Trigonometric Functions - II

sin 5θ + sin θ

Step 1: Use the product-to-sum formula: 2sinA cosB = sin(A+B) + sin(A-B). Step 2: Here A = 3θ and B = 2θ. Step 3: 2sin3θ cos2θ = sin(3θ + 2θ) + sin(3θ - 2θ). Step 4: = sin(5θ) + sin(θ). Final Answer: 2sin3θ cos2θ = sin5θ + sinθ. A common mistake is writing sin5θ - sinθ, which is the result for 2cos3θ sin2θ (note: A and B are swapped).

2 sin((C+D)/2) cos((C-D)/2)

Step 1: The sum-to-product formula for sinC + sinD is derived by setting C = A+B and D = A-B. Step 2: Then A = (C+D)/2 and B = (C-D)/2. Step 3: From 2sinA cosB = sin(A+B) + sin(A-B), we have sinC + sinD = 2sinA cosB. Step 4: Substituting back: sinC + sinD = 2sin((C+D)/2)cos((C-D)/2). Final Answer: sinC + sinD = 2sin((C+D)/2)cos((C-D)/2). Do not confuse this with cosC + cosD, which uses 2cos·cos.

2tanA / (1 + tan²A)

Step 1: Start with sin2A = 2sinA cosA. Step 2: Write this as 2sinA cosA × (cos²A + sin²A) in the denominator (since cos²A + sin²A = 1). Step 3: Divide numerator and denominator by cos²A: sin2A = 2(sinA/cosA) / (1 + sin²A/cos²A). Step 4: = 2tanA / (1 + tan²A). Final Answer: sin2A = 2tanA/(1 + tan²A). Do not confuse with tan2A = 2tanA/(1 - tan²A). The signs in the denominator are different.

1 - 2cos²A

Step 1: The standard forms of cos2A are: cos²A - sin²A, 2cos²A - 1, and 1 - 2sin²A. Step 2: Check '1 - 2cos²A': this would be -(2cos²A - 1) = -(cos2A) = -cos2A. Step 3: So 1 - 2cos²A = -cos2A, NOT cos2A. Step 4: The correct form involving only cosA is 2cos²A - 1 (not 1 - 2cos²A). Final Answer: '1 - 2cos²A' is NOT equal to cos2A. It equals -cos2A. The form 1 - 2sin²A IS correct because replacing sin²A = 1 - cos²A in cos²A - sin²A gives 2cos²A - 1, and using cos²A = 1 - sin²A gives 1 - 2sin²A.