State whether the following are true or false. Justify your answer. (i) . (ii) The value of increases as increases. (iii) The value of increases as increases. (iv) for all values of . (v) is not defined for .

Understand the Question In trigonometry, trigonometric functions operate on angles as arguments, not as algebraic multipliers; thus, standard algebraic distributive laws do not apply directly to trigonometric functions. For acute angles (): increases from to . decreases from to . holds only at . Trigonometric ratios involving division by zero (e.g., ) are undefined. Part (i) . Step 1 · Test with Counterexample Let and . Evaluating LHS Evaluating RHS Since , the statement is false. Answer > (i) False Part (ii) The value of increases as increases. Step 1 · Evaluate Values of Evaluating for standard angles in : Since , the value of increases as increases. Answer > (ii) True Part (iii) The value of increases as increases. Step 1 · Evaluate Values of Evaluating for standard angles in : Since , the value of decreases as increases. Answer > (iii) False Part (iv) for all values of . Step 1 · Test with Counterexample Let . Since , the statement is not true for all values of (it is only true for in ). Answer > (iv) False Part (v) is not defined for . Step 1 · Evaluate Using the quotient relation : For Division by zero is undefined, so is not defined for . Answer > (v) True Common Mistakes Distributive Property Fallacy: Treating as . Trigonometric operators cannot be distributed over addition. Opposing Trends of Sine and Cosine: Confusing the variation of (which increases from to ) with (which decreases from to ) in the first quadrant. Generalizing a Single Match: Assuming holds generally simply because . Zero in the Denominator: Forgetting that , which is undefined, not .

Introduction to Trigonometry | Exercise 8.2

Question 4

State whether the following are true or false. Justify your answer.

(i) $\sin (A + B) = \sin A + \sin B$ .

(ii) The value of $\sin \theta$ increases as $\theta$ increases.

(iii) The value of $\cos \theta$ increases as $\theta$ increases.

(iv) $\sin \theta = \cos \theta$ for all values of $\theta$ .

(v) $\cot A$ is not defined for $A = 0^\circ$ .

Solution

Understand the Question

In trigonometry, trigonometric functions operate on angles as arguments, not as algebraic multipliers; thus, standard algebraic distributive laws do not apply directly to trigonometric functions.
For acute angles ( $0^\circ \le \theta \le 90^\circ$ $0^{\circ} \leq θ \leq 9 0^{\circ}$ ):
- $\sin \theta$ increases from $0$ to $1$ .
- $\cos \theta$ decreases from $1$ to $0$ .
- $\sin \theta = \cos \theta$ holds only at $\theta = 45^\circ$ .
- Trigonometric ratios involving division by zero (e.g., $\cot 0^\circ = \dfrac{\cos 0^\circ}{\sin 0^\circ} = \dfrac{1}{0}$ ) are undefined.

(i) $\sin (A + B) = \sin A + \sin B$ .

Step 1 · Test with Counterexample

Let $A = 30^\circ$ and $B = 60^\circ$ . Diagram 1

Evaluating LHS

\begin{aligned} \sin (A + B) &= \sin (30^\circ + 60^\circ) \\[0.6em] &= \sin 90^\circ \\[0.6em] &= 1 \end{aligned}

Evaluating RHS

\begin{aligned} \sin A + \sin B &= \sin 30^\circ + \sin 60^\circ \\[0.6em] &= \dfrac{1}{2} + \dfrac{\sqrt{3}}{2} \\[0.6em] &= \dfrac{1 + \sqrt{3}}{2} \end{aligned}

Since $\text{LHS} \ne \text{RHS}$ , the statement is false.

Answer

(i) False

(ii) The value of $\sin \theta$ increases as $\theta$ increases.

Step 1 · Evaluate Values of $\sin \theta$

Evaluating $\sin \theta$ for standard angles in $0^\circ \le \theta \le 90^\circ$ :* $\sin 0^\circ = 0$

$\sin 30^\circ = \dfrac{1}{2}$
$\sin 45^\circ = \dfrac{1}{\sqrt{2}}$
$\sin 60^\circ = \dfrac{\sqrt{3}}{2}$
$\sin 90^\circ = 1$

Since $0 < \dfrac{1}{2} < \dfrac{1}{\sqrt{2}} < \dfrac{\sqrt{3}}{2} < 1$ , the value of $\sin \theta$ increases as $\theta$ increases.

Answer

(ii) True

(iii) The value of $\cos \theta$ increases as $\theta$ increases.

Step 1 · Evaluate Values of $\cos \theta$

Evaluating $\cos \theta$ for standard angles in $0^\circ \le \theta \le 90^\circ$ : Diagram 3

$\cos 0^\circ = 1$
$\cos 30^\circ = \dfrac{\sqrt{3}}{2}$
$\cos 45^\circ = \dfrac{1}{\sqrt{2}}$
$\cos 60^\circ = \dfrac{1}{2}$
$\cos 90^\circ = 0$

Since $1 > \dfrac{\sqrt{3}}{2} > \dfrac{1}{\sqrt{2}} > \dfrac{1}{2} > 0$ , the value of $\cos \theta$ decreases as $\theta$ increases.

Answer

(iii) False

(iv) $\sin \theta = \cos \theta$ for all values of $\theta$ .

Step 1 · Test with Counterexample

Let $\theta = 30^\circ$ . $\sin 30^\circ = \dfrac{1}{2}$

$\cos 30^\circ = \dfrac{\sqrt{3}}{2}$

Since $\sin 30^\circ \ne \cos 30^\circ$ , the statement is not true for all values of $\theta$ (it is only true for $\theta = 45^\circ$ in $0^\circ \le \theta \le 90^\circ$ ).

Answer

(iv) False

(v) $\cot A$ is not defined for $A = 0^\circ$ .

Step 1 · Evaluate $\cot 0^\circ$

Using the quotient relation $\cot A = \dfrac{\cos A}{\sin A}$ : Diagram 5

For $A = 0^\circ$

\begin{aligned} \cot 0^\circ &= \dfrac{\cos 0^\circ}{\sin 0^\circ} \\[0.6em] &= \dfrac{1}{0} \end{aligned}

Division by zero is undefined, so $\cot A$ is not defined for $A = 0^\circ$ .

Answer

(v) True

Common Mistakes

Distributive Property Fallacy: Treating $\sin(A + B)$ as $\sin \times (A + B) = \sin A + \sin B$ . Trigonometric operators cannot be distributed over addition.
Opposing Trends of Sine and Cosine: Confusing the variation of $\sin \theta$ (which increases from $0$ to $1$ ) with $\cos \theta$ (which decreases from $1$ to $0$ ) in the first quadrant.
Generalizing a Single Match: Assuming $\sin \theta = \cos \theta$ holds generally simply because $\sin 45^\circ = \cos 45^\circ$ .
Zero in the Denominator: Forgetting that $\cot 0^\circ = \dfrac{\cos 0^\circ}{\sin 0^\circ} = \dfrac{1}{0}$ , which is undefined, not $0$ .

Question 4

More questions in Exercise 8.2