Trigonometric basics

This chapter establishes the foundational principles of trigonometry, encompassing fundamental identities, quadrant sign conventions, reduction formulas, and standard trigonometric values. A thorough understanding of these concepts is indispensable for addressing various problems in the CMI examination and serves as a prerequisite for advanced topics in trigonometry and complex numbers.

---

Chapter Contents

|

| Topic |

|---|-------| | 1 | Fundamental identities | | 2 | Signs in quadrants | | 3 | Reduction formulas | | 4 | Standard values |

---

We begin with Fundamental identities.

Part 1: Fundamental identities

Fundamental Identities

Overview

Fundamental identities are the backbone of all trigonometric simplification. These identities connect the six trigonometric functions and let us move between different forms of an expression. In exam problems, they are used not only for direct simplification, but also for proving equations, solving identities, and reducing complicated expressions to a single function. ---

Learning Objectives

---

Core Identities

---

Why These Identities Matter

---

Standard Simplification Moves

---

Minimal Worked Examples

Example 1 Simplify $\qquad \dfrac{1-\cos^2\theta}{\sin\theta}$ Using $\qquad 1-\cos^2\theta=\sin^2\theta$, we get $\qquad \dfrac{\sin^2\theta}{\sin\theta}=\sin\theta$ provided $\sin\theta\ne 0$. --- Example 2 Simplify $\qquad \sec^2\theta-\tan^2\theta$ Using $\qquad \sec^2\theta=1+\tan^2\theta$, we get $\qquad \sec^2\theta-\tan^2\theta=1$ ---

Identity-Proof Strategy

---

Common Mistakes

---

CMI Strategy

---

Practice Questions

:::question type="MCQ" question="Which of the following is always equal to $1$?" options=["$\sin^2\theta+\cos^2\theta$","$\tan^2\theta+\cot^2\theta$","$\sec^2\theta-\csc^2\theta$","$\sin\theta+\cos\theta$"] answer="A" hint="Recall the basic Pythagorean identity." solution="The standard identity is $\qquad \sin^2\theta+\cos^2\theta=1$ Hence the correct option is $\boxed{A}$." ::: :::question type="NAT" question="Simplify $\sec^2\theta-\tan^2\theta$." answer="1" hint="Use $1+\tan^2\theta=\sec^2\theta$." solution="Using $\qquad \sec^2\theta=1+\tan^2\theta$, we get $\qquad \sec^2\theta-\tan^2\theta=1$ So the answer is $\boxed{1}$." ::: :::question type="MSQ" question="Which of the following are true wherever both sides are defined?" options=["$\tan\theta=\dfrac{\sin\theta}{\cos\theta}$","$1+\tan^2\theta=\sec^2\theta$","$1+\cot^2\theta=\csc^2\theta$","$\sin^2\theta-\cos^2\theta=1$"] answer="A,B,C" hint="Three are standard identities and one is false." solution="1. True.

True.

True.

False. The correct basic identity is

$\qquad \sin^2\theta+\cos^2\theta=1$ Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="Prove that $\dfrac{1-\cos^2\theta}{\sin\theta}=\sin\theta$ wherever both sides are defined." answer="$\sin\theta$" hint="Use $1-\cos^2\theta=\sin^2\theta$." solution="Using the Pythagorean identity $\qquad \sin^2\theta+\cos^2\theta=1$, we get $\qquad 1-\cos^2\theta=\sin^2\theta$ Therefore $\qquad \dfrac{1-\cos^2\theta}{\sin\theta}=\dfrac{\sin^2\theta}{\sin\theta}$ where $\sin\theta\ne 0$. So $\qquad \dfrac{\sin^2\theta}{\sin\theta}=\sin\theta$ Hence the identity is proved wherever both sides are defined." ::: ---

Summary

---

---

Part 2: Signs in quadrants

Signs in Quadrants

Overview

Quadrant-based sign reasoning is one of the first structural ideas in trigonometry. It tells us when $\sin\theta$, $\cos\theta$, and $\tan\theta$ are positive or negative, and it is essential in simplification, equation solving, inverse-trigonometric reasoning, and graph interpretation. In CMI-style questions, this topic is usually not asked as a memory test alone; it is used inside angle constraints, coordinate interpretation, and sign-based elimination. ---

Learning Objectives

---

Core Idea

---

Quadrant Sign Table

---

Reciprocal Functions

---

Signs on the Axes

Examples:

