Exercise 8.2 Practice
Evaluating Trigonometric Ratios of Specific Angles
Reference Triangles
Use these standard triangles to recall values for 30°, 45°, and 60° without memorizing the table.
30°-60°-90° Triangle
45°-45°-90° Triangle
Q1: Evaluate Expressions
Evaluate the following:
(i) $\sin 30^\circ \cos 60^\circ + \cos 30^\circ \sin 60^\circ$
(ii) $3 \tan^2 45^\circ - \sin^2 60^\circ - \frac{1}{2}\cot^2 30^\circ$
(iii) $\frac{\cos 60^\circ}{\sec 45^\circ + \text{cosec } 45^\circ}$
(iv) $\frac{4 \sin^2 60^\circ + 3 \tan^2 30^\circ - 8 \sin 45^\circ \cos 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}$
(i) $\sin 30^\circ \cos 60^\circ + \cos 30^\circ \sin 60^\circ$
(ii) $3 \tan^2 45^\circ - \sin^2 60^\circ - \frac{1}{2}\cot^2 30^\circ$
(iii) $\frac{\cos 60^\circ}{\sec 45^\circ + \text{cosec } 45^\circ}$
(iv) $\frac{4 \sin^2 60^\circ + 3 \tan^2 30^\circ - 8 \sin 45^\circ \cos 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}$
(i) $\sin 30^\circ \cos 60^\circ + \cos 30^\circ \sin 60^\circ$
$= (\frac{1}{2})(\frac{1}{2}) + (\frac{\sqrt{3}}{2})(\frac{\sqrt{3}}{2})$
$= \frac{1}{4} + \frac{3}{4} = \frac{4}{4} = 1$
$= (\frac{1}{2})(\frac{1}{2}) + (\frac{\sqrt{3}}{2})(\frac{\sqrt{3}}{2})$
$= \frac{1}{4} + \frac{3}{4} = \frac{4}{4} = 1$
(ii) $3 \tan^2 45^\circ - \sin^2 60^\circ - \frac{1}{2}\cot^2 30^\circ$
$= 3(1)^2 - (\frac{\sqrt{3}}{2})^2 - \frac{1}{2}(\sqrt{3})^2$
$= 3(1) - \frac{3}{4} - \frac{1}{2}(3) = 3 - \frac{3}{4} - \frac{3}{2}$
$= \frac{12 - 3 - 6}{4} = \frac{3}{4}$
$= 3(1)^2 - (\frac{\sqrt{3}}{2})^2 - \frac{1}{2}(\sqrt{3})^2$
$= 3(1) - \frac{3}{4} - \frac{1}{2}(3) = 3 - \frac{3}{4} - \frac{3}{2}$
$= \frac{12 - 3 - 6}{4} = \frac{3}{4}$
(iii) $\frac{\cos 60^\circ}{\sec 45^\circ + \text{cosec } 45^\circ}$
$= \frac{1/2}{\sqrt{2} + \sqrt{2}} = \frac{1/2}{2\sqrt{2}} = \frac{1}{4\sqrt{2}}$
Rationalizing: $\frac{1 \times \sqrt{2}}{4\sqrt{2} \times \sqrt{2}} = \frac{\sqrt{2}}{8}$
$= \frac{1/2}{\sqrt{2} + \sqrt{2}} = \frac{1/2}{2\sqrt{2}} = \frac{1}{4\sqrt{2}}$
Rationalizing: $\frac{1 \times \sqrt{2}}{4\sqrt{2} \times \sqrt{2}} = \frac{\sqrt{2}}{8}$
(iv) Denominator: $\sin^2 30^\circ + \cos^2 30^\circ = 1$ (Identity)
Numerator: $4(\frac{\sqrt{3}}{2})^2 + 3(\frac{1}{\sqrt{3}})^2 - 8(\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}})$
$= 4(\frac{3}{4}) + 3(\frac{1}{3}) - 8(\frac{1}{2})$
$= 3 + 1 - 4 = 0$
Result: $0/1 = 0$
Numerator: $4(\frac{\sqrt{3}}{2})^2 + 3(\frac{1}{\sqrt{3}})^2 - 8(\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}})$
$= 4(\frac{3}{4}) + 3(\frac{1}{3}) - 8(\frac{1}{2})$
$= 3 + 1 - 4 = 0$
Result: $0/1 = 0$
Q2: Multiple Choice Questions
Choose the correct option and justify your choice:
(i) $\frac{2 \tan 60^\circ}{1 + \tan^2 60^\circ} =$
(A) $\sin 60^\circ$ (B) $\cos 60^\circ$ (C) $\tan 60^\circ$ (D) $\sin 30^\circ$
(ii) $\frac{1 - \tan^2 60^\circ}{1 + \tan^2 60^\circ} =$
(A) $\tan 60^\circ$ (B) $1$ (C) $\sin 45^\circ$ (D) $-1/2$
(iii) $\sin 2A = 2 \sin A$ is true when $A =$
(A) $0^\circ$ (B) $30^\circ$ (C) $45^\circ$ (D) $60^\circ$
(iv) $\frac{2 \tan 60^\circ}{1 - \tan^2 60^\circ} =$
(A) $\cos 60^\circ$ (B) $\sin 60^\circ$ (C) $\tan 60^\circ$ (D) $-\sqrt{3}$
(i) $\frac{2 \tan 60^\circ}{1 + \tan^2 60^\circ} =$
(A) $\sin 60^\circ$ (B) $\cos 60^\circ$ (C) $\tan 60^\circ$ (D) $\sin 30^\circ$
(ii) $\frac{1 - \tan^2 60^\circ}{1 + \tan^2 60^\circ} =$
(A) $\tan 60^\circ$ (B) $1$ (C) $\sin 45^\circ$ (D) $-1/2$
(iii) $\sin 2A = 2 \sin A$ is true when $A =$
(A) $0^\circ$ (B) $30^\circ$ (C) $45^\circ$ (D) $60^\circ$
(iv) $\frac{2 \tan 60^\circ}{1 - \tan^2 60^\circ} =$
(A) $\cos 60^\circ$ (B) $\sin 60^\circ$ (C) $\tan 60^\circ$ (D) $-\sqrt{3}$
(i) Answer: [Use formula $\sin 2\theta = \frac{2 \tan \theta}{1+\tan^2 \theta}$]
Value method: $\frac{2\sqrt{3}}{1+3} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2} = \sin 60^\circ$.
