EXERCISE 1 \( \begin{array}{l}\text { (a) If }-5 \sin \theta=4 \text { and } \theta \in\left(90^{\circ} ; 270^{\circ}\right) \text {, determine without the use of a calculator and with the } \\ \text { aid of a diagram the value of: } \\ \begin{array}{lll}\text { (1) } 5 \cos \theta-3 \tan \theta & \text { (2) } \frac{4}{\sin \theta}-\frac{3}{\cos \theta}\end{array} \\ \text { (b) If } 2 \cos \alpha+1=0 \text { and } \sin \alpha<0 \text {, determine without the use of a calculator and with the aid of } \\ \text { a diagram the value of: } \\ \begin{array}{lll}\text { (1) } \tan \alpha-\sin \alpha & \text { (2) } 2 \sin ^{2} \alpha-\cos ^{2} \alpha\end{array} \\ \text { (c) If } 3 \tan \beta=-2 \text { and } \beta \in\left(0^{\circ} ; 180^{\circ}\right) \text {, determine without the use of a calculator and with the aid } \\ \text { of a diagram the value of: } \\ \begin{array}{lll}\text { (1) } \cos \beta & \text { (2) } 2 \sin ^{2} \beta-1 & \text { (3) } \sqrt{13} \cos \beta-13 \sin ^{2} \beta\end{array}\end{array} \)

We are given several trigonometric equations and asked to determine certain expressions without using a calculator, but with the help of a diagram. It is important to decide in which quadrant the angle lies so that the signs of sine and cosine are chosen correctly. ───────────────────────────── Part (a) Problem Statement: • –5 sin θ = 4 • θ ∈ (90° , 270°) Step 1. Solve for sin θ:   –5 sin θ = 4 ⟹ sin θ = –4/5 Since sin θ is negative and the angle is between 90° and 270°, θ must be in Quadrant III where both sine and cosine are negative. Step 2. Determine cos θ: Using the Pythagorean identity   sin²θ + cos²θ = 1 we have:   (–4/5)² + cos²θ = 1 ⟹ 16/25 + cos²θ = 1 ⟹ cos²θ = 1 – 16/25 = 9/25 Since θ is in Quadrant III, cos θ is negative:   cos θ = –3/5 Step 3. Part (a)(1): Evaluate 5 cos θ – 3 tan θ  a. Find tan θ:   tan θ = sin θ/cos θ = (–4/5)/(–3/5) = 4/3  b. Now compute:   5 cos θ = 5(–3/5) = –3   3 tan θ = 3(4/3) = 4 Thus,   5 cos θ – 3 tan θ = –3 – 4 = –7 Step 4. Part (a)(2): Evaluate (4/ sin θ) – (3/ cos θ)  a. Compute each term:   4/ sin θ = 4/(–4/5) = –5   3/ cos θ = 3/(–3/5) = –5  b. Thus,   (4/ sin θ) – (3/ cos θ) = –5 – (–5) = 0 ───────────────────────────── Part (b) Problem Statement: • 2 cos α + 1 = 0 • sin α < 0 Step 1. Solve for cos α:   2 cos α + 1 = 0 ⟹ cos α = –1/2 Since cos α is negative and we are given sin α < 0, α must be in Quadrant III. Step 2. Determine sin α: Using sin²α + cos²α = 1,   sin²α = 1 – cos²α = 1 – (1/4) = 3/4 Since α is in Quadrant III, sin α is negative:   sin α = –√3/2 Step 3. Part (b)(1): Evaluate tan α – sin α  a. Compute tan α:   tan α = sin α/cos α = (–√3/2)/(–1/2) = √3  b. Thus,   tan α – sin α = √3 – (–√3/2) = √3 + (√3/2) = (3√3)/2 Step 4. Part (b)(2): Evaluate 2 sin²α – cos²α We already have sin²α = 3/4 and cos²α = 1/4, so:   2 sin²α – cos²α = 2(3/4) – (1/4) = (6/4) – (1/4) = 5/4 ───────────────────────────── Part (c) Problem Statement: • 3 tan β = –2 ⟹ tan β = –2/3 • β ∈ (0° , 180°) Step 1. Determine the quadrant for β: Since β lies between 0° and 180°, it is either in Quadrant I or II. Tan β is negative, which means that sin β and cos β have opposite signs. In Quadrant II, sine is positive and cosine is negative. Therefore, β is in Quadrant II. Step 2. Find sin β and cos β: Assume   sin β = 2k and cos β = –3k Using the identity sin²β + cos²β = 1:   (2k)² + (–3k)² = 1 ⟹ 4k² + 9k² = 1 ⟹ 13k² = 1 ⟹ k = 1/√13 Thus,   sin β = 2/√13  and cos β = –3/√13 Step 3. Part (c)(1): Evaluate cos β This is already found:   cos β = –3/√13 Step 4. Part (c)(2): Evaluate 2 sin²β – 1 First, calculate sin²β:   sin²β = (2/√13)² = 4/13 Then,   2 sin²β – 1 = 2(4/13) – 1 = 8/13 – 13/13 = –5/13 Step 5. Part (c)(3): Evaluate √13 cos β – 13 sin²β Compute each term:  a. √13 cos β = √13 (–3/√13) = –3  b. 13 sin²β = 13 (4/13) = 4 Therefore,   √13 cos β – 13 sin²β = –3 – 4 = –7 ───────────────────────────── Final Answers (a) Given –5 sin θ = 4 and θ ∈ (90°, 270°):  (1) 5 cos θ – 3 tan θ = –7  (2) (4/ sin θ) – (3/ cos θ) = 0 (b) Given 2 cos α + 1 = 0 and sin α < 0:  (1) tan α – sin α = (3√3)/2  (2) 2 sin²α – cos²α = 5/4 (c) Given 3 tan β = –2 and β ∈ (0°, 180°):  (1) cos β = –3/√13  (2) 2 sin²β – 1 = –5/13  (3) √13 cos β – 13 sin²β = –7 Any answer which is algebraically equivalent is correct.