Day -3: Right Triangle Definitions & Special Angles & The Unit Circle & Radian Mastery & Fundamental Identities & Graphs of Trig Functions & Addition, Double-Angle & Half-Angle Formulas & Law of Sines & Cosines & Reduction Formulas & Inverse Trig Functions (Arcsine/Arccos/Arctan)

Right Triangle Definitions & Special Angles

Before the unit circle, we define trig ratios using right triangles (SOH-CAH-TOA) and memorize key values used in nearly every Calculus problem.

Core Theorem

1. Right Triangle (Acute Angles)

* \sin \theta = \frac{Opposite}{Hypotenuse}

*\cos \theta = \frac{Adjacent}{Hypotenuse}

*\tan \theta = \frac{Opposite}{Adjacent}

*\cot \theta = \frac{Adjacent}{Opposite}, \sec \theta = \frac{Hypotenuse}{Adjacent}, \csc \theta = \frac{Hypotenuse}{Opposite}

2. Special Angle Values

Angle (rad)	sinθ	cosθ	tanθ	secθ
0	0	1	0	1
π/6 (30°)	1/2	√3/2	1/√3	2/√3
π/4 (45°)	√2/2	√2/2	1	√2
π/3 (60°)	√3/2	1/2	√3	2
π/2 (90°)	1	0	und.	und.

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Easy

A right triangle has a hypotenuse of 10 and an angle of

\pi/6

. Find the lengths of the other two sides.

How to think about it

Key idea: Use $\sin(\pi/6) = 1/2$ and $\cos(\pi/6) = \sqrt{3}/2$ .

Full solution

1. Opposite Side

10 \cdot \sin(\pi/6) = 10 \cdot (1/2) = 5

.

2. Adjacent Side

10 \cdot \cos(\pi/6) = 10 \cdot (\sqrt{3}/2) = 5\sqrt{3}

.

Step-by-Step SOP

1
Angle Conversion
Always convert degrees to radians first: $\text{rad} = \text{deg} \cdot \frac{\pi}{180}$ .

Common Pitfalls

⚠
The 'Undefined' TrapAt $\pi/2$ ( $90^\circ$ ), $\cos\theta = 0$ . Since $\tan\theta = \sin/\cos$ and $\sec\theta = 1/\cos$ , both are undefined (X) at this angle.

The Unit Circle & Radian Mastery

The foundation of all trig in Calculus. Coordinates on the unit circle are $(\cos \theta, \sin \theta)$ .

Core Theorem

The ASTC Rule (All Students Take Calculus): In Quadrant I, All are positive; QII, Sine is positive; QIII, Tangent is positive; QIV, Cosine is positive.

Trigonometric Signs by Quadrant (ASTC)

Quadrant	sin θ	cos θ	tan θ	Positive Functions
I (0 to π/2)	＋	＋	＋	All
II (π/2 to π)	＋	－	－	Sin (csc)
III (π to 3π/2)	－	－	＋	Tan (cot)
IV (3π/2 to 2π)	－	＋	－	Cos(sec)

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Easy

Evaluate

\sin(\frac{7\pi}{6})

and

\cos(\frac{7\pi}{6})

.

How to think about it

Key idea: Identify the quadrant and the reference angle.

Full solution

1. Step 1: Locate the Quadrant

\frac{7\pi}{6}

is slightly more than

\pi

(

6\pi/6

), so it is in Quadrant III.

2. Step 2: Apply ASTC

In QIII, both Sine and Cosine are negative.

3. Step 3: Reference Angle

The reference angle is

\pi/6

(

30^circ

).

\sin(\pi/6) = 1/2

and

\cos(\pi/6) = \sqrt{3}/2

.

4. Step 4: Final Result

\sin(7\pi/6) = -1/2

and

\cos(7\pi/6) = -\sqrt{3}/2

.

