c^{2}}{2ab}}$.
* If you are given any two angles A and B and the length a side not in between these two angles - either a or b, then you can use the sine rule to calculate one of the missing sides: ${\frac {a}{\sin {A}}}={\frac {b}{\sin {B}}}={\frac {c}{\sin {C}}}$.
* If you are given the lengths of any two sides a and b, and the angle A or B that is not in between the two sides, then this is known as the ambiguous case as the sine rule: ${\frac {\sin {A}}{a}}={\frac {\sin {B}}{b}}={\frac {\sin {C}}{c}}$ will result in two possible sizes for one of the missing angles: $A=\sin ^{-1}{\frac {a\sin {B}}{b}},\pi -\sin ^{-1}{\frac {a\sin {B}}{b}}$. In this case, determine which one of the angles is not possible from the information given in the question.
The area of any triangle, given the lengths of any two sides a and b and the included angle C, is: $Area={\frac {1}{2}}ab\sin {C}$
These rules are given in the formula booklet.