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Show that the equation $$3 \, \sin^2 \theta - 2 \, \cos^2 \theta = 1$$ can be written as $$5 \, \sin^2 \theta = 3.$$ (b) Hence solve, for $0^\circ < \theta < 360^\circ$, the equation $$3 \, \sin^2 \theta - 2 \, \cos^2 \theta = 1,$$ giving your answers to 1 decimal place. - Edexcel - A-Level Maths Pure - Question 5 - 2008 - Paper 2
Question 5
Show that the equation $$3 \, \sin^2 \theta - 2 \, \cos^2 \theta = 1$$ can be written as $$5 \, \sin^2 \theta = 3.$$ (b) Hence solve, for $0^\circ < \theta < 3... show full transcript
Worked Solution & Example Answer:Show that the equation $$3 \, \sin^2 \theta - 2 \, \cos^2 \theta = 1$$ can be written as $$5 \, \sin^2 \theta = 3.$$ (b) Hence solve, for $0^\circ < \theta < 360^\circ$, the equation $$3 \, \sin^2 \theta - 2 \, \cos^2 \theta = 1,$$ giving your answers to 1 decimal place. - Edexcel - A-Level Maths Pure - Question 5 - 2008 - Paper 2
Step 1
Show that the equation can be written as $5 \sin^2 \theta = 3$
Answer
To transform the given equation, we start with:
Using the Pythagorean identity, we replace ( \cos^2 \theta ) with ( 1 - \sin^2 \theta ):
Expanding this gives:
Combine like terms:
Adding 2 to both sides results in:
This confirms the original equation can be rewritten as required.
Step 2
Solve, for $0^\circ < \theta < 360^\circ$, the equation $3 \sin^2 \theta - 2 \cos^2 \theta = 1$
Answer
From part (a), we have:
which simplifies to:
Taking the square root gives us two potential solutions:
Calculating ( \sqrt{\frac{3}{5}} \approx 0.7746 ).
To find angles corresponding to these sine values, we consider:
- ( \theta = \sin^{-1}(0.7746) ) which gives approximately ( \theta \approx 51.8^\circ ).
- For the negative value:
- The angles in the fourth quadrant where ( \sin \theta < 0 ) gives:
- ( \theta = 180^\circ + \angle ) gives approximately ( \theta \approx 308.2^\circ ).
Thus, the solutions to 1 decimal place are:
- ( \theta \approx 51.8^\circ )
- ( \theta \approx 308.2^\circ )