-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 $[-1, 1]$ .

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