• at $\theta=0^\circ$, $\sin\theta=0$, $\cos\theta=1$

• at $\theta=90^\circ$, $\sin\theta=1$, $\cos\theta=0$

---

Signs from the Unit Circle

---

Minimal Worked Examples

Example 1 If $\theta$ is in Quadrant II, determine the signs of $\sin\theta,\cos\theta,\tan\theta$. In Quadrant II:

• $\sin\theta>0$

• $\cos\theta<0$

• $\tan\theta<0$

--- Example 2 Suppose $\cos\theta<0$ and $\tan\theta>0$. Which quadrant can $\theta$ lie in? For $\tan\theta>0$, sine and cosine must have the same sign. Since $\cos\theta<0$, we must also have $\sin\theta<0$. So $\theta$ lies in Quadrant III. ---

Standard Elimination Logic

This is often faster than drawing a full diagram. ---

Common Patterns

---

Common Mistakes

---

CMI Strategy

---

Practice Questions

:::question type="MCQ" question="If $\theta$ lies in Quadrant III, then which of the following is positive?" options=["$\sin\theta$","$\cos\theta$","$\tan\theta$","$\sec\theta$"] answer="C" hint="Recall the sign pattern in Quadrant III." solution="In Quadrant III, $\qquad \sin\theta<0,\ \cos\theta<0$ so $\qquad \tan\theta=\dfrac{\sin\theta}{\cos\theta}>0$. Hence the correct option is $\boxed{C}$." ::: :::question type="NAT" question="If $\sin\theta>0$ and $\cos\theta<0$, then $\theta$ lies in quadrant number __." answer="2" hint="Use the sign table." solution="Positive sine means the angle is above the $x$-axis. Negative cosine means the angle is in the left half-plane. The only quadrant satisfying both is Quadrant II. Therefore the answer is $\boxed{2}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["In Quadrant II, $\sin\theta>0$","In Quadrant IV, $\cos\theta>0$","In Quadrant III, $\tan\theta<0$","If $\sec\theta<0$, then $\cos\theta<0$"] answer="A,B,D" hint="Use the quadrant sign chart and reciprocal-sign rule." solution="1. True. Sine is positive in Quadrant II.

True. Cosine is positive in Quadrant IV.

False. Tangent is positive in Quadrant III because sine and cosine are both negative.

True. $\sec\theta$ has the same sign as $\cos\theta$.

Hence the correct answer is $\boxed{A,B,D}$." ::: :::question type="SUB" question="Suppose $\cos\theta<0$ and $\tan\theta>0$. Determine the quadrant of $\theta$ and the sign of $\sin\theta$." answer="Quadrant III and $\sin\theta<0$" hint="Use the fact that $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$." solution="Since $\qquad \tan\theta>0$, the sine and cosine of $\theta$ must have the same sign. We are also given $\qquad \cos\theta<0$. Therefore $\sin\theta$ must also be negative. So both sine and cosine are negative, which means $\theta$ lies in Quadrant III. Hence the answer is: $\qquad \boxed{\text{Quadrant III and }\sin\theta<0}$." ::: ---

Summary

---

---

Part 3: Reduction formulas

Reduction Formulas

Overview

Reduction formulas in elementary trigonometry are the identities that reduce angles like $\pi-\theta$, $\pi+\theta$, $\dfrac{\pi}{2}-\theta$, and $2\pi-\theta$ to expressions involving just $\theta$. These formulas are essential for evaluating trigonometric functions in different quadrants and for simplifying exact-value problems. In exam problems, the main challenge is controlling the sign correctly. ---

Learning Objectives

---

Core Reduction Formulas

---

Periodicity

These formulas are often used before reduction. ::: ---

Quadrant Sign Rule

This is the quickest way to check signs after reducing the angle. ::: ---

Related-Angle Rule

---

Minimal Worked Examples

Example 1 Evaluate $\qquad \sin\left(\pi-\dfrac{\pi}{6}\right)$ Using $\qquad \sin(\pi-\theta)=\sin\theta$, we get $\qquad \sin\left(\pi-\dfrac{\pi}{6}\right)=\sin\dfrac{\pi}{6}=\dfrac{1}{2}$ --- Example 2 Evaluate $\qquad \cos\left(\dfrac{\pi}{2}+\theta\right)$ Using the reduction formula, $\qquad \cos\left(\dfrac{\pi}{2}+\theta\right)=-\sin\theta$ ---

