Introduction to Trigonometry | Exercise 8.2

Question 1

Evaluate the following :

(i) sin⁡60∘cos⁡30∘+sin⁡30∘cos⁡60∘\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ

(ii) 2tan⁡245∘+cos⁡230∘−sin⁡260∘2 \tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ

(iii) cos⁡45∘sec⁡30∘+csc⁡30∘\dfrac{\cos 45^\circ}{\sec 30^\circ + \csc 30^\circ}

(iv) sin⁡30∘+tan⁡45∘−csc⁡60∘sec⁡30∘+cos⁡60∘+cot⁡45∘\dfrac{\sin 30^\circ + \tan 45^\circ - \csc 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}

(v) 5cos⁡260∘+4sec⁡230∘−tan⁡245∘sin⁡230∘+cos⁡230∘\dfrac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}

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Solution
Understand the Question
  • To evaluate trigonometric expressions, substitute standard values for specific angles (0∘,30∘,45∘,60∘,90∘0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ) and simplify algebraically.
  • Key standard trigonometric values:
Anglesin⁡cos⁡tan⁡0∘01030∘12321345∘1212160∘3212390∘10undefined\begin{array}{|c|c|c|c|} \hline \text{Angle} & \sin & \cos & \tan \\ \hline 0^\circ & 0 & 1 & 0 \\ \hline 30^\circ & \dfrac{1}{2} & \dfrac{\sqrt{3}}{2} & \dfrac{1}{\sqrt{3}} \\ \hline 45^\circ & \dfrac{1}{\sqrt{2}} & \dfrac{1}{\sqrt{2}} & 1 \\ \hline 60^\circ & \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} & \sqrt{3} \\ \hline 90^\circ & 1 & 0 & \text{undefined} \\ \hline \end{array}
  • Reciprocal identities: sec⁡θ=1cos⁡θ\sec \theta = \dfrac{1}{\cos \theta}, csc⁡θ=1sin⁡θ\csc \theta = \dfrac{1}{\sin \theta}, and cot⁡θ=1tan⁡θ\cot \theta = \dfrac{1}{\tan \theta}.

(i) Evaluate sin⁡60∘cos⁡30∘+sin⁡30∘cos⁡60∘\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ

Step 1 · Substitute Values and Evaluate

Diagram 1

Substitute sin⁡60∘=32\sin 60^\circ = \dfrac{\sqrt{3}}{2}, cos⁡30∘=32\cos 30^\circ = \dfrac{\sqrt{3}}{2}, sin⁡30∘=12\sin 30^\circ = \dfrac{1}{2}, and cos⁡60∘=12\cos 60^\circ = \dfrac{1}{2}

sin⁡60∘cos⁡30∘+sin⁡30∘cos⁡60∘=(32)(32)+(12)(12)=34+14=44=1\begin{aligned} \sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ &= \left(\dfrac{\sqrt{3}}{2}\right) \left(\dfrac{\sqrt{3}}{2}\right) + \left(\dfrac{1}{2}\right) \left(\dfrac{1}{2}\right) \\[0.6em] &= \dfrac{3}{4} + \dfrac{1}{4} \\[0.6em] &= \dfrac{4}{4} \\[0.6em] &= 1 \end{aligned}
Answer

(i) 11

(ii) Evaluate 2tan⁡245∘+cos⁡230∘−sin⁡260∘2 \tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ

Step 1 · Substitute Values and Evaluate

Substitute tan⁡45∘=1\tan 45^\circ = 1, cos⁡30∘=32\cos 30^\circ = \dfrac{\sqrt{3}}{2}, and sin⁡60∘=32\sin 60^\circ = \dfrac{\sqrt{3}}{2}

2tan⁡245∘+cos⁡230∘−sin⁡260∘=2(1)2+(32)2−(32)2=2(1)+34−34=2+0=2\begin{aligned} 2 \tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ &= 2(1)^2 + \left(\dfrac{\sqrt{3}}{2}\right)^2 - \left(\dfrac{\sqrt{3}}{2}\right)^2 \\[0.6em] &= 2(1) + \dfrac{3}{4} - \dfrac{3}{4} \\[0.6em] &= 2 + 0 \\[0.6em] &= 2 \end{aligned}
Answer

(ii) 22

(iii) Evaluate cos⁡45∘sec⁡30∘+csc⁡30∘\dfrac{\cos 45^\circ}{\sec 30^\circ + \csc 30^\circ}

Step 1 · Substitute Values and Simplify

Substitute cos⁡45∘=12\cos 45^\circ = \dfrac{1}{\sqrt{2}}, sec⁡30∘=23\sec 30^\circ = \dfrac{2}{\sqrt{3}}, and csc⁡30∘=2\csc 30^\circ = 2

cos⁡45∘sec⁡30∘+csc⁡30∘=1223+2=122+233=12×32+23=322+26\begin{aligned} \dfrac{\cos 45^\circ}{\sec 30^\circ + \csc 30^\circ} &= \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{2}{\sqrt{3}} + 2} \\[1.1em] &= \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{2 + 2\sqrt{3}}{\sqrt{3}}} \\[1.1em] &= \dfrac{1}{\sqrt{2}} \times \dfrac{\sqrt{3}}{2 + 2\sqrt{3}} \\[0.6em] &= \dfrac{\sqrt{3}}{2\sqrt{2} + 2\sqrt{6}} \end{aligned}

