Class 11 Mathematics Chapter 3 Trigonometric Functions

Key Concept: Origin of Term

c) Measuring the sides of a triangle

[Solution Description] The word ‘trigonometry’ comes from the Greek words ‘trigon’, meaning triangle, and ‘metron’, meaning measure. This term reflects the primary objective of trigonometry: measuring the sides of a triangle.

Key Concept: Basic Identities

c) $-\cos x$

[Solution Description] According to the basic trigonometric identity, we know that the cosine of an angle plus $\pi$ is equal to the negative cosine of that angle: $\cos(\pi + x) = -\cos x$. This can be directly applied to get the answer.

Key Concept: Basic Values

b) $\frac{1}{2}$

[Solution Description] The cosine of $60^\circ$ can be directly recalled from standard trigonometric values. For $\cos 60^\circ$, the value is $\frac{1}{2}$.

Key Concept: Reciprocal Functions

c) $2$

[Solution Description]

$$\csc(x) = \frac{1}{\sin(x)}$$

The sine of $\pi/6$ is known:

$\sin(\pi/6) = \frac{1}{2}.$

Using the reciprocal identity,

$\csc(\pi/6) = \frac{1}{\sin(\pi/6)} = \frac{1}{1/2} = 2.$

Thus, the value of $\csc(\pi/6)$ is $2$.

Key Concept: Basic Identity

a) $$sin x \cdot cos y + cos x \cdot sin y$$

[Solution Description] The formula for the sine of the sum of two angles is given as:

$$sin(x + y) = sin x \cdot cos y + cos x \cdot sin y$$

This directly follows from the trigonometric identity.

Key Concept: Radian Measure

c) $2\pi$

[Solution Description] A complete revolution corresponds to $360^\circ$. In radians, a full circle measures $2\pi$.

Key Concept: Direct Calculation

a) $\frac{\sqrt{3}}{2}$

[Solution Description] Using the identity $\cos x + \cos y = 2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}$:

$$\cos 90^\circ + \cos 30^\circ = 2 \cos \left(\frac{90^\circ+30^\circ}{2}\right) \cos \left(\frac{90^\circ-30^\circ}{2}\right)$$

Simplifying further,

$$= 2 \cos(60^\circ) \cos(30^\circ)$$

$$= 2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2}$$

$$= \frac{\sqrt{3}}{2}$$

So the result is $\frac{\sqrt{3}}{2}$.

Key Concept: Angle Definition

a) Initial side

[Solution Description] The initial side of an angle is the fixed ray from which the rotation begins.

Key Concept: Function Range

b) [-1, 1]

[Solution Description]

The sine function oscillates between -1 and 1. Therefore, over any interval spanning full cycles such as $[-\pi, \pi]$, its complete range can be covered, which is $[-1, 1]$.

Key Concept: Quadrant Signs

b) Second

[Solution Description]

The sine function is positive in the first and second quadrants, while the cosine function is negative in the second and third quadrants. Therefore, the only quadrant where $\sin x$ is positive and $\cos x$ is negative is the second quadrant.

Key Concept: Angle Values

b) 1

[Solution Description] At $45^\circ$, both sine and cosine have the same value, leading to $\tan 45^\circ = \frac{\sin 45^\circ}{\cos 45^\circ} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1$.

Key Concept: Identity Recognition

c) $\cos 3x = 4 \cos^3 x – 3 \cos x$

[Solution Description]

The identity for $\cos 3x$ is given by: $\cos 3x = 4 \cos^3 x – 3 \cos x$.

Key Concept: Degree to Radian

c) $\frac{\pi}{4}$

[Solution Description] To convert degrees to radians, use the formula $Radian = \frac{\pi}{180} \times Degree$. So for 45°, we have:

$$Radian = \frac{\pi}{180} \times 45 = \frac{\pi}{4}.$$

The radian measure of 45° is $\frac{\pi}{4}$.

Key Concept: Degree to Radian Formula

d) $\frac{\pi}{4}$ radians

[Solution Description]

To convert degrees to radians, we use the formula $1 \text{ degree} = \frac{\pi}{180} \text{ radians}$. Therefore, for 45°, the radian measure will be:

$$45^\circ \times \frac{\pi}{180} = \frac{\pi}{4} \text{ radians}$$

Key Concept: Basic Definition

b) Measuring the sides of a triangle

[Solution Description] The word ‘trigonometry’ is derived from the Greek words ‘trigon’ and ‘metron’, which mean ‘measuring the sides of a triangle’.

