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EXERCISE 6 1. Evaluate, without using a calculator. (a) \( \cos 330^{\circ} \cdot \sin 60^{\circ} \) (c) \( \tan 315^{\circ}-2 \cos 60^{\circ}+\sin 210^{\circ} \) (e) \( \frac{\tan 330^{\circ}}{\sin 330^{\circ}} \) (g) \( \sin 150^{\circ}-\tan 240^{\circ} \cdot \cos 210^{\circ} \) (i) \( \frac{\tan 150^{\circ}}{\tan 240^{\circ}}-\frac{\sin 300^{\circ}}{\sin 120^{\circ}} \) (k) \( \frac{\tan 225^{\circ}-\sin x}{\sin x+\sin 270^{\circ}} \)

Ask by Clark Carlson. in South Africa
Feb 25,2025

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(a) 3/4 (c) –5/2 (e) 2/√3 (g) 2 (i) 2/3 (k) –1

Solución

Below are one set of detailed solutions for each part. ────────────────────────────── (a) Evaluate cos 330° · sin 60° Step 1. Recall:   cos 330° = cos(360° – 30°) = cos 30° = (√3)/2   sin 60° = (√3)/2 Step 2. Multiply:   (cos 330°)(sin 60°) = (√3/2) · (√3/2) = 3/4 ────────────────────────────── (c) Evaluate tan 315° – 2cos 60° + sin 210° Step 1. Find each value:   tan 315° = tan(360° – 45°) = –tan 45° = –1     cos 60° = 1/2   so 2cos 60° = 2·(1/2) = 1   sin 210° = sin(180° + 30°) = –sin 30° = –1/2 Step 2. Substitute and sum:   tan 315° – 2cos 60° + sin 210° = (–1) – 1 + (–1/2)   = –2 – 1/2 = –(4/2 + 1/2) = –5/2 ────────────────────────────── (e) Evaluate (tan 330°)/(sin 330°) Step 1. Write tan 330° as sin 330°/cos 330°:   sin 330° = –sin 30° = –1/2   cos 330° = cos 30° = (√3)/2   ⇒ tan 330° = (–1/2)/(√3/2) = –1/√3 Step 2. Divide by sin 330°:   (tan 330°)/(sin 330°) = (–1/√3)/(–1/2) = (–1/√3) · (–2/1) = 2/√3 Step 3. (Optional) Rationalize the denominator:   2/√3 = (2√3)/3 ────────────────────────────── (g) Evaluate sin 150° – (tan 240° · cos 210°) Step 1. Find each value:   sin 150° = sin(180° – 30°) = sin 30° = 1/2   tan 240° = tan(240° – 180°) = tan 60°    In quadrant III, tangent is positive, so tan 240° = √3   cos 210° = cos(180° + 30°) = –cos 30° = –(√3)/2 Step 2. Compute the product:   tan 240° · cos 210° = (√3) · [–(√3)/2] = –(3/2) Step 3. Substitute back:   sin 150° – (tan 240° · cos 210°) = 1/2 – [–3/2] = 1/2 + 3/2 = 4/2 = 2 ────────────────────────────── (i) Evaluate [tan 150°/tan 240°] – [sin 300°/sin 120°] Step 1. Compute the tangent ratios:   tan 150° = tan(180° – 30°) = –tan 30° = –1/√3   tan 240° = tan(240° – 180°) = tan 60° = √3   Thus, (tan 150°)/(tan 240°) = (–1/√3)/(√3) = –1/(√3·√3) = –1/3 Step 2. Compute the sine ratios:   sin 300° = sin(360° – 60°) = –sin 60° = –(√3)/2   sin 120° = sin(180° – 60°) = sin 60° = (√3)/2   Thus, (sin 300°)/(sin 120°) = [–(√3)/2] / [(√3)/2] = –1 Step 3. Combine the two parts:   [–1/3] – [–1] = –1/3 + 1 = (–1 + 3)/3 = 2/3 ────────────────────────────── (k) Evaluate (tan 225° – sin x)/(sin x + sin 270°) Step 1. Find specific values:   tan 225° = tan(180° + 45°) = tan 45° = 1   sin 270° = –1 Step 2. Substitute into the expression:   (tan 225° – sin x)/(sin x + sin 270°)    = (1 – sin x)/(sin x – 1) Step 3. Notice that the denominator is just –(1 – sin x). Therefore,   (1 – sin x)/(–(1 – sin x)) = –1          (provided sin x ≠ 1) ────────────────────────────── Final Answers: (a) 3/4 (c) –5/2 (e) 2/√3     (or, equivalently, 2√3/3) (g) 2 (i) 2/3 (k) –1     (for sin x ≠ 1)

