Chapter 8 — Introduction to Trigonometry Notes

1. What is Trigonometry?

Trigonometry deals with the relationship between the angles and sides of a triangle, especially a right-angled triangle.

It is useful for finding quantities such as heights, distances and angles that cannot be measured directly.

For an acute angle $A$ in a right triangle:

• Hypotenuse → side opposite the $90^\circ$ angle; it is the longest side.
• Opposite side → side facing angle $A$.
• Adjacent side → side next to angle $A$, other than the hypotenuse.

Important: Opposite and adjacent depend on which angle you are considering.

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2. Six Trigonometric Ratios

For angle $A$:$$\boxed{\sin A=\frac{\text{Opposite}}{\text{Hypotenuse}}}$$$$\boxed{\cos A=\frac{\text{Adjacent}}{\text{Hypotenuse}}}$$$$\boxed{\tan A=\frac{\text{Opposite}}{\text{Adjacent}}}$$

The remaining three are reciprocals:$$\boxed{\cosec A=\frac{1}{\sin A} =\frac{\text{Hypotenuse}}{\text{Opposite}}}$$$$\boxed{\sec A=\frac{1}{\cos A} =\frac{\text{Hypotenuse}}{\text{Adjacent}}}$$$$\boxed{\cot A=\frac{1}{\tan A} =\frac{\text{Adjacent}}{\text{Opposite}}}$$

Useful relationships:$$\boxed{\tan A=\frac{\sin A}{\cos A}}$$$$\boxed{\cot A=\frac{\cos A}{\sin A}}$$

Easy memory aid

SOH–CAH–TOA

• Sin = Opposite / Hypotenuse
• Cos = Adjacent / Hypotenuse
• Tan = Opposite / Adjacent

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3. Ratios Depend Only on the Angle

If two right triangles have the same acute angle, their corresponding sides are proportional because the triangles are similar.

Therefore, the value of a trigonometric ratio does not depend on the actual size of the triangle. It depends only on the angle.

This is a fundamental idea behind trigonometry.

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4. Finding Other Ratios from One Ratio

If one ratio is given, construct a suitable right triangle and use Pythagoras’ theorem to find the missing side.

Example pattern

If$$\sin A=\frac13$$

then$$\frac{\text{Opposite}}{\text{Hypotenuse}}=\frac13$$

Take:$$\text{Opposite}=k,\qquad \text{Hypotenuse}=3k$$

Using Pythagoras:$$\text{Adjacent}^2=(3k)^2-k^2=8k^2$$

so$$\text{Adjacent}=2\sqrt2k$$

Hence:$$\cos A=\frac{2\sqrt2}{3}$$

and the remaining ratios can similarly be obtained.

Key restriction

For acute angles:$$\boxed{0<\sin A\leq1,\qquad 0<\cos A\leq1}$$

because the hypotenuse is the longest side.

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5. Standard Trigonometric Values

These values are extremely important for calculations.

Angle	$0^\circ$	$30^\circ$	$45^\circ$	$60^\circ$	$90^\circ$
$\sin A$	$0$	$\frac12$	$\frac1{\sqrt2}$	$\frac{\sqrt3}{2}$	$1$
$\cos A$	$1$	$\frac{\sqrt3}{2}$	$\frac1{\sqrt2}$	$\frac12$	$0$
$\tan A$	$0$	$\frac1{\sqrt3}$	$1$	$\sqrt3$	Not defined
$\cosec A$	Not defined	$2$	$\sqrt2$	$\frac2{\sqrt3}$	$1$
$\sec A$	$1$	$\frac2{\sqrt3}$	$\sqrt2$	$2$	Not defined
$\cot A$	Not defined	$\sqrt3$	$1$	$\frac1{\sqrt3}$	$0$

Pattern worth remembering

As $A$ increases from $0^\circ$ to $90^\circ$:$$\boxed{\sin A\text{ increases from }0\text{ to }1}$$

while$$\boxed{\cos A\text{ decreases from }1\text{ to }0}$$

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6. How the Standard Values Are Obtained

$45^\circ$

A right triangle with angles $45^\circ,45^\circ,90^\circ$ has its two perpendicular sides equal.

If each is $a$, then:$$H=a\sqrt2$$

Therefore:$$\boxed{\sin45^\circ=\cos45^\circ=\frac1{\sqrt2}}$$$$\boxed{\tan45^\circ=1}$$

$30^\circ$ and $60^\circ$

They can be obtained by dividing an equilateral triangle into two equal right triangles.

The resulting important values are:$$\boxed{\sin30^\circ=\frac12,\quad \cos30^\circ=\frac{\sqrt3}{2},\quad \tan30^\circ=\frac1{\sqrt3}}$$

and$$\boxed{\sin60^\circ=\frac{\sqrt3}{2},\quad \cos60^\circ=\frac12,\quad \tan60^\circ=\sqrt3}$$

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7. Ratios at $0^\circ$ and $90^\circ$

At the limiting values:

At $0^\circ$

$$\boxed{\sin0^\circ=0}$$$$\boxed{\cos0^\circ=1}$$$$\boxed{\tan0^\circ=0}$$$$\boxed{\sec0^\circ=1}$$

But:$$\boxed{\cosec0^\circ,\ \cot0^\circ\text{ are not defined}}$$

because their denominators become zero.