Key Concept: Zero Values

c) $\pi$

[Solution Description] The sine function $\sin(x)$ equals zero at angles $x = n\pi$, where $n$ is an integer. Therefore, the smallest positive angle where this occurs is $x = \pi$.

Key Concept: Quadrantal Angles

b) $90^{\circ}$

[Solution Description] A quadrantal angle is an angle that is an integral multiple of $\frac{\pi}{2}$. Here, we will evaluate each option to determine which one satisfies this condition.

Option A: $45^{\circ}$ = $\frac{\pi}{4}$ radians.

Option B: $90^{\circ}$ = $\frac{\pi}{2}$ radians.

Option C: $60^{\circ}$ = $\frac{\pi}{3}$ radians.

Option D: $30^{\circ}$ = $\frac{\pi}{6}$ radians.

The only option that is an integral multiple of $\frac{\pi}{2}$ is Option B ($90^{\circ}$), which is indeed a quadrantal angle.

Key Concept: Basic Identities

a) $\sin^2 x + \cos^2 x = 1$

[Solution Description] The identity $\sin^2 x + \cos^2 x = 1$ is one of the fundamental trigonometric identities, which holds true for any real value of $x$. This identity comes from the Pythagorean theorem applied to the unit circle.

Key Concept: Radian Definition

a) 1 radian

[Solution Description]

By definition, the radian measure of an angle subtended by an arc of length 1 unit in a unit circle is 1 radian.

Key Concept: Quadrant Signs

c) Quadrant III

[Solution Description] To determine where $\tan x$ is positive and $\sin x$ is negative, consider the signs of trigonometric functions in each quadrant:

– Quadrant I: All functions positive

– Quadrant II: Sine positive, others negative

– Quadrant III: Tangent positive, others negative

– Quadrant IV: Cosine positive, others negative

Therefore, in Quadrant III, $\tan x$ is positive and $\sin x$ is negative.

Key Concept: Periodicity

b) $2\pi$

[Solution Description] The periodicity of a trigonometric function is the smallest positive value for which the function repeats its values. Both $\sec(x)$ and $cosec(x)$ have a period of $2\pi$. Therefore, the period of $y = cosec(x)$ is $2\pi$.

Key Concept: Basic Identity

b) $-0.34$

[Solution Description] Using the identity $\cos(x+y) = \cos x \cos y – \sin x \sin y$, we substitute the given values:

$$\cos(x+y) = (0.5)(0.6) – (0.8)(0.8)$$

Calculate step by step:

$$\cos(x+y) = 0.3 – 0.64 = -0.34$$

Thus, $\cos(x+y) = -0.34$.

Key Concept: Simplification

b) $\frac{1 – \cos(x)}{2}$

[Solution Description] The half-angle identity for sine squared is:

$$\sin^2\left(\frac{x}{2}\right) = \left(\pm \sqrt{\frac{1 – \cos(x)}{2}}\right)^2 = \frac{1 – \cos(x)}{2}$$

Therefore, $\sin^2(\frac{x}{2})$ simplifies to:

$$\frac{1 – \cos(x)}{2}$$

Key Concept: Quadrant Analysis

b) Second Quadrant

[Solution Description] $\sin x$ is positive in the first and second quadrants. $\tan x$ is negative in the second quadrant. Thus, both conditions are satisfied only in the second quadrant.

Key Concept: Angle Sum and Difference

c) $\frac{\sqrt{6} + \sqrt{2}}{4}$

[Solution Description]

The angle difference formula for cosine is: $\cos(a – b) = \cos a \cdot \cos b + \sin a \cdot \sin b$. Let’s choose $a = 45^\circ$ and $b = 30^\circ$. So,

$$\cos(75^\circ) = \cos(45^\circ – 30^\circ) = \cos 45^\circ \cdot \cos 30^\circ + \sin 45^\circ \cdot \sin 30^\circ$$

Calculating each term:

$\cos 45^\circ = \frac{\sqrt{2}}{2}$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$

$\sin 45^\circ = \frac{\sqrt{2}}{2}$, $\sin 30^\circ = \frac{1}{2}$

$$\cos(75^\circ) = \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}$$

Key Concept: Function Signs

b) Negative

[Solution Description]

The angle $210^\circ$ lies in the third quadrant where the sine function is negative. Thus, $\sin(210^\circ)$ is negative.