Common Structure

---

Common Mistakes

---

CMI Strategy

---

Practice Questions

:::question type="MCQ" question="Which of the following is equal to $\cos(\pi-\theta)$?" options=["$\cos\theta$","$-\cos\theta$","$\sin\theta$","$-\sin\theta$"] answer="B" hint="Use the reduction formula for $\pi-\theta$." solution="The standard identity is $\qquad \cos(\pi-\theta)=-\cos\theta$ Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="Find the value of $\sin\left(\pi-\dfrac{\pi}{4}\right)$." answer="sqrt(2)/2" hint="Use $\sin(\pi-\theta)=\sin\theta$." solution="Using $\qquad \sin(\pi-\theta)=\sin\theta$, we get $\qquad \sin\left(\pi-\dfrac{\pi}{4}\right)=\sin\dfrac{\pi}{4}=\dfrac{\sqrt{2}}{2}$ So the answer is $\boxed{\dfrac{\sqrt{2}}{2}}$." ::: :::question type="MSQ" question="Which of the following are true?" options=["$\sin(\pi+\theta)=-\sin\theta$","$\cos(\pi+\theta)=-\cos\theta$","$\tan(\pi+\theta)=\tan\theta$","$\sin\left(\dfrac{\pi}{2}-\theta\right)=\sin\theta$"] answer="A,B,C" hint="Three are standard reduction formulas and one is false." solution="1. True.

True.

True.

False. The correct identity is

$\qquad \sin\left(\dfrac{\pi}{2}-\theta\right)=\cos\theta$ Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="Simplify $\cos\left(\dfrac{\pi}{2}+\theta\right)\sin(\pi-\theta)+\tan(\pi+\theta)$." answer="$-\sin^2\theta+\tan\theta$" hint="Reduce each term separately." solution="Use standard reduction formulas: $\qquad \cos\left(\dfrac{\pi}{2}+\theta\right)=-\sin\theta$ $\qquad \sin(\pi-\theta)=\sin\theta$ $\qquad \tan(\pi+\theta)=\tan\theta$ So the expression becomes $\qquad (-\sin\theta)(\sin\theta)+\tan\theta$ $\qquad = -\sin^2\theta+\tan\theta$ Hence the simplified form is $\boxed{-\sin^2\theta+\tan\theta}$." ::: ---

Summary

---

---

Part 4: Standard values

Standard Values

Overview

Standard trigonometric values are the exact values of trigonometric functions at the most important special angles. These values appear everywhere: simplification, equation solving, geometric ratios, inverse trigonometry, and graph interpretation. In CMI-style problems, the real challenge is not memorizing a table, but knowing why the values are what they are and how to use them together with quadrant information. ---

Learning Objectives

---

Core Angles

---

Exact Values of Sine and Cosine

---

Exact Values of Tangent

---

Reciprocal Values

---

Where the Values Come From

---

Reference-Angle Method

---

Minimal Worked Examples

Example 1 Find $\sin 45^\circ$ and $\cos 45^\circ$. From the isosceles right triangle, $\qquad \sin 45^\circ=\cos 45^\circ=\dfrac{\sqrt2}{2}$ --- Example 2 Find $\tan 120^\circ$. Reference angle is $60^\circ$. In Quadrant II, tangent is negative. So $\qquad \tan 120^\circ=-\tan 60^\circ=-\sqrt3$ ---

Standard Identity Checks

---

Common Patterns

---

Common Mistakes

---

CMI Strategy

---

Practice Questions

:::question type="MCQ" question="The value of $\sin 60^\circ$ is" options=["$\dfrac12$","$\dfrac{\sqrt2}{2}$","$\dfrac{\sqrt3}{2}$","$\sqrt3$"] answer="C" hint="Use the $30^\circ\!-\!60^\circ\!-\!90^\circ$ triangle." solution="From the standard-angle table, $\qquad \sin 60^\circ=\dfrac{\sqrt3}{2}$. Hence the correct option is $\boxed{C}$." ::: :::question type="NAT" question="Find $\cos 45^\circ$." answer="sqrt(2)/2" hint="Use the $45^\circ\!-\!45^\circ\!-\!90^\circ$ triangle." solution="In a $45^\circ\!-\!45^\circ\!-\!90^\circ$ triangle, the side ratio is $\qquad 1:1:\sqrt2$. So $\qquad \cos 45^\circ=\dfrac{1}{\sqrt2}=\dfrac{\sqrt2}{2}$ Therefore the answer is $\boxed{\dfrac{\sqrt2}{2}}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["$\sin 30^\circ=\dfrac12$","$\cos 60^\circ=\dfrac12$","$\tan 45^\circ=1$","$\tan 90^\circ=0$"] answer="A,B,C" hint="Use the exact-value table." solution="1. True.