Step 2 · Rationalise the Denominator

Rationalising the denominator by multiplying numerator and denominator by (26−22)(2\sqrt{6} - 2\sqrt{2})

322+26=326+22×26−2226−22=218−26(26)2−(22)2=2×32−2624−8=62−2616=2(32−6)16=32−68\begin{aligned} \dfrac{\sqrt{3}}{2\sqrt{2} + 2\sqrt{6}} &= \dfrac{\sqrt{3}}{2\sqrt{6} + 2\sqrt{2}} \times \dfrac{2\sqrt{6} - 2\sqrt{2}}{2\sqrt{6} - 2\sqrt{2}} \\[0.6em] &= \dfrac{2\sqrt{18} - 2\sqrt{6}}{(2\sqrt{6})^2 - (2\sqrt{2})^2} \\[0.6em] &= \dfrac{2 \times 3\sqrt{2} - 2\sqrt{6}}{24 - 8} \\[0.6em] &= \dfrac{6\sqrt{2} - 2\sqrt{6}}{16} \\[0.6em] &= \dfrac{2(3\sqrt{2} - \sqrt{6})}{16} \\[0.6em] &= \dfrac{3\sqrt{2} - \sqrt{6}}{8} \end{aligned}
Answer

(iii) 32−68\dfrac{3\sqrt{2} - \sqrt{6}}{8}

(iv) Evaluate sin⁡30∘+tan⁡45∘−csc⁡60∘sec⁡30∘+cos⁡60∘+cot⁡45∘\dfrac{\sin 30^\circ + \tan 45^\circ - \csc 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}

Step 1 · Substitute Values and Simplify

Substitute sin⁡30∘=12\sin 30^\circ = \dfrac{1}{2}, tan⁡45∘=1\tan 45^\circ = 1, csc⁡60∘=23\csc 60^\circ = \dfrac{2}{\sqrt{3}}, sec⁡30∘=23\sec 30^\circ = \dfrac{2}{\sqrt{3}}, cos⁡60∘=12\cos 60^\circ = \dfrac{1}{2}, and cot⁡45∘=1\cot 45^\circ = 1

sin⁡30∘+tan⁡45∘−csc⁡60∘sec⁡30∘+cos⁡60∘+cot⁡45∘=12+1−2323+12+1=32−2323+32=33−4234+3323=33−433+4\begin{aligned} \dfrac{\sin 30^\circ + \tan 45^\circ - \csc 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ} &= \dfrac{\dfrac{1}{2} + 1 - \dfrac{2}{\sqrt{3}}}{\dfrac{2}{\sqrt{3}} + \dfrac{1}{2} + 1} \\[1.1em] &= \dfrac{\dfrac{3}{2} - \dfrac{2}{\sqrt{3}}}{\dfrac{2}{\sqrt{3}} + \dfrac{3}{2}} \\[1.1em] &= \dfrac{\dfrac{3\sqrt{3} - 4}{2\sqrt{3}}}{\dfrac{4 + 3\sqrt{3}}{2\sqrt{3}}} \\[1.1em] &= \dfrac{3\sqrt{3} - 4}{3\sqrt{3} + 4} \end{aligned}

Step 2 · Rationalise the Denominator

Rationalising the denominator by multiplying numerator and denominator by (33−4)(3\sqrt{3} - 4)

33−433+4=33−433+4×33−433−4=(33)2−2(33)(4)+(4)2(33)2−(4)2=27−243+1627−16=43−24311\begin{aligned} \dfrac{3\sqrt{3} - 4}{3\sqrt{3} + 4} &= \dfrac{3\sqrt{3} - 4}{3\sqrt{3} + 4} \times \dfrac{3\sqrt{3} - 4}{3\sqrt{3} - 4} \\[0.6em] &= \dfrac{(3\sqrt{3})^2 - 2(3\sqrt{3})(4) + (4)^2}{(3\sqrt{3})^2 - (4)^2} \\[0.6em] &= \dfrac{27 - 24\sqrt{3} + 16}{27 - 16} \\[0.6em] &= \dfrac{43 - 24\sqrt{3}}{11} \end{aligned}
Answer

(iv) 43−24311\dfrac{43 - 24\sqrt{3}}{11}

(v) Evaluate 5cos⁡260∘+4sec⁡230∘−tan⁡245∘sin⁡230∘+cos⁡230∘\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 · Evaluate Numerator and Denominator