At $90^\circ$

$$\boxed{\sin90^\circ=1}$$$$\boxed{\cos90^\circ=0}$$

Consequently:$$\boxed{\tan90^\circ,\ \sec90^\circ\text{ are not defined}}$$

while$$\boxed{\cosec90^\circ=1,\quad\cot90^\circ=0}$$

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8. Solving Right-Triangle Problems

A practical method:

Step 1

Identify the given angle.

Step 2

Label the sides as:

• Opposite
• Adjacent
• Hypotenuse

Step 3

Choose the ratio containing the known and required sides.

Step 4

Substitute the known trigonometric value.

Step 5

Solve for the unknown.

For example, if a side is opposite $30^\circ$ and the hypotenuse is known:$$\sin30^\circ=\frac{\text{opposite}}{\text{hypotenuse}}$$

This approach is illustrated in the chapter’s worked examples.

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9. Trigonometric Identities

An identity is an equation that remains true for every value of the angle for which the expressions are defined.

The three most important identities in this chapter are:

Identity 1

$$\boxed{\sin^2 A+\cos^2 A=1}$$

Identity 2

$$\boxed{1+\tan^2 A=\sec^2 A}$$

or$$\boxed{\sec^2 A-\tan^2 A=1}$$

Identity 3

$$\boxed{1+\cot^2 A=\cosec^2 A}$$

or$$\boxed{\cosec^2 A-\cot^2 A=1}$$

These come directly from applying Pythagoras’ theorem and dividing by an appropriate side squared.

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10. Reciprocal Relationships

Memorise these:$$\boxed{\sin A=\frac1{\cosec A}}$$$$\boxed{\cos A=\frac1{\sec A}}$$$$\boxed{\tan A=\frac1{\cot A}}$$

and therefore:$$\boxed{\cosec A=\frac1{\sin A},\quad \sec A=\frac1{\cos A},\quad \cot A=\frac1{\tan A}}$$

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11. Converting One Ratio into Others

If one ratio is known, identities allow the remaining ratios to be found.

For example, if$$\tan A=\frac1{\sqrt3}$$

then$$\cot A=\sqrt3$$

and$$\sec^2A=1+\tan^2A =1+\frac13 =\frac43$$

so$$\sec A=\frac2{\sqrt3}$$

and$$\cos A=\frac{\sqrt3}{2}.$$

The same strategy works when any other ratio is given.

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12. Important Rules for Proving Identities

When proving a trigonometric identity:

1. Work on one side at a time, preferably the more complicated side.
2. Convert everything to sin⁡\sin and cos⁡\cos when useful.
3. Use: $$\sin^2A+\cos^2A=1$$
4. Or use: $$1+\tan^2A=\sec^2A$$ and $$1+\cot^2A=\cosec^2A.$$
5. Use reciprocal relations when necessary.
6. Simplify until the two sides become identical.

The chapter demonstrates this technique through several identity proofs.

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13. Common Traps

Don’t confuse:

$$\boxed{\cosec A\neq \sin^{-1}A}$$

Here,$$\cosec A=(\sin A)^{-1}$$

whereas $\sin^{-1}A$ represents an inverse-trigonometric function in higher mathematics.

Similarly, sin⁡A\sin A means “sine of angle $A$”; it is not multiplication of “sin” by $A$.

Also remember:

• $\tan90^\circ$ is not defined.
• $\sec90^\circ$ is not defined.
• $\cot0^\circ$ is not defined.
• $\cosec0^\circ$ is not defined.
• $\sin A$ and $\cos A$ cannot exceed $1$.

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Formula Sheet

$$\boxed{\sin A=\frac OH} \qquad \boxed{\cos A=\frac AH} \qquad \boxed{\tan A=\frac OA}$$$$\boxed{\cosec A=\frac HO} \qquad \boxed{\sec A=\frac HA} \qquad \boxed{\cot A=\frac AO}$$$$\boxed{\tan A=\frac{\sin A}{\cos A}} \qquad \boxed{\cot A=\frac{\cos A}{\sin A}}$$$$\boxed{\sin^2A+\cos^2A=1}$$$$\boxed{1+\tan^2A=\sec^2A}$$$$\boxed{1+\cot^2A=\cosec^2A}$$

Standard values to memorise

$$\boxed{ \begin{array}{c|ccccc} A&0^\circ&30^\circ&45^\circ&60^\circ&90^\circ\\ \hline \sin A&0&\frac12&\frac1{\sqrt2}&\frac{\sqrt3}{2}&1\\ \cos A&1&\frac{\sqrt3}{2}&\frac1{\sqrt2}&\frac12&0\\ \tan A&0&\frac1{\sqrt3}&1&\sqrt3&\text{ND} \end{array}}$$