Trigonometry

• Convert $150^\circ$ to radians: $150 \times \frac{\pi}{180} = \frac{5\pi}{6}$ rad
• Convert $\frac{7\pi}{4}$ rad to degrees: $\frac{7\pi}{4} \times \frac{180}{\pi} = 315^\circ$
• Convert $1.2$ rad to degrees: $1.2 \times \frac{180}{\pi} \approx 68.75^\circ$

Arc Length

For an arc subtending an angle $\theta$ (in radians) at the centre of a circle of radius $r$:

This formula only applies when $\theta$ is in radians. If the angle is given in degrees, convert to Radians first. This formula is also covered in the context of circles and sectors.

Area of a Sector

For a sector of a circle of radius $r$ with central angle $\theta$ (in radians):

The area of a segment (the region between a chord and its arc) is:

• In a right-angled triangle, if the hypotenuse is $13$ and one leg is $5$ Then \sin\theta = \frac{5}{13}$$\cos\theta = \frac{12}{13}$$\tan\theta = \frac{5}{12}.
• Given $\tan\theta = 3$ and $\theta$ is acute: construct a triangle with opposite $= 3$ Adjacent $= 1$ Hypotenuse $= \sqrt{10}$. Then $\sin\theta = \frac{3}{\sqrt{10}}$ and $\cos\theta = \frac{1}{\sqrt{10}}$.
• If $\sec\theta = \frac{5}{3}$ and $\theta$ is in the first quadrant, then \cos\theta = \frac{3}{5}$$\sin\theta = \frac{4}{5} And $\tan\theta = \frac{4}{3}$.

Trigonometric Identities

Pythagorean Identity

For any angle $\theta$:

This follows directly from the Pythagorean identity applied to a Right-angled triangle with hypotenuse $1$. Two useful corollaries are obtained by dividing through by $\cos^2\theta$ or $\sin^2\theta$:

Quotient Identity

These hold whenever the denominators are non-zero.

Related Identities

Additional identities useful for simplification and solving equations:

Double angle formulas (extension for stronger students):

Angle sum and difference (extension):

• Evaluate $2\sin 30^\circ \cos 60^\circ = 2 \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2}$
• Evaluate $\sin^2 45^\circ + \cos^2 45^\circ = \frac{1}{2} + \frac{1}{2} = 1$ (confirms the Pythagorean identity)
• Evaluate $\frac{\tan 60^\circ}{\sin 30^\circ} = \frac{\sqrt{3}}{1/2} = 2\sqrt{3}$
• Simplify $\sin\frac{\pi}{3}\cos\frac{\pi}{6} + \cos\frac{\pi}{3}\sin\frac{\pi}{6} = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} \cdot \frac{1}{2} = \frac{3}{4} + \frac{1}{4} = 1$

Graphs of Trigonometric Functions

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Use the sliders to adjust amplitude, period, and phase shift, and observe how each parameter affects the graph.

$y = \sin x$

The graph is an odd function ($\sin(-x) = -\sin x$), symmetric about the origin.

$y = \cos x$

The graph is an even function ($\cos(-x) = \cos x$), symmetric about the $y$-axis.

$y = \tan x$

The graph is an odd function ($\tan(-x) = -\tan x$). The tangent function has vertical asymptotes where $\cos x = 0$.

General Transformations

For transformed trigonometric functions of the form:

• Solve $\sin\theta = \frac{1}{2}$ for $0^\circ \leq \theta < 360^\circ$: $\alpha = 30^\circ$. Solutions: $\theta = 30^\circ$ or $\theta = 180^\circ - 30^\circ = 150^\circ$.
• Solve $2\cos^2\theta - \cos\theta - 1 = 0$ for $0 \leq \theta < 2\pi$: Let $u = \cos\theta$. Then $2u^2 - u - 1 = 0 \implies (2u + 1)(u - 1) = 0$. So $\cos\theta = -\frac{1}{2}$ or $\cos\theta = 1$. Solutions: \theta = \frac{2\pi}{3}$$\theta = \frac{4\pi}{3}$$\theta = 0.
• Solve $\sin 2\theta = \cos\theta$ for $0^\circ \leq \theta < 360^\circ$: $2\sin\theta\cos\theta = \cos\theta \implies \cos\theta(2\sin\theta - 1) = 0$. Either $\cos\theta = 0 \implies \theta = 90^\circ, 270^\circ$ Or $\sin\theta = \frac{1}{2} \implies \theta = 30^\circ, 150^\circ$. Solutions: $30^\circ, 90^\circ, 150^\circ, 270^\circ$.
• Solve $\tan^2\theta = 3$ for $0 \leq \theta < 2\pi$: $\tan\theta = \pm\sqrt{3}$. $\tan\theta = \sqrt{3} \implies \theta = \frac{\pi}{3}, \frac{4\pi}{3}$. $\tan\theta = -\sqrt{3} \implies \theta = \frac{2\pi}{3}, \frac{5\pi}{3}$.

