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4. (a) Show that the equation 3 sin² θ - 2 cos² θ = 1 can be written as 5 sin² θ = 3 - Edexcel - A-Level Maths Pure - Question 6 - 2008 - Paper 2
Question 6
4. (a) Show that the equation 3 sin² θ - 2 cos² θ = 1 can be written as 5 sin² θ = 3. (b) Hence solve, for 0° < θ < 360°, the equation 3 sin² θ - 2 cos² θ = 1, ... show full transcript
Worked Solution & Example Answer:4. (a) Show that the equation 3 sin² θ - 2 cos² θ = 1 can be written as 5 sin² θ = 3 - Edexcel - A-Level Maths Pure - Question 6 - 2008 - Paper 2
Step 1
Show that the equation can be written as 5 sin² θ = 3.
Answer
To transform the equation, we start from:
Using the Pythagorean identity, we know that:
Substituting this into the equation gives:
Simplifying this step-by-step:
- Distributing the -2 gives:
- Combine like terms:
- Adding 2 to both sides results in: $$5 ext{sin}^2 θ = 3.$
Step 2
Hence solve, for 0° < θ < 360°, the equation 3 sin² θ - 2 cos² θ = 1.
Answer
From part (a), we have:
To find sin² θ, divide both sides by 5:
ext{sin}^2 θ = rac{3}{5}
Taking the square root gives:
ext{sin} θ = ± rac{ ext{sqrt}3}{ ext{sqrt}5} = ± rac{ ext{sqrt}15}{5}
Calculating the angles:
- For the positive case: θ = ext{sin}^{-1}igg( rac{ ext{sqrt}15}{5}igg) = 0.726 ext{ radians} ext{ (approximately)} o 41.4°
- For the negative case, using: θ = 180° - ext{sin}^{-1}igg( rac{ ext{sqrt}15}{5}igg) o 180° - 41.4° = 138.6°
- The second quadrant solution: θ = 360° + ext{sin}^{-1}igg(- rac{ ext{sqrt}15}{5}igg) = 360° - 41.4° = 318.6°
Thus, the solutions for 0° < θ < 360° are:
- θ ≈ 41.4°
- θ ≈ 138.6°
- θ ≈ 318.6° (rounded to 1 decimal place)