True.

True.

False. $\tan 90^\circ$ is undefined because $\cos 90^\circ=0$.

Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="Evaluate $\sin 150^\circ+\cos 150^\circ+\tan 135^\circ$ exactly." answer="$-\dfrac{\sqrt3}{2}-\dfrac12$" hint="Use reference angles and quadrant signs." solution="Use reference angles: $\qquad \sin 150^\circ=\sin 30^\circ=\dfrac12$ $\qquad \cos 150^\circ=-\cos 30^\circ=-\dfrac{\sqrt3}{2}$ $\qquad \tan 135^\circ=-\tan 45^\circ=-1$ So the sum is $\qquad \dfrac12-\dfrac{\sqrt3}{2}-1 = -\dfrac{\sqrt3}{2}-\dfrac12$ Hence the exact value is $\qquad \boxed{-\dfrac{\sqrt3}{2}-\dfrac12}$." ::: ---

Summary

Chapter Summary

Chapter Review Questions

:::question type="MCQ" question="If $\sin \theta = 3/5$ and $\theta$ lies in the second quadrant, what is the value of $\tan(\pi/2 + \theta)$?" options=["$3/4$","$-3/4$","$4/3$","$-4/3$"] answer="$4/3$" hint="First determine $\cos \theta$ using the Pythagorean identity, considering the quadrant. Then apply the reduction formula for $\tan(\pi/2 + \theta)$." solution="Given $\sin \theta = 3/5$ and $\theta$ is in the second quadrant.
In the second quadrant, $\cos \theta$ is negative.
Using $\sin^2 \theta + \cos^2 \theta = 1$:
$\cos \theta = -\sqrt{1 - \sin^2 \theta} = -\sqrt{1 - (3/5)^2} = -\sqrt{1 - 9/25} = -\sqrt{16/25} = -4/5$.
The reduction formula for $\tan(\pi/2 + \theta)$ is $-\cot \theta$.
$\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{-4/5}{3/5} = -4/3$.
Therefore, $\tan(\pi/2 + \theta) = -(-4/3) = 4/3$."
:::

:::question type="NAT" question="Evaluate the expression $4 \sin(\pi/6) \cos(\pi/3) \tan(\pi/4)$." answer="1" hint="Recall the standard trigonometric values for common angles." solution="The standard values are:
$\sin(\pi/6) = 1/2$
$\cos(\pi/3) = 1/2$
$\tan(\pi/4) = 1$
Substitute these values into the expression:
$4 \times (1/2) \times (1/2) \times 1 = 4 \times (1/4) \times 1 = 1$."
:::

:::question type="MCQ" question="If $\tan \alpha > 0$ and $\cos \alpha < 0$, in which quadrant does the angle $\alpha$ lie?" options=["First Quadrant","Second Quadrant","Third Quadrant","Fourth Quadrant"] answer="Third Quadrant" hint="Apply the ASTC rule to determine where tangent is positive and where cosine is negative." solution="For $\tan \alpha > 0$, $\alpha$ must lie in the First Quadrant or the Third Quadrant.
For $\cos \alpha < 0$, $\alpha$ must lie in the Second Quadrant or the Third Quadrant.
For both conditions to be satisfied simultaneously, $\alpha$ must lie in the Third Quadrant."
:::

:::question type="NAT" question="Simplify the expression $\frac{\sin(\pi - \theta)}{\cos(\pi/2 - \theta)}$." answer="1" hint="Apply the relevant reduction formulas for $\sin(\pi - \theta)$ and $\cos(\pi/2 - \theta)$." solution="Using reduction formulas:
$\sin(\pi - \theta) = \sin \theta$
$\cos(\pi/2 - \theta) = \sin \theta$
Substituting these into the expression:
$\frac{\sin \theta}{\sin \theta} = 1$ (assuming $\sin \theta \neq 0$)"
:::

What's Next?