Applications

Sine Rule

For any triangle $ABC$ with sides a$$b$$c opposite the respective angles:

The sine rule is most useful when one side-side-angle (SSA) or angle-angle-side (AAS) configuration is known. In the ambiguous SSA case, two distinct triangles may satisfy the given conditions (the “ambiguous case”).

Cosine Rule

For any triangle $ABC$ with sides a$$b$$c opposite the respective angles:

The cosine rule generalises the Pythagorean theorem and is most useful for side-side-side (SSS) or Side-angle-side (SAS) configurations.

Area of a Triangle Using Trigonometry

This is derived from the standard formula $\mathrm{Area} = \frac{1}{2} \times \mathrm{base} \times \mathrm{height}$ Where the height is Expressed using a trigonometric ratio.

Bearings

A bearing is the angle measured clockwise from north. Key conventions:

• Bearings are always given as three-digit numbers from $000^\circ$ to $360^\circ$.
• The bearing of $B$ from $A$ is generally different from the bearing of $A$ from $B$ (they differ by $180^\circ$ unless the points are collinear with north).

Angles of Elevation and Depression

• Angle of elevation: the angle above the horizontal from the observer to the object.
• Angle of depression: the angle below the horizontal from the observer to the object.

When the observer and the object are at different heights, these two angles are equal if measured from corresponding horizontal lines (alternate angles).

3D Problems

Three-dimensional trigonometry problems often involve finding the angle between a line and a plane. The standard approach:

1. Identify the relevant right-angled triangle in the 3D figure.
2. Drop a perpendicular from a point to the plane to create a right angle.
3. The angle between the line and the plane is the angle between the line and its projection onto the plane. If $\alpha$ is the angle between the line and the normal to the plane, then the angle $\phi$ between the line and the plane satisfies $\phi = 90^\circ - \alpha$.

This connects to the vector formulation described in the 3D geometry section.

1. Question: Convert $210^\circ$ to radians and find the exact values of $\sin 210^\circ$ $\cos 210^\circ$ And $\tan 210^\circ$.

Details

• $210^\circ = 210 \times \frac{\pi}{180} = \frac{7\pi}{6}$ rad.
• Reference angle: $210^\circ - 180^\circ = 30^\circ$. Since $210^\circ$ is in the third quadrant, both sine and cosine are negative.
• $\sin 210^\circ = -\sin 30^\circ = -\frac{1}{2}$
• $\cos 210^\circ = -\cos 30^\circ = -\frac{\sqrt{3}}{2}$
• $\tan 210^\circ = \frac{\sin 210^\circ}{\cos 210^\circ} = \frac{-1/2}{-\sqrt{3}/2} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

1. Question: A sector of a circle of radius $8$ cm has an area of $32\pi$ cm$^2$. Find the arc Length and the perimeter of the sector.

• Let $u = \sin\theta$: $3u^2 - 2u - 1 = 0$.
• $(3u + 1)(u - 1) = 0 \implies u = -\frac{1}{3}$ or $u = 1$.
• Case 1: $\sin\theta = 1 \implies \theta = 90^\circ$.
• Case 2: $\sin\theta = -\frac{1}{3}$. Principal value $\alpha = \arcsin(-\frac{1}{3}) \approx -19.47^\circ$.
• $\theta = 360^\circ + (-19.47^\circ) = 340.53^\circ$ or $\theta = 180^\circ - (-19.47^\circ) = 199.47^\circ$.
• Solutions: $\theta \approx 90^\circ, 199.5^\circ, 340.5^\circ$.

1. Question: In \triangle ABC$$a = 10$$b = 7$$c = 8. Find the largest angle of the Triangle.

• Let the height be $h$ and let $B$ be $x$ m from the base. Then $A$ is $(x + 30)$ m from the base.
• $\tan 55^\circ = \frac{h}{x} \implies h = x\tan 55^\circ$.
• $\tan 40^\circ = \frac{h}{x + 30} \implies h = (x + 30)\tan 40^\circ$.
• Equating: $x\tan 55^\circ = (x + 30)\tan 40^\circ$.
• $x(\tan 55^\circ - \tan 40^\circ) = 30\tan 40^\circ$.
• $x = \frac{30\tan 40^\circ}{\tan 55^\circ - \tan 40^\circ} = \frac{30(0.8391)}{1.4281 - 0.8391} = \frac{25.17}{0.589} \approx 42.74$ m.
• $h = 42.74 \times \tan 55^\circ \approx 42.74 \times 1.4281 \approx 61.1$ m.

1. Question: A ship $S$ is observed from two lighthouses $A$ and $B$ which are $5$ km apart. The Bearing of $B$ from $A$ is $090^\circ$ (due east), the bearing of $S$ from $A$ is $050^\circ$ And The bearing of $S$ from $B$ is $320^\circ$. Find the distance of $S$ from $A$.

Starting from the LHS, multiply numerator and denominator by $(1 + \cos\theta)$:

$\frac{\sin\theta}{1 - \cos\theta} \cdot \frac{1 + \cos\theta}{1 + \cos\theta} = \frac{\sin\theta(1 + \cos\theta)}{1 - \cos^2\theta}$.