Substitute cos⁡60∘=12\cos 60^\circ = \dfrac{1}{2}, sec⁡30∘=23\sec 30^\circ = \dfrac{2}{\sqrt{3}}, tan⁡45∘=1\tan 45^\circ = 1, sin⁡30∘=12\sin 30^\circ = \dfrac{1}{2}, and cos⁡30∘=32\cos 30^\circ = \dfrac{\sqrt{3}}{2}

Numerator:

5(12)2+4(23)2−(1)2=5(14)+4(43)−1=54+163−1=1512+6412−1212=15+64−1212=6712\begin{aligned} 5 \left(\dfrac{1}{2}\right)^2 + 4 \left(\dfrac{2}{\sqrt{3}}\right)^2 - (1)^2 &= 5 \left(\dfrac{1}{4}\right) + 4 \left(\dfrac{4}{3}\right) - 1 \\[0.6em] &= \dfrac{5}{4} + \dfrac{16}{3} - 1 \\[0.6em] &= \dfrac{15}{12} + \dfrac{64}{12} - \dfrac{12}{12} \\[0.6em] &= \dfrac{15 + 64 - 12}{12} \\[0.6em] &= \dfrac{67}{12} \end{aligned}

Denominator:

sin⁡230∘+cos⁡230∘=(12)2+(32)2=14+34=44=1\begin{aligned} \sin^2 30^\circ + \cos^2 30^\circ &= \left(\dfrac{1}{2}\right)^2 + \left(\dfrac{\sqrt{3}}{2}\right)^2 \\[0.6em] &= \dfrac{1}{4} + \dfrac{3}{4} \\[0.6em] &= \dfrac{4}{4} \\[0.6em] &= 1 \end{aligned}

Divide Numerator by Denominator: 67121=6712\dfrac{\dfrac{67}{12}}{1} = \dfrac{67}{12}

Answer

(v) 6712\dfrac{67}{12}

Common Mistakes
  • Value Swapping: Confusing values of complementary angles, such as writing sin⁡60∘=12\sin 60^\circ = \dfrac{1}{2} instead of 32\dfrac{\sqrt{3}}{2}, or cos⁡60∘=32\cos 60^\circ = \dfrac{\sqrt{3}}{2} instead of 12\dfrac{1}{2}.
  • Rationalisation Errors: When rationalising 322+26\dfrac{\sqrt{3}}{2\sqrt{2} + 2\sqrt{6}}, rearrange terms to 26+222\sqrt{6} + 2\sqrt{2} first to keep the denominator positive and avoid negative sign errors.
  • Identity Shortcut: In part (v), the denominator sin⁡230∘+cos⁡230∘\sin^2 30^\circ + \cos^2 30^\circ simplifies directly to 11 using the fundamental identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1.

More questions in Exercise 8.2

Q1

Evaluate the following :

(i) sin⁡60∘cos⁡30∘+sin⁡30∘cos⁡60∘\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ

(ii) 2tan⁡245∘+cos⁡230∘−sin⁡260∘2 \tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ

(iii) cos⁡45∘sec⁡30∘+csc⁡30∘\dfrac{\cos 45^\circ}{\sec 30^\circ + \csc 30^\circ}

(iv) sin⁡30∘+tan⁡45∘−csc⁡60∘sec⁡30∘+cos⁡60∘+cot⁡45∘\dfrac{\sin 30^\circ + \tan 45^\circ - \csc 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}

(v) 5cos⁡260∘+4sec⁡230∘−tan⁡245∘sin⁡230∘+cos⁡230∘\dfrac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}

Q2

Choose the correct option and justify your choice :

(i) 2tan⁡30∘1+tan⁡230∘=\dfrac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} = (A) sin⁡60∘\sin 60^\circ (B) cos⁡60∘\cos 60^\circ (C) tan⁡60∘\tan 60^\circ (D) sin⁡30∘\sin 30^\circ

(ii) 1−tan⁡245∘1+tan⁡245∘=\dfrac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ} = (A) tan⁡90∘\tan 90^\circ (B) 11 (C) sin⁡45∘\sin 45^\circ (D) 00

(iii) sin⁡2A=2sin⁡A\sin 2A = 2 \sin A is true when A=A = (A) 0∘0^\circ (B) 30∘30^\circ (C) 45∘45^\circ (D) 60∘60^\circ

(iv) 2tan⁡30∘1−tan⁡230∘=\dfrac{2 \tan 30^\circ}{1 - \tan^2 30^\circ} = (A) cos⁡60∘\cos 60^\circ (B) sin⁡60∘\sin 60^\circ (C) tan⁡60∘\tan 60^\circ (D) sin⁡30∘\sin 30^\circ

Q3

If tan⁡(A+B)=3\tan (A + B) = \sqrt{3} and tan⁡(A−B)=13\tan (A - B) = \dfrac{1}{\sqrt{3}}; 0∘<A+B≤90∘0^\circ < A + B \le 90^\circ; A>BA > B, find AA and BB.

Q4

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

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

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

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

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

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

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