Step-by-Step SOP

1
Visualize the Circle
Always sketch a quick circle to confirm the +/- sign before writing an answer.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Easy

Find the value of

\tan(\frac{3\pi}{4})

.

Need a hint?

\tan \theta = \frac{\sin \theta}{\cos \theta}

.

Show solution

1. Step 1: Quadrant

3\pi/4

is in QII. Sine is (+), Cosine is (-).

2. Step 2: Values

At

\pi/4

, both magnitude are

\sqrt{2}/2

.

3. Step 3: Solve

\tan(3\pi/4) = \frac{\sqrt{2}/2}{-\sqrt{2}/2} = -1

.

Common Pitfalls

⚠
Degree TrapCalculus formulas like $\frac{d}{dx}\sin x = \cos x$ ONLY work in radians. If you use degrees, your answers will be off by a factor of $\pi/180$ .
⚠
Mixing up Cos and SinRemember: Alphabetical order. (x, y) corresponds to (cos, sin). C comes before S.

Fundamental Identities

The most common identities used to simplify complex derivatives and integrals.

Core Theorem

1. Reciprocal Identities:
-

\csc x = \frac{1}{\sin x}, \quad \sec x = \frac{1}{\cos x}, \quad \cot x = \frac{1}{\tan x}

2. Pythagorean Identities:
-

\sin^2 x + \cos^2 x = 1

-

1 + \tan^2 x = \sec^2 x

-

1 + \cot^2 x = \csc^2 x

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Easy

Simplify the expression:

f(x) = \frac{1 - \cos^2 x}{\sin x}

.

How to think about it

Key idea: Use the Pythagorean identity to replace the numerator.

Full solution

1. Step 1: Substitute

Since

\sin^2 x + \cos^2 x = 1

, we know

1 - \cos^2 x = \sin^2 x

.

2. Step 2: Reduce

\frac{\sin^2 x}{\sin x} = \sin x

.

Worked Example 02Easy

Simplify the expression:

f(x) = \tan x \cdot \cos x

.

How to think about it

Key idea: Convert $\tan x$ into its sine and cosine components.

Full solution

1. Step 1: Rewrite Tangent

Using the quotient identity,

\tan x = \frac{\sin x}{\cos x}

.

2. Step 2: Simplify

\frac{\sin x}{\cos x} \cdot \cos x = \sin x

.

Step-by-Step SOP

1
Look for Squares
Whenever you see trig functions squared, look for a Pythagorean substitution.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

Show that

\sec^2 x - \tan^2 x = 1

.

Need a hint?

Start with the Pythagorean identity

1 + \tan^2 x = \sec^2 x

.

Show solution

1. Step 1: Rearrange Identity

From the theorem

1 + \tan^2 x = \sec^2 x

, subtract

\tan^2 x

from both sides.

2. Step 2: Conclusion

1 = \sec^2 x - \tan^2 x

. This is a very useful identity for simplifying integrals involving secant and tangent.

Practice 02Medium

Simplify

g(x) = \frac{\csc x}{\cot x}

.

Need a hint?

Convert both terms to

\sin x

and

\cos x

.

Show solution

1. Step 1: Use Reciprocal and Quotient Identities

\csc x = \frac{1}{\sin x}

and

\cot x = \frac{\cos x}{\sin x}

.

2. Step 2: Divide Fractions

\frac{1}{\sin x} \div \frac{\cos x}{\sin x} = \frac{1}{\sin x} \cdot \frac{\sin x}{\cos x} = \frac{1}{\cos x}

.

3. Step 3: Final Form

Using the reciprocal identity:

\frac{1}{\cos x} = \sec x

.

Common Pitfalls

⚠
The Square Notation $\sin^2 x$ means $(\sin x)^2$ . It is NOT the same as $\sin(x^2)$ .

Graphs of Trig Functions

Understanding the 'shape' of functions is key for limits and continuity.

Core Theorem

Sine and Cosine are continuous everywhere and oscillate between -1 and 1. Tangent has vertical asymptotes at

x = \frac{\pi}{2} + n\pi

.