Since $1 - \cos^2\theta = \sin^2\theta$ (Pythagorean identity):

$= \frac{\sin\theta(1 + \cos\theta)}{\sin^2\theta} = \frac{1 + \cos\theta}{\sin\theta}$.

This equals the RHS, completing the ./1-number-and-algebra/3_proof-and-logic.

1. Question: A pyramid has a square base $ABCD$ of side $6$ cm and vertex $V$. The slant edges $VA = VB = VC = VD = 5\sqrt{2}$ cm. Find (a) the height of the pyramid, and (b) the angle between the slant edge $VA$ and the base.

• Amplitude $= 2$ Period $= 2\pi$ Phase shift $= \frac{\pi}{4}$ to the right, vertical shift $= 1$.
• Maximum $= 2(1) + 1 = 3$ Minimum $= 2(-1) + 1 = -1$.
• For $x$-intercepts: $2\sin\left(x - \frac{\pi}{4}\right) + 1 = 0 \implies \sin\left(x - \frac{\pi}{4}\right) = -\frac{1}{2}$.
• Let $u = x - \frac{\pi}{4}$. Since $x \in [0, 2\pi]$ We have $u \in \left[-\frac{\pi}{4}, \frac{7\pi}{4}\right]$.
• General solutions for $\sin u = -\frac{1}{2}$: $u = \frac{7\pi}{6} + 2k\pi$ or $u = \frac{11\pi}{6} + 2k\pi$.
• Also $u = -\frac{\pi}{6}$ (which equals $\frac{11\pi}{6} - 2\pi$) lies in $\left[-\frac{\pi}{4}, \frac{7\pi}{4}\right]$.
• $u = \frac{7\pi}{6}$ lies in the interval.
• $u = \frac{11\pi}{6}$ lies in the interval.
• $x = u + \frac{\pi}{4}$ So $x = -\frac{\pi}{6} + \frac{\pi}{4} = \frac{\pi}{12}$ $x = \frac{7\pi}{6} + \frac{\pi}{4} = \frac{17\pi}{12}$ $x = \frac{11\pi}{6} + \frac{\pi}{4} = \frac{25\pi}{12}$. Since $\frac{25\pi}{12} \gt 2\pi$ Only two $x$-intercepts lie in $[0, 2\pi]$.
• $x$-intercepts: $\left(\frac{\pi}{12}, 0\right)$ and $\left(\frac{17\pi}{12}, 0\right)$.

1. Question: Two observers $A$ and $B$ are on opposite sides of a vertical tower. $A$ and $B$ Are $100$ m apart on level ground. The angle of elevation of the top of the tower from $A$ is $30^\circ$ and from $B$ is $45^\circ$. Find the height of the tower.

• Using $\cos 2\theta = 1 - 2\sin^2\theta$: $1 - 2\sin^2\theta = \sin\theta$.
• $2\sin^2\theta + \sin\theta - 1 = 0$.
• Let $u = \sin\theta$: $2u^2 + u - 1 = 0 \implies (2u - 1)(u + 1) = 0$.
• $\sin\theta = \frac{1}{2}$ or $\sin\theta = -1$.
• $\sin\theta = \frac{1}{2} \implies \theta = \frac{\pi}{6}, \frac{5\pi}{6}$.
• $\sin\theta = -1 \implies \theta = \frac{3\pi}{2}$.
• Solutions: $\theta = \frac{\pi}{6}, \frac{3\pi}{2}, \frac{5\pi}{6}$.

1. Question: In \triangle ABC$$a = 5$$b = 7$$A = 40^\circ. Determine whether two Distinct triangles exist, and find all possible values of $B$.

Starting from the LHS, using $\cos 2\theta = 1 - 2\sin^2\theta = 2\cos^2\theta - 1$:

This equals the RHS. $\qed$

$BH = \sqrt{6^2 + 8^2 + 4^2} = \sqrt{36 + 64 + 16} = \sqrt{116} = 2\sqrt{29}$.

The projection of $BH$ onto the base $ABCD$ is the diagonal $BD = \sqrt{6^2 + 8^2} = 10$.

The angle between $BH$ and the base is $\phi$ where $\cos\phi = \dfrac{BD}{BH} = \dfrac{10}{2\sqrt{29}} = \dfrac{5}{\sqrt{29}}$.

Using $\cos 2\theta = 2\cos^2\theta - 1$:

Let $u = \cos\theta$: $2u^2 + 3u + 1 = 0 \implies (2u + 1)(u + 1) = 0$.

$u = -\dfrac{1}{2}$ or $u = -1$.

$\cos\theta = -\dfrac{1}{2} \implies \theta = \dfrac{2\pi}{3}, \dfrac{4\pi}{3}$.

$\cos\theta = -1 \implies \theta = \pi$.

Solutions: $\theta = \dfrac{2\pi}{3},\; \pi,\; \dfrac{4\pi}{3}$.

By the sine rule: $\dfrac{\sin B}{10} = \dfrac{\sin 40^\circ}{8}$.

Since $\sin B < 1$ and $b > a$ (i.e., $10 > 8$), there is exactly one solution: