Trigonometry

7/4

Step 1: Find each value — cos 0°=1, sin 45°=1/√2, sin 30°=1/2, sin 90°=1, cos 45°=1/√2, cos 60°=1/2. Step 2: First bracket = 1 + 1/√2 + 1/2 = (2√2 + 2 + √2)/(2√2) = (3√2 + 2)/(2√2). Step 3: Second bracket = 1 - 1/√2 + 1/2 = (3√2 - 2)/(2√2). Step 4: Product = (3√2 + 2)(3√2 - 2)/(4×2) = (18 - 4)/8 = 14/8 = 7/4. This uses the identity (a+b)(a-b) = a² - b² where a = 3√2 and b = 2.

2/7

Step 1: 3 cot A = 2, so cot A = 2/3, meaning adjacent/opposite = 2/3. Step 2: Let opposite = 3, adjacent = 2. Hypotenuse = √(9+4) = √13. Step 3: sin A = 3/√13, cos A = 2/√13. Step 4: Numerator = 4(3/√13) - 3(2/√13) = (12-6)/√13 = 6/√13. Step 5: Denominator = 2(3/√13) + 3(2/√13) = (6+6)/√13 = 12/√13. Step 6: Result = 6/12 = 1/2. Rechecking: 4×3=12, 3×2=6, 12-6=6; 2×3=6, 3×2=6, 6+6=12; 6/12 = 1/2. The answer is 1/2, making 2/7 incorrect — the correct answer should be 1/2. Students should verify by re-reading the problem carefully.

1

Step 1: Pair each angle with its complement: tan 1° with tan 89°, tan 2° with tan 88°, etc. Step 2: Note that tan(90° - θ) = cot θ, so tan 89° = cot 1°, tan 88° = cot 2°, and so on. Step 3: Each pair gives: tan θ × cot θ = tan θ × (1/tan θ) = 1. Step 4: There are 44 such pairs (1°,89°), (2°,88°),...,(44°,46°), and the middle term is tan 45° = 1. Step 5: Product = 1 × 1 × ... × 1 × tan 45° = 1. This is a classic complementary angle problem.

90°

Step 1: sin α = 1/2. From the table of standard values, sin 30° = 1/2, so α = 30°. Step 2: cos β = 1/2. From the table of standard values, cos 60° = 1/2, so β = 60°. Step 3: α + β = 30° + 60° = 90°. Step 4: This result also makes sense because sin α = cos β means sin α = sin(90° - β), giving α = 90° - β, so α + β = 90°. This shows how complementary angle relationships connect standard angle values.