Key Concept: Angle Definition

b) Counterclockwise

[Solution Description] Positive rotation for an angle is defined as counterclockwise according to standard mathematical convention.

Key Concept: Unit Circle Basics

b) 1

[Solution Description]

On the unit circle, the sine of $\frac{\pi}{2}$ radians (which corresponds to $90^\circ$) is at the topmost point on the unit circle. This point has coordinates $(0, 1)$. Therefore, $\sin(\frac{\pi}{2})$ is equal to 1.

Key Concept: Quadrantal Angles

c) Undefined

[Solution Description]

To find $\tan \frac{3\pi}{2}$, we use the relation $\tan x = \frac{\sin x}{\cos x}$. At $x = \frac{3\pi}{2}$, $\sin \frac{3\pi}{2} = -1$ and $\cos \frac{3\pi}{2} = 0$. Therefore, $\tan \frac{3\pi}{2} = \frac{-1}{0}$ which is undefined.

Key Concept: Angle Definition

c) The ray after rotation

[Solution Description]

The terminal side of an angle is the ray that has been rotated from its initial position to form the angle. It is distinguished from the initial side, which remains fixed.

Key Concept: Angle Addition

a) $\frac{\tan x + \tan y}{1 – \tan x \tan y}$

[Solution Description]

The formula for the tangent of the sum of two angles is given by:

$\tan(x + y) = \frac{\tan x + \tan y}{1 – \tan x \tan y}$

This formula is used to find the tangent of the sum of two angles using their individual tangents.

Key Concept: Angle Basics

b) Positive x-axis

[Solution Description] The initial side of an angle is the ray’s position before rotation begins. In this case, it starts in the positive x-axis direction.

Key Concept: Radian Definition

a) 1 radian

[Solution Description] According to the definition of radian, if an arc of length 1 unit subtends an angle at the center of a circle with a radius of 1 unit (unit circle), then the angle measures 1 radian.

Key Concept: Complex Identities, Function Composition

c) $\frac{828}{2197}$

[Solution Description]

We start by finding $\cos A$. Since $\sin^2 A + \cos^2 A = 1$, we can write:

$$\cos^2 A = 1 – \sin^2 A = 1 – \left(\frac{5}{13}\right)^2 = \frac{169 – 25}{169} = \frac{144}{169}.$$

Therefore,

$$\cos A = \frac{\sqrt{144}}{13} = \frac{12}{13}.$$

Using the triple angle formula for cosine, $\cos 3A = 4\cos^3 A – 3\cos A$, we find:

$$\cos 3A = 4\left( \frac{12}{13} \right)^3 – 3 \times \frac{12}{13}.$$

Calculating $\left( \frac{12}{13} \right)^3$:

$$\left( \frac{12}{13} \right)^3 = \frac{12 \times 12 \times 12}{2197} = \frac{1728}{2197}.$$

So, substituting back we have:

$$\cos 3A = 4 \times \frac{1728}{2197} – \frac{36}{13}.$$

Converting $\frac{36}{13}$ to a denominator of 2197:

$$\frac{36}{13} = \frac{36 \times 169}{2197} = \frac{6084}{2197}.$$

Thus:

$$\cos 3A = \frac{6912}{2197} – \frac{6084}{2197} = \frac{828}{2197}.$$

Simplifying gives us the result for $\cos 3A$.

Key Concept: Compound Angles, Trigonometric Equations

a) $\frac{4 – 3\sqrt{3}}{10}$

[Solution Description]

To solve this, we will use the compound angle formula for cosine:

$$\cos(x + y) = \cos x \cos y – \sin x \sin y$$

Given $\sin x = \frac{3}{5}$, we can find $\cos x$ using the identity:

$$\cos^2 x = 1 – \sin^2 x = 1 – \left(\frac{3}{5}\right)^2 = \frac{16}{25}$$

Thus, $\cos x = \frac{4}{5}$ because $x$ is in the first quadrant where cosine is positive.

