Evaluate the following:
(i) \(1\)
(ii) \(2\)
(iii) \(\dfrac{3\sqrt{2} - \sqrt{6}}{8}\)
(iv) \(\dfrac{43 - 24\sqrt{3}}{11}\)
(v) \(\dfrac{67}{12}\)
To evaluate such expressions quickly, it helps to keep the standard trigonometric values in a table.
Standard Trigonometric Values Table (0°, 30°, 45°, 60°, 90°)
| Trigonometric Function | \(0^\circ\) | \(30^\circ\) | \(45^\circ\) | \(60^\circ\) | \(90^\circ\) |
|---|---|---|---|---|---|
| \(\sin\theta\) | 0 | \(\dfrac{1}{2}\) | \(\dfrac{\sqrt{2}}{2}\) | \(\dfrac{\sqrt{3}}{2}\) | 1 |
| \(\cos\theta\) | 1 | \(\dfrac{\sqrt{3}}{2}\) | \(\dfrac{\sqrt{2}}{2}\) | \(\dfrac{1}{2}\) | 0 |
| \(\tan\theta\) | 0 | \(\dfrac{1}{\sqrt{3}}\) | 1 | \(\sqrt{3}\) | Not defined |
| \(\text{cosec }\theta\) | Not defined | 2 | \(\sqrt{2}\) | \(\dfrac{2}{\sqrt{3}}\) | 1 |
| \(\sec\theta\) | 1 | \(\dfrac{2}{\sqrt{3}}\) | \(\sqrt{2}\) | 2 | Not defined |
| \(\cot\theta\) | Not defined | \(\sqrt{3}\) | 1 | \(\dfrac{1}{\sqrt{3}}\) | 0 |
Note: \(\text{cosec }\theta = \dfrac{1}{\sin\theta}\), \(\sec\theta = \dfrac{1}{\cos\theta}\), \(\cot\theta = \dfrac{1}{\tan\theta}\). Values become “Not defined” when the denominator is 0.
(i) \(\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ\)
Using the table:
\(\sin 60^\circ = \dfrac{\sqrt{3}}{2},\; \cos 30^\circ = \dfrac{\sqrt{3}}{2},\; \sin 30^\circ = \dfrac{1}{2},\; \cos 60^\circ = \dfrac{1}{2}\)
\(= \left(\dfrac{\sqrt{3}}{2}\right)\left(\dfrac{\sqrt{3}}{2}\right) + \left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)\)
\(= \dfrac{3}{4} + \dfrac{1}{4} = 1\)
Answer: \(1\)
(ii) \(2 \tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ\)
From the table: \(\tan 45^\circ = 1\), \(\cos 30^\circ = \dfrac{\sqrt{3}}{2}\), \(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\).
So, \(\tan^2 45^\circ = 1\), \(\cos^2 30^\circ = \dfrac{3}{4}\), \(\sin^2 60^\circ = \dfrac{3}{4}\).
\(= 2(1) + \dfrac{3}{4} - \dfrac{3}{4} = 2\)
Answer: \(2\)
(iii) \(\dfrac{\cos 45^\circ}{\sec 30^\circ + \text{cosec } 30^\circ}\)
From the table: \(\cos 45^\circ = \dfrac{\sqrt{2}}{2}\), \(\sec 30^\circ = \dfrac{2}{\sqrt{3}}\), \(\text{cosec } 30^\circ = 2\).
\(= \dfrac{\frac{\sqrt{2}}{2}}{\frac{2}{\sqrt{3}} + 2}\)
Step 1: Simplify denominator:
\(\frac{2}{\sqrt{3}} + 2 = 2\left(\frac{1}{\sqrt{3}} + 1\right) = 2\left(\frac{1+\sqrt{3}}{\sqrt{3}}\right) = \frac{2(1+\sqrt{3})}{\sqrt{3}}\)
So,
\(= \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2(1+\sqrt{3})} = \frac{\sqrt{6}}{4(1+\sqrt{3})}\)
Step 2: Rationalize:
\(\frac{\sqrt{6}}{4(1+\sqrt{3})} \times \frac{1-\sqrt{3}}{1-\sqrt{3}} = \frac{\sqrt{6}(1-\sqrt{3})}{4(1-3)}\)
\(= \frac{\sqrt{6}(\sqrt{3}-1)}{8} = \frac{\sqrt{18}-\sqrt{6}}{8} = \frac{3\sqrt{2}-\sqrt{6}}{8}\)
Answer: \(\dfrac{3\sqrt{2}-\sqrt{6}}{8}\)
(iv) \(\dfrac{\sin 30^\circ + \tan 45^\circ - \text{cosec } 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}\)
From the table: \(\sin 30^\circ = \dfrac{1}{2}\), \(\tan 45^\circ = 1\), \(\text{cosec } 60^\circ = \dfrac{2}{\sqrt{3}}\), \(\sec 30^\circ = \dfrac{2}{\sqrt{3}}\), \(\cos 60^\circ = \dfrac{1}{2}\), \(\cot 45^\circ = 1\).
Numerator \(= \dfrac{1}{2} + 1 - \dfrac{2}{\sqrt{3}} = \dfrac{3}{2} - \dfrac{2}{\sqrt{3}}\)
Denominator \(= \dfrac{2}{\sqrt{3}} + \dfrac{1}{2} + 1 = \dfrac{2}{\sqrt{3}} + \dfrac{3}{2}\)
So, \(= \dfrac{\frac{3}{2} - \frac{2}{\sqrt{3}}}{\frac{3}{2} + \frac{2}{\sqrt{3}}}\)
Step 1: Multiply numerator and denominator by \(2\sqrt{3}\):
\(= \dfrac{3\sqrt{3}-4}{3\sqrt{3}+4}\)
Step 2: Multiply by conjugate \((3\sqrt{3}-4)\):
\(= \frac{(3\sqrt{3}-4)^2}{(3\sqrt{3})^2-4^2}\)
Denominator: \(27-16=11\)
Numerator: \(27-24\sqrt{3}+16 = 43-24\sqrt{3}\)
\(= \dfrac{43 - 24\sqrt{3}}{11}\)
Answer: \(\dfrac{43 - 24\sqrt{3}}{11}\)
(v) \(\dfrac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}\)
Step 1: Numerator
From the table:
\(\cos 60^\circ = \dfrac{1}{2} \Rightarrow \cos^2 60^\circ = \dfrac{1}{4}\)
So \(5\cos^2 60^\circ = \dfrac{5}{4}\)
\(\sec 30^\circ = \dfrac{2}{\sqrt{3}} \Rightarrow \sec^2 30^\circ = \dfrac{4}{3}\)
So \(4\sec^2 30^\circ = \dfrac{16}{3}\)
\(\tan 45^\circ = 1 \Rightarrow \tan^2 45^\circ = 1\)
Numerator \(= \dfrac{5}{4} + \dfrac{16}{3} - 1\)
LCM \(=12\): \(= \dfrac{15}{12} + \dfrac{64}{12} - \dfrac{12}{12} = \dfrac{67}{12}\)
Step 2: Denominator
\(\sin^2 30^\circ + \cos^2 30^\circ = \left(\dfrac{1}{2}\right)^2 + \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{1}{4} + \dfrac{3}{4} = 1\)
Step 3: Final value
\(= \dfrac{\frac{67}{12}}{1} = \dfrac{67}{12}\)
Answer: \(\dfrac{67}{12}\)
Student Tip: If you memorize the table and also remember \(\sin^2\theta+\cos^2\theta=1\), most evaluation questions become very fast.