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

Find the vertical asymptotes of

y = \tan x

on the interval

[0, 2\pi]

.

How to think about it

Key idea: Tangent is undefined where the denominator (Cosine) is zero.

Full solution

1. Step 1: Set Denominator to Zero

\tan x = \frac{\sin x}{\cos x}

. Find where

\cos x = 0

.

2. Step 2: Solve on Interval

On the unit circle,

\cos x = 0

at

x = \pi/2

and

x = 3\pi/2

.

Step-by-Step SOP

1
Point-Check Method
If you forget the graph, plot values for $0, \pi/2, \pi$ to quickly reconstruct the shape.

Common Pitfalls

⚠
Amplitude vs. PeriodDon't confuse vertical stretch with horizontal frequency. In $\sin(2x)$ , the '2' makes it go twice as fast, shortening the period to $\pi$ .

Addition, Double-Angle & Half-Angle Formulas

These formulas are the 'power tools' of Calculus, used to transform complex trig integrals into manageable forms.

Core Theorem

1. Addition/Subtraction:
-

\sin(\alpha \pm \beta) = \sin\alpha\cos\beta \pm \cos\alpha\sin\beta

-

\cos(\alpha \pm \beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta

2. Double-Angle:
-

\sin 2\alpha = 2\sin\alpha\cos\alpha

-

\cos 2\alpha = \cos^2\alpha - \sin^2\alpha = 2\cos^2\alpha - 1 = 1 - 2\sin^2\alpha

3. Half-Angle (Power Reduction):
-

\sin^2 \frac{\alpha}{2} = \frac{1-\cos\alpha}{2}

-

\cos^2 \frac{\alpha}{2} = \frac{1+\cos\alpha}{2}

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

Prove that

\tan x \sin x + \cos x = \sec x

.

How to think about it

Key idea: Convert everything to sine and cosine first.

Full solution

1. Substitution

\frac{\sin x}{\cos x} \cdot \sin x + \cos x = \frac{\sin^2 x}{\cos x} + \cos x

.

2. Common Denominator

\frac{\sin^2 x + \cos^2 x}{\cos x}

.

3. Pythagorean Identity

Since

\sin^2 x + \cos^2 x = 1

, we get

\frac{1}{\cos x}

, which is

\sec x

.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Hard

Solve

2 + \cos 2x = 3 \cos x

for

x \in [0, 2\pi]

.

Need a hint?

Use the double-angle formula to create a quadratic equation in terms of

\cos x

.

Show solution

1. Double-Angle Substitution

Replace

\cos 2x

with

2\cos^2 x - 1

. The equation becomes

2 + (2\cos^2 x - 1) = 3 \cos x

.

2. Rearrange to Quadratic

2\cos^2 x - 3\cos x + 1 = 0

.

3. Factor

(2\cos x - 1)(\cos x - 1) = 0

. So

\cos x = 1/2

or

\cos x = 1

.

4. Final Angles

x = \pi/3, 5\pi/3

(from

1/2

) and

x = 0, 2\pi

(from

1

).

Common Pitfalls

⚠
The cos2x ChoiceThere are three versions of the $\cos 2x$ formula. Always pick the one that matches the other trig terms in your equation to simplify things.
⚠
The $\cos(\alpha \pm \beta)$ SignWatch out! The cosine addition formula flips the sign: $\cos(A+B)$ uses a minus ( $-$ ), and $\cos(A-B)$ uses a plus ( $+$ ).

Law of Sines & Cosines

Used in 'Related Rates' problems where triangles are not necessarily right-angled.

Core Theorem

1. Law of Sines:

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

.
2. Law of Cosines:

c^2 = a^2 + b^2 - 2ab \cos C

.
3. Area:

\text{Area} = \frac{1}{2}ab \sin C

.