We know:

$$\cos 60^\circ = \frac{1}{2}, \quad \sin 60^\circ = \frac{\sqrt{3}}{2}$$

Using these, we find:

$$\cos(x + 60^\circ) = \left(\frac{4}{5} \cdot \frac{1}{2}\right) – \left(\frac{3}{5} \cdot \frac{\sqrt{3}}{2}\right)$$

$$= \frac{4}{10} – \frac{3\sqrt{3}}{10}$$

$$= \frac{4 – 3\sqrt{3}}{10}$$

Therefore, the answer is $\frac{4 – 3\sqrt{3}}{10}$.

Key Concept: Angle Direction

b) Negative

[Solution Description] When an angle is described as rotating clockwise from the initial side, it is considered a negative angle. Clockwise rotations are designated with a negative sign.

Key Concept: Direct Substitution

d) $\frac{\sqrt{3} + 1}{\sqrt{2}}$

[Solution Description] Using $\cos x + \cos y = 2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}$, we substitute $x = 60^\circ$ and $$y = 30^\circ$.

First, calculate:

$$\frac{x+y}{2} = \frac{60^\circ + 30^\circ}{2} = 45^\circ$$

Next, calculate:

$$\frac{x-y}{2} = \frac{60^\circ – 30^\circ}{2} = 15^\circ$$

Thus,

$$\cos x + \cos y = 2 \cos 45^\circ \cos 15^\circ = 2 \times \frac{\sqrt{2}}{2} \times \left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) = \frac{\sqrt{3} + 1}{\sqrt{2}}$$

Key Concept: Degree Basics

c) 3600 seconds

[Solution Description] One degree is divided into 60 minutes, and each minute is further divided into 60 seconds. Thus, the number of seconds in one degree can be calculated as $60$ minutes $\times 60$ seconds/minute = $3600$ seconds.

Key Concept: Ratio of Radii, Complex Conversion Problems

a) $2:3$

[Solution Description]

Convert the angles into radians:

$$90^\circ = \frac{\pi}{2} \text{ radians}, \quad 60^\circ = \frac{\pi}{3} \text{ radians}$$

Using the arc length formula $l = r\theta$, and knowing $l_1 = l_2$, we can write:

$$r_1 \cdot \frac{\pi}{2} = r_2 \cdot \frac{\pi}{3}$$

Simplifying,

$$r_1 \cdot 3 = r_2 \cdot 2$$

Therefore, the ratio of the radii is:

$$\frac{r_1}{r_2} = \frac{2}{3}$$

Key Concept: Degree Basics

c) Minute

[Solution Description] One degree ($1^\circ$) is divided into 60 equal parts known as minutes. Therefore, one sixtieth of a degree is called a minute.

Key Concept: Identity Transformation

b) $\sqrt{1-\cos^2 x}(4\cos^2 x – 1)$

[Solution Description] Start with the identity $\sin 3x = 3\sin x – 4\sin^3 x$. Use the Pythagorean identity $\sin^2 x = 1 – \cos^2 x$ to express $\sin^3 x$ in terms of $\cos x$:

$$\sin^3 x = \sin x(1-\cos^2 x) = \sin x – \sin x \cos^2 x$$

Substitute back,

$$\sin 3x = 3\sin x – 4(\sin x – \sin x \cos^2 x)$$

Simplify,

$$\sin 3x = 3\sin x – 4\sin x + 4\sin x \cos^2 x$$

$$= -\sin x + 4\sin x \cos^2 x$$

$$= \sin x (-1 + 4\cos^2 x)$$

Substituting $\sin x = \sqrt{1-\cos^2 x}$ gives:

$$\sin 3x = \sqrt{1-\cos^2 x}(4\cos^2 x – 1)$$

This is the expression for $\sin 3x$ in terms of $\cos x$.

Key Concept: Angle Addition/Subtraction

a) Positive

[Solution Description]

$\hspace{10pt}$ Given that $\cos x$ is negative, it implies that x is either in the second quadrant or the third quadrant.

