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6. (i) Use an appropriate double angle formula to show that cosec2x = λ cosec x sec x, and state the value of the constant λ - Edexcel - A-Level Maths Pure - Question 8 - 2013 - Paper 8
Question 8
6. (i) Use an appropriate double angle formula to show that cosec2x = λ cosec x sec x, and state the value of the constant λ. (ii) Solve, for 0 ≤ θ < 2π, the equat... show full transcript
Worked Solution & Example Answer:6. (i) Use an appropriate double angle formula to show that cosec2x = λ cosec x sec x, and state the value of the constant λ - Edexcel - A-Level Maths Pure - Question 8 - 2013 - Paper 8
Step 1
Use an appropriate double angle formula to show that cosec2x = λ cosec x sec x, and state the value of the constant λ.
Answer
To show that ( cosec(2x) = \lambda \cdot cosec(x) \cdot sec(x) ), we start by using the double angle identity:
[ cosec(2x) = \frac{1}{sin(2x)} = \frac{1}{2sin(x)cos(x)} ]\
From the double angle formula, we know that ( sin(2x) = 2sin(x)cos(x) ). Therefore:
[ cosec(2x) = \frac{1}{2sin(x)cos(x)} ]\
This can be rewritten as follows:
[ cosec(2x) = \frac{1}{2} \cdot \frac{1}{sin(x)} \cdot \frac{1}{cos(x)} = \frac{1}{2} cosec(x) sec(x) ]
Thus, we identify ( \lambda = \frac{1}{2} ).
Step 2
Solve, for 0 ≤ θ < 2π, the equation 3sec²θ + 3secθ = 2 tan² θ.
Answer
Starting with the equation:
[ 3sec²θ + 3secθ - 2tan²θ = 0 ]
Using the identity ( tan²θ = sec²θ - 1 ), we can substitute:
[ 3sec²θ + 3secθ - 2(sec²θ - 1) = 0 ]
Simplifying gives:
[ 3sec²θ + 3secθ - 2sec²θ + 2 = 0 ]
Or:
[ sec²θ + 3secθ + 2 = 0 ]
This can be factored as:
[ (secθ + 1)(secθ + 2) = 0 ]
This yields solutions: [ secθ = -1 \quad ext{or} \quad secθ = -2 ]
Finding ( θ ):
- For ( secθ = -1 ):
[ θ = \frac{3π}{2} ] - For ( secθ = -2 ):
[ cosθ = -\frac{1}{2} \implies θ = \frac{2π}{3}, \frac{4π}{3} ]\
Thus, the complete set of solutions in the range ( 0 ≤ θ < 2π ) is: [ θ = \frac{2π}{3}, \frac{3π}{2}, \frac{4π}{3} ]