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

A triangle has sides

a=3, b=5

and angle

C=60^\circ

. Find the length of side

c

.

How to think about it

Key idea: Use the Law of Cosines.

Full solution

1. Formula

c^2 = 3^2 + 5^2 - 2(3)(5)\cos(60^\circ)

.

2. Calculation

c^2 = 9 + 25 - 30(0.5) = 34 - 15 = 19

.

3. Result

c = \sqrt{19}

.

Reduction Formulas

How to shift any trig function back to the first quadrant using the 'Odd-Even' rule.

Core Theorem

1. If using

n\pi \pm \theta

(even multiples of

\pi/2

): The function remains the same.
2. If using

(n + 1/2)\pi \pm \theta

(odd multiples of

\pi/2

): The function changes to its co-function (sin

\leftrightarrow

cos, tan

\leftrightarrow

cot).
3. Sign Rule: The

\pm

sign is determined by the quadrant of the original function.

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

Simplify

\cos(\frac{\pi}{2} + \theta)

.

How to think about it

Key idea: Is $\pi/2$ an odd or even multiple? Which quadrant is it in?

Full solution

1. Function Change

Since

\pi/2

is

1 \times \frac{\pi}{2}

(odd), Cosine changes to its co-function: Sine.

2. Quadrant Check

Assuming

\theta

is small,

\frac{\pi}{2} + \theta

is in Quadrant II. In QII, the *original* function (Cosine) is **negative**.

3. Final Answer

Combining both:

\cos(\frac{\pi}{2} + \theta) = -\sin \theta

.

Step-by-Step SOP

1
The 3-Step Process
1. Identify the multiple of $\pi/2$ . 2. Determine if the function changes. 3. Use ASTC to find the sign of the original function in the target quadrant.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Easy

Simplify

\sin(\pi - \theta)

.

Need a hint?

\pi

is

2 \times \frac{\pi}{2}

.

Show solution

1. Step 1: Function Type

\pi

is an even multiple of

\pi/2

, so Sine stays as Sine.

2. Step 2: Quadrant

\pi - \theta

is in Quadrant II. In QII, Sine is **positive**.

3. Step 3: Result

\sin(\pi - \theta) = \sin \theta

.

Inverse Trig Functions (Arcsine/Arccos/Arctan)

To make trig functions invertible, we restrict their domains. Inverse trig functions return the angle $\theta$ for a given value $x$ .

Core Theorem

1. Domain and Range (Principal Values):
-

\sin^{-1} x: [-1, 1] \to [-\frac{\pi}{2}, \frac{\pi}{2}]

-

\cos^{-1} x: [-1, 1] \to [0, \pi]

-

\tan^{-1} x: (-\infty, \infty) \to (-\frac{\pi}{2}, \frac{\pi}{2})

-

\sec^{-1} x: (-\infty, -1] \cup [1, \infty) \to [0, \frac{\pi}{2}) \cup [\pi, \frac{3\pi}{2})

2. Cancellation Identities
-

\sin(\sin^{-1} x) = x

for

x \in [-1, 1]

-

\sin^{-1}(\sin x) = x

ONLY if

x \in [-\frac{\pi}{2}, \frac{\pi}{2}]

3. Negative Arguments:
-

\sin^{-1}(-x) = -\sin^{-1} x

-

\cos^{-1}(-x) = \pi - \cos^{-1} x

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

Evaluate the following expressions:
(1)

\sin^{-1}(\frac{\sqrt{3}}{2})

(2)

\cos^{-1}(-\frac{1}{2})

(3)

\tan(\sin^{-1} \frac{1}{3})

How to think about it

Key idea: For (1) and (2), find the angle in the restricted range. For (3), draw a reference right triangle.

Full solution

1. Part 1: Arcsine Evaluation

For

\sin^{-1}(\frac{\sqrt{3}}{2})

, we need

\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]

such that

\sin \theta = \frac{\sqrt{3}}{2}

.
Since

\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}

and

\frac{\pi}{3}

is in the valid range,

\sin^{-1}(\frac{\sqrt{3}}{2}) = \frac{\pi}{3}

.