Option (A) is correct. (Note: $\sin 120^\circ$ also works but $\sin 60^\circ$ is standard).
Value method: $\frac{2\sqrt{3}}{1+3} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2} = \sin 60^\circ$.
Option (A) is correct. (Note: $\sin 120^\circ$ also works but $\sin 60^\circ$ is standard).
(ii) Answer: [Use formula $\cos 2\theta = \frac{1-\tan^2 \theta}{1+\tan^2 \theta}$]
Value method: $\frac{1 - 3}{1 + 3} = \frac{-2}{4} = -1/2$.
Option (D) is correct. (Also equals $\cos 120^\circ$).
Value method: $\frac{1 - 3}{1 + 3} = \frac{-2}{4} = -1/2$.
Option (D) is correct. (Also equals $\cos 120^\circ$).
(iii) Answer: Check A = 0°:
LHS = $\sin 0 = 0$; RHS = $2 \sin 0 = 0$. Matches.
Check A = 30°: $\sin 60 = \sqrt{3}/2 \neq 2(1/2) = 1$.
Option (A) is correct.
LHS = $\sin 0 = 0$; RHS = $2 \sin 0 = 0$. Matches.
Check A = 30°: $\sin 60 = \sqrt{3}/2 \neq 2(1/2) = 1$.
Option (A) is correct.
(iv) Answer: [Use formula $\tan 2\theta = \frac{2 \tan \theta}{1-\tan^2 \theta}$]
Value method: $\frac{2\sqrt{3}}{1-3} = \frac{2\sqrt{3}}{-2} = -\sqrt{3}$.
Option (D) is $-\sqrt{3}$. Also $\tan 120^\circ$.
Option (D) is correct.
Value method: $\frac{2\sqrt{3}}{1-3} = \frac{2\sqrt{3}}{-2} = -\sqrt{3}$.
Option (D) is $-\sqrt{3}$. Also $\tan 120^\circ$.
Option (D) is correct.
Q3: Finding Angles
If $\sin(A + B) = 1$ and $\cos(A - B) = \frac{\sqrt{3}}{2}$;
$0^\circ < A + B \leq 90^\circ$; $A > B$, find A and B.
Equation 1: $\sin(A + B) = 1$
Since $\sin 90^\circ = 1$, we have: $A + B = 90^\circ$ ... (i)
Since $\sin 90^\circ = 1$, we have: $A + B = 90^\circ$ ... (i)
Equation 2: $\cos(A - B) = \frac{\sqrt{3}}{2}$
Since $\cos 30^\circ = \frac{\sqrt{3}}{2}$, we have: $A - B = 30^\circ$ ... (ii)
Since $\cos 30^\circ = \frac{\sqrt{3}}{2}$, we have: $A - B = 30^\circ$ ... (ii)
Solve: Add (i) and (ii):
$(A + B) + (A - B) = 90 + 30$
$2A = 120 \Rightarrow A = 60^\circ$
$(A + B) + (A - B) = 90 + 30$
$2A = 120 \Rightarrow A = 60^\circ$
Substitute A in (i):
$60 + B = 90 \Rightarrow B = 30^\circ$
$60 + B = 90 \Rightarrow B = 30^\circ$
A = 60°, B = 30°
Q4: True or False
State whether the following are true or false. Justify your answer.
(i) $\tan(A + B) = \tan A + \tan B$.
(ii) The value of $\cos \theta$ increases as $\theta$ increases.
(iii) $\sin \theta = \cos \theta$ for all values of $\theta$.
(iv) $\cot A$ is not defined for $A = 0^\circ$.
(v) $\sec \theta$ is always greater than or equal to 1.
(i) $\tan(A + B) = \tan A + \tan B$.
(ii) The value of $\cos \theta$ increases as $\theta$ increases.
(iii) $\sin \theta = \cos \theta$ for all values of $\theta$.
(iv) $\cot A$ is not defined for $A = 0^\circ$.
(v) $\sec \theta$ is always greater than or equal to 1.
(i) False. Check $A=30, B=30$. LHS $\tan 60 = \sqrt{3}$. RHS $\tan 30 + \tan 30 = 2/\sqrt{3}$. Not equal.
(ii) False. $\cos 0 = 1, \cos 90 = 0$. Value decreases as $\theta$ increases.
(iii) False. Only true for $\theta = 45^\circ$. Example: $\sin 30 = 0.5, \cos 30 = 0.866$.
(iv) True. $\cot 0^\circ = \frac{\cos 0}{\sin 0} = \frac{1}{0}$ (undefined).
(v) True. $\sec \theta = \frac{\text{Hypotenuse}}{\text{Adjacent}}$. Hypotenuse $\geq$ Adjacent, so ratio $\geq 1$.