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Sure! Let's break it down and evaluate each expression step by step: **(a)** \( \cos 330^{\circ} \cdot \sin 60^{\circ} \) First, recall that \( \cos 330^{\circ} = \cos(-30^{\circ}) = \cos 30^{\circ} = \frac{\sqrt{3}}{2} \). For \( \sin 60^{\circ} \), we know that \( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \). Thus, the expression becomes: \[ \cos 330^{\circ} \cdot \sin 60^{\circ} = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{3}{4} \] --- **(c)** \( \tan 315^{\circ} - 2 \cos 60^{\circ} + \sin 210^{\circ} \) Here, \( \tan 315^{\circ} = -1 \) (since \( 315^{\circ} \) is in the fourth quadrant), \( \cos 60^{\circ} = \frac{1}{2} \) leads to \( 2 \cos 60^{\circ} = 1 \), and \( \sin 210^{\circ} = -\frac{1}{2} \). Therefore, we have: \[ \tan 315^{\circ} - 2 \cos 60^{\circ} + \sin 210^{\circ} = -1 - 1 - \frac{1}{2} = -2.5 \] --- **(e)** \( \frac{\tan 330^{\circ}}{\sin 330^{\circ}} \) First, \( \tan 330^{\circ} = -\frac{1}{\sqrt{3}} \) (because \( \tan 330^{\circ} = \tan(-30^{\circ}) \)), and \( \sin 330^{\circ} = -\frac{1}{2} \), hence: \[ \frac{\tan 330^{\circ}}{\sin 330^{\circ}} = \frac{-\frac{1}{\sqrt{3}}}{-\frac{1}{2}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \] --- **(g)** \( \sin 150^{\circ} - \tan 240^{\circ} \cdot \cos 210^{\circ} \) Here \( \sin 150^{\circ} = \frac{1}{2} \), \( \tan 240^{\circ} = \frac{1}{\sqrt{3}} \), and \( \cos 210^{\circ} = -\frac{\sqrt{3}}{2} \). Plugging these in, we get: \[ \tan 240^{\circ} \cdot \cos 210^{\circ} = \frac{1}{\sqrt{3}} \cdot -\frac{\sqrt{3}}{2} = -\frac{1}{2} \] So: \[ \sin 150^{\circ} - \tan 240^{\circ} \cdot \cos 210^{\circ} = \frac{1}{2} - (-\frac{1}{2}) = 1 \] --- **(i)** \( \frac{\tan 150^{\circ}}{\tan 240^{\circ}} - \frac{\sin 300^{\circ}}{\sin 120^{\circ}} \) Calculating each part, we find \( \tan 150^{\circ} = -\frac{1}{\sqrt{3}} \) and \( \tan 240^{\circ} = \frac{1}{\sqrt{3}} \), giving us: \[ \frac{\tan 150^{\circ}}{\tan 240^{\circ}} = \frac{-\frac{1}{\sqrt{3}}}{\frac{1}{\sqrt{3}}} = -1 \] Also, \( \sin 300^{\circ} = -\frac{\sqrt{3}}{2} \) and \( \sin 120^{\circ} = \frac{\sqrt{3}}{2} \), so: \[ \frac{\sin 300^{\circ}}{\sin 120^{\circ}} = \frac{-\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}} =

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