2. Part 2: Arccos Evaluation

For

\cos^{-1}(-\frac{1}{2})

, we need

\theta \in [0, \pi]

such that

\cos \theta = -\frac{1}{2}

.
We know

\cos(\frac{\pi}{3}) = \frac{1}{2}

. In Quadrant II (where

\cos

is negative), the angle is

\pi - \frac{\pi}{3} = \frac{2\pi}{3}

.
Since

\frac{2\pi}{3}

is in

[0, \pi]

,

\cos^{-1}(-\frac{1}{2}) = \frac{2\pi}{3}

.

3. Part 3: Composite Function (Triangle Method)

Let

\theta = \sin^{-1}(\frac{1}{3})

, so

\sin \theta = \frac{1}{3}

.
Draw a right triangle with Opposite side

= 1

and Hypotenuse

= 3

.
Adjacent side

= \sqrt{3^2 - 1^2} = \sqrt{8} = 2\sqrt{2}

.
Therefore,

\tan(\sin^{-1} \frac{1}{3}) = \tan \theta = \frac{\text{opp}}{\text{adj}} = \frac{1}{2\sqrt{2}}

or

\frac{\sqrt{2}}{4}

.

Worked Example 02Medium

Evaluate

\sec(2\tan^{-1}\frac{1}{3})

.

How to think about it

Key idea: Let $\theta = \tan^{-1}\frac{1}{3}$ . Use the double-angle formula $\cos 2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta}$ or find $\sin\theta, \cos\theta$ via a triangle.

Full solution

1. Step 1: Set up the Angle

Let

\theta = \tan^{-1}\frac{1}{3}

, which means

\tan \theta = \frac{1}{3}

. We need to find

\sec(2\theta)

, which is equivalent to

\frac{1}{\cos 2\theta}

.

2. Step 2: Triangle Calculation

Draw a right triangle with Opposite

= 1

and Adjacent

= 3

.
Hypotenuse

= \sqrt{1^2 + 3^2} = \sqrt{10}

.
Thus,

\sin \theta = \frac{1}{\sqrt{10}}

and

\cos \theta = \frac{3}{\sqrt{10}}

.

3. Step 3: Apply Double-Angle Formula

Using

\cos 2\theta = \cos^2 \theta - \sin^2 \theta

:

\cos 2\theta = (\frac{3}{\sqrt{10}})^2 - (\frac{1}{\sqrt{10}})^2 = \frac{9}{10} - \frac{1}{10} = \frac{8}{10} = \frac{4}{5}

.

4. Step 4: Final Evaluation

Since

\sec 2\theta = \frac{1}{\cos 2\theta}

, we have:

\sec(2\tan^{-1}\frac{1}{3}) = \frac{1}{4/5} = \frac{5}{4}

.

Step-by-Step SOP

1
The Triangle Method
For composite functions like $\sin(\cos^{-1} x)$ , always draw a right triangle to find the missing side lengths.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Hard

Find the domain of

f(x) = (\sin^{-1}(x^{-1}))^{-1}

.

Need a hint?

Analyze the inner term

1/x

for Arcsine's domain, then ensure the final denominator is not zero.

Show solution

1. Step 1: Arcsine Input Restriction

For

\sin^{-1}(\frac{1}{x})

to be defined, its input must satisfy:

-1 \le \frac{1}{x} \le 1

.

2. Step 2: Solve the Inequality

The inequality

|\frac{1}{x}| \le 1

is equivalent to

|x| \ge 1

(provided

x \neq 0

).
This gives

x \in (-\infty, -1] \cup [1, \infty)

.