– For $\cos(\pi + x)$, we know from the properties of cosine: $\cos(\pi + x) = -\cos x$.

– Since $\cos x$ is negative, $-\cos x$ will be positive.

Therefore, the sign of cos($\pi + x$) is positive.

Key Concept: Arc Length to Angle

c) 57.3 degrees

[Solution Description] The formula for finding the angle in radians is $\theta = \frac{s}{r}$ where $s$ is the arc length and $r$ is the radius. Thus, $\theta = \frac{7}{7} = 1$ radian. Converting to degrees: $1 \text{ radian} = \frac{180}{\pi}$ degrees, so $1 \times \frac{180}{\pi} \approx 57.3$ degrees.

Key Concept: Conversion Basics

c) $$\frac{2\pi}{3}$$

[Solution Description] To convert from degrees to radians, use the formula $$1° = \frac{\pi}{180} \text{ radians}$$. For 120°, we have:

$$120° \times \frac{\pi}{180} = \frac{120\pi}{180} = \frac{2\pi}{3}.$$

Therefore, 120° is equivalent to $$\frac{2\pi}{3}$$ radians.

Key Concept: Trigonometric Function Signs

b) Negative

[Solution Description]

In the third quadrant, both $x$-coordinate ($a$) and $y$-coordinate ($b$) are negative. Since cosine corresponds to the $x$-coordinate, $\cos(x)$ will be negative.

Key Concept: Compound Angle

b) $$2 – \sqrt{3}$$

[Solution Description] To find $\tan(x – y)$, we use the formula:

$$\tan(x – y) = \frac{\tan x – \tan y}{1 + \tan x \tan y}$$

We know that $\tan(\frac{\pi}{4}) = 1$ and $\tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$. Substituting these values into the formula gives us:

$$\tan(x – y) = \frac{1 – \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}}$$

Simplifying the numerator:

$$1 – \frac{1}{\sqrt{3}} = \frac{\sqrt{3} – 1}{\sqrt{3}}$$

Simplifying the denominator:

$$1 + \frac{1}{\sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{3}}$$

Therefore,

$$\tan(x – y) = \frac{\frac{\sqrt{3} – 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} = \frac{\sqrt{3} – 1}{\sqrt{3} + 1}$$

Multiplying numerator and denominator by $$\sqrt{3} – 1$$ to rationalize:

$$\frac{(\sqrt{3} – 1)^2}{(\sqrt{3} + 1)(\sqrt{3} – 1)} = \frac{3 – 2\sqrt{3} + 1}{3 – 1} = \frac{4 – 2\sqrt{3}}{2} = 2 – \sqrt{3}$$

Key Concept: Basic Periodicity

c) $\pi$

[Solution Description] The periodicity of $\cot(x)$ is given by the formula that $\cot(\pi + x) = \cot(x)$. Therefore, the period of $\cot(x)$ is $\pi$.

Key Concept: Double Angle

c) $\frac{\sqrt{3}}{2}$

[Solution Description]

Since $\cos x = \frac{\sqrt{3}}{2}$, we can find $\sin x$ using the identity $\sin^2 x + \cos^2 x = 1$.

$$\sin^2 x = 1 – \cos^2 x = 1 – \left(\frac{\sqrt{3}}{2}\right)^2$$

$$\sin^2 x = 1 – \frac{3}{4} = \frac{1}{4}$$

$$\sin x = \frac{1}{2} \quad (\text{since } x \text{ is in the first quadrant})$$

Now, use the double angle formula for sine:

$$\sin 2x = 2\sin x \cos x = 2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$$

Key Concept: Angle Definition

b) Negative 90°

[Solution Description] Rotating a ray clockwise results in a negative angle. Therefore, a 90 degree clockwise rotation results in an angle of $-90^\circ$.

Key Concept: Reciprocal Function Domain

b) $x \neq n\pi, n \in \mathbb{Z}$

[Solution Description]

The domain of $y = \cot x$ is all real numbers except where $\sin x = 0$. This happens when $x = n\pi$, $n \in \mathbb{Z}$. Therefore, the domain is $\{ x : x \in \mathbb{R} \text{ and } x \neq n\pi, n \in \mathbb{Z} \}$.