-3

Day -3

Prep Trigonometry Essentials

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.

Step-by-Step SOP

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

Practice Exercises

Example 01Easy

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

\pi/6

. Find the lengths of the other two sides.

NEED A HINT?

Use

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

and

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

SHOW DETAILED EXPLANATION

Opposite Side

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

Adjacent Side

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

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)

Step-by-Step SOP

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

Practice Exercises

Example 01Easy

Evaluate

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

and

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

NEED A HINT?

Identify the quadrant and the reference angle.

SHOW DETAILED EXPLANATION

Step 1: Locate the Quadrant

\frac{7\pi}{6}

is slightly more than

\pi

(

6\pi/6

), so it is in Quadrant III.

Step 2: Apply ASTC

In QIII, both Sine and Cosine are negative.

Step 3: Reference Angle

The reference angle is

\pi/6

(

30^circ

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

and

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

Step 4: Final Result

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

and

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

Example 02Easy

Find the value of

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

NEED A HINT?

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

SHOW DETAILED EXPLANATION

Step 1: Quadrant

3\pi/4

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

Step 2: Values

\pi/4

, both magnitude are

\sqrt{2}/2

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

Step-by-Step SOP

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

Practice Exercises

Example 01Easy

Simplify the expression:

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

NEED A HINT?

Use the Pythagorean identity to replace the numerator.

SHOW DETAILED EXPLANATION

Step 1: Substitute

Since

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

, we know

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

Step 2: Reduce

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

Example 02Easy

Simplify the expression:

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

NEED A HINT?

Convert

\tan x

into its sine and cosine components.

SHOW DETAILED EXPLANATION

Step 1: Rewrite Tangent

Using the quotient identity,

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

Step 2: Simplify

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

Example 03Medium

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 DETAILED EXPLANATION

Step 1: Rearrange Identity

From the theorem

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

, subtract

\tan^2 x

from both sides.

Step 2: Conclusion

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

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

Example 04Medium

Simplify

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

NEED A HINT?

Convert both terms to

\sin x

and

\cos x

SHOW DETAILED EXPLANATION

Step 1: Use Reciprocal and Quotient Identities

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

and

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

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}

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

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.

Practice Exercises

Example 01Medium

Find the vertical asymptotes of

y = \tan x

on the interval

[0, 2\pi]

NEED A HINT?

Tangent is undefined where the denominator (Cosine) is zero.

SHOW DETAILED EXPLANATION

Step 1: Set Denominator to Zero

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

. Find where

\cos x = 0

Step 2: Solve on Interval

On the unit circle,

\cos x = 0

x = \pi/2

and

x = 3\pi/2

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}

Practice Exercises

Example 01Medium

Prove that

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

NEED A HINT?

Convert everything to sine and cosine first.

SHOW DETAILED EXPLANATION

Substitution

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

Common Denominator

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

Pythagorean Identity

Since

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

, we get

\frac{1}{\cos x}

, which is

\sec x

Example 02Hard

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 DETAILED EXPLANATION

Double-Angle Substitution

Replace

\cos 2x

with

2\cos^2 x - 1

. The equation becomes

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

Rearrange to Quadratic

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

Factor

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

. So

\cos x = 1/2

\cos x = 1

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

Practice Exercises

Example 01Medium

A triangle has sides

a=3, b=5

and angle

C=60^\circ

. Find the length of side

c

NEED A HINT?

Use the Law of Cosines.

SHOW DETAILED EXPLANATION

Formula

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

Calculation

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

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.

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 Exercises

Example 01Medium

Simplify

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

NEED A HINT?

\pi/2

an odd or even multiple? Which quadrant is it in?

SHOW DETAILED EXPLANATION

Function Change

Since

\pi/2

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

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

Quadrant Check

Assuming

\theta

is small,

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

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

Final Answer

Combining both:

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

Example 02Easy

Simplify

\sin(\pi - \theta)

NEED A HINT?

\pi

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

SHOW DETAILED EXPLANATION

Step 1: Function Type

\pi

is an even multiple of

\pi/2

, so Sine stays as Sine.

Step 2: Quadrant

\pi - \theta

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

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

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 Exercises

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})

NEED A HINT?

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

SHOW DETAILED EXPLANATION

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}

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}

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}}

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

Example 02Medium

Evaluate

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

NEED A HINT?

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.

SHOW DETAILED EXPLANATION

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}

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}}

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}

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}

Example 03Hard

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 DETAILED EXPLANATION

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

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)

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.

Final Answer

Combining all conditions, the domain is

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

Example 04Hard

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 DETAILED EXPLANATION

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

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.

Graph Comparison

f(x)

is a finite line segment from

(-1, -1)

(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 $[-1, 1]$ .

Gary Chang

Calculus Educator

5+ Years of Calculus Teaching Experience | AP Calculus Specialist. Dedicated to helping students master calculus through step-by-step logic and clear visualizations.

Discuss with Gary on Reddit

2. Special Angle Values

Angle Conversion

Practice Exercises

Opposite Side

Adjacent Side

Trigonometric Signs by Quadrant (ASTC)

Visualize the Circle

Practice Exercises

Step 1: Locate the Quadrant

Step 2: Apply ASTC

Step 3: Reference Angle

Step 4: Final Result

Step 1: Quadrant

Step 2: Values

Step 3: Solve

Look for Squares

Practice Exercises

Step 1: Substitute

Step 2: Reduce

Step 1: Rewrite Tangent

Step 2: Simplify

Step 1: Rearrange Identity

Step 2: Conclusion

Step 1: Use Reciprocal and Quotient Identities

Step 2: Divide Fractions

Step 3: Final Form

Point-Check Method

Practice Exercises

Step 1: Set Denominator to Zero

Step 2: Solve on Interval

Practice Exercises

Substitution

Common Denominator

Pythagorean Identity

Double-Angle Substitution

Rearrange to Quadratic

Factor

Final Angles

Practice Exercises

Formula

Calculation

Result

The 3-Step Process

Practice Exercises

Function Change

Quadrant Check

Final Answer

Step 1: Function Type

Step 2: Quadrant

Step 3: Result

The Triangle Method

Practice Exercises

Part 1: Arcsine Evaluation

Part 2: Arccos Evaluation

Part 3: Composite Function (Triangle Method)

Step 1: Set up the Angle

Step 2: Triangle Calculation

Step 3: Apply Double-Angle Formula

Step 4: Final Evaluation

Step 1: Arcsine Input Restriction

Step 2: Solve the Inequality

Step 3: Non-zero Denominator

Final Answer

Analyze f(x)

Analyze g(x)

Graph Comparison

Gary Chang