3. Step 3: Non-zero Denominator

The entire expression is a fraction

1/\sin^{-1}(\frac{1}{x})

, so the denominator cannot be zero:

\sin^{-1}(\frac{1}{x}) \neq 0 \implies \frac{1}{x} \neq \sin(0) = 0

.
Since

\frac{1}{x}

can never be

0

for any real

x

, this condition is always satisfied within the existing domain.

4. Final Answer

Combining all conditions, the domain is

(-\infty, -1] \cup [1, \infty)

.

Practice 02Hard

Compare

f(x) = \sin(\sin^{-1} x)

and

g(x) = \sin^{-1}(\sin x)

.

Need a hint?

Focus on the domain of the INNER function first.

Show solution

1. Analyze f(x)

For

f(x) = \sin(\sin^{-1} x)

, the inner function

\sin^{-1} x

is only defined for

x \in [-1, 1]

. Within this domain, they cancel out perfectly:

f(x) = x

.

2. Analyze g(x)

For

g(x) = \sin^{-1}(\sin x)

, the inner function

\sin x

is defined for all

x \in \mathbb{R}

. However, the outer function

\sin^{-1}

always outputs values in

[-\frac{\pi}{2}, \frac{\pi}{2}]

.
Thus,

g(x) = x

ONLY when

x \in [-\frac{\pi}{2}, \frac{\pi}{2}]

; for other

x

, it creates a 'sawtooth' pattern.

3. Graph Comparison

f(x)

is a finite line segment from

(-1, -1)

to

(1, 1)

.

g(x)

is a continuous periodic zigzag wave.

Common Pitfalls

⚠
Notation Confusion $(\sin x)^{-1}$ means $\frac{1}{\sin x}$ (Cosecant), whereas $\sin^{-1} x$ is the inverse function (Arcsine). They are completely different!
⚠
Domain ErrorInputting $x = 2$ into $\sin^{-1}(x)$ will result in an error because the domain is restricted to $[-1, 1]$ .

2. Special Angle Values

Worked Examples

1. Opposite Side

2. Adjacent Side

Angle Conversion

Trigonometric Signs by Quadrant (ASTC)

Worked Examples

1. Step 1: Locate the Quadrant

2. Step 2: Apply ASTC

3. Step 3: Reference Angle

4. Step 4: Final Result

Visualize the Circle

Practice — Your Turn

1. Step 1: Quadrant

2. Step 2: Values

3. Step 3: Solve

Worked Examples

1. Step 1: Substitute

2. Step 2: Reduce

1. Step 1: Rewrite Tangent

2. Step 2: Simplify

Look for Squares

Practice — Your Turn

1. Step 1: Rearrange Identity

2. Step 2: Conclusion

1. Step 1: Use Reciprocal and Quotient Identities

2. Step 2: Divide Fractions

3. Step 3: Final Form

Worked Examples

1. Step 1: Set Denominator to Zero

2. Step 2: Solve on Interval

Point-Check Method

Worked Examples

1. Substitution

2. Common Denominator

3. Pythagorean Identity

Practice — Your Turn

1. Double-Angle Substitution

2. Rearrange to Quadratic

3. Factor

4. Final Angles

Worked Examples

1. Formula

2. Calculation

3. Result

Worked Examples

1. Function Change

2. Quadrant Check

3. Final Answer

The 3-Step Process

Practice — Your Turn

1. Step 1: Function Type

2. Step 2: Quadrant

3. Step 3: Result

Worked Examples

1. Part 1: Arcsine Evaluation

2. Part 2: Arccos Evaluation

3. Part 3: Composite Function (Triangle Method)

1. Step 1: Set up the Angle

2. Step 2: Triangle Calculation

3. Step 3: Apply Double-Angle Formula

4. Step 4: Final Evaluation

The Triangle Method

Practice — Your Turn

1. Step 1: Arcsine Input Restriction

2. Step 2: Solve the Inequality

3. Step 3: Non-zero Denominator

4. Final Answer

1. Analyze f(x)

2. Analyze g(x)

3. Graph Comparison

Gary Chang