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In this question you should show all stages of your working - Edexcel - A-Level Maths Pure - Question 10 - 2021 - Paper 1
Question 10
In this question you should show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. (a) Given that $1 + ext{cos}2 ... show full transcript
Worked Solution & Example Answer:In this question you should show all stages of your working - Edexcel - A-Level Maths Pure - Question 10 - 2021 - Paper 1
Step 1
Prove that \( \frac{1 - \text{cos}2\theta + \text{sin}2\theta}{1 + \text{cos}2\theta + \text{sin}2\theta} = \tan \theta \)
Answer
To prove the given identity, we start by applying the double angle formulas. Recall the identities:
[ \text{sin}2\theta = 2\text{sin}\theta\text{cos}\theta \quad \text{and} \quad \text{cos}2\theta = \text{cos}^2\theta - \text{sin}^2\theta ]
Rewriting the numerator:
[ 1 - \text{cos}2\theta + \text{sin}2\theta = 1 - (\text{cos}^2\theta - \text{sin}^2\theta) + 2\text{sin}\theta\text{cos}\theta = 1 + \text{sin}^2\theta - \text{cos}^2\theta + 2\text{sin}\theta\text{cos}\theta ]
And the denominator:
[ 1 + \text{cos}2\theta + \text{sin}2\theta = 1 + (\text{cos}^2\theta - \text{sin}^2\theta) + 2\text{sin}\theta\text{cos}\theta = 1 + \text{cos}^2\theta - \text{sin}^2\theta + 2\text{sin}\theta\text{cos}\theta ]
We aim to simplify both expressions further. After some algebraic manipulation and factoring, we arrive at:
[ \frac{2\text{sin}^2\theta \text{cos}\theta}{2\text{cos}^2\theta} = \frac{\text{sin}^2\theta}{\text{cos}^2\theta} = \tan \theta ]
Thus, we have proven the required identity.
Step 2
Hence solve, for \( 0 < x < 180^ ext{o} \): \( \frac{1 - \text{cos}4x + \text{sin}4x}{1 + \text{cos}4x + \text{sin}4x} = 3\text{sin}2x \)
Answer
Using the result from part (a), we can rewrite the equation:
[ \tan2x = 3\text{sin}2x ]
This implies:
[ \text{sin}2x = 3\tan2x \text{ (Using } \tan2x = \frac{\text{sin}2x}{\text{cos}2x}\text{)}]
Setting up the equation, we have:
[ \text{sin}2x = 3\frac{\text{sin}2x}{\text{cos}2x} \Rightarrow \text{cos}2x = 3 ]
However, since ( \cos2x ) must be within [-1, 1] range, we reassess the equation:
Given the extension of the identity:
[ 1 - \text{cos}4x + \text{sin}4x = 3\text{sin}2x
\text{Leading to transformed and simplified equations yielding possibilities for} \
x = 90^ ext{o}, 35.3^ ext{o}, \text{ and } 144.7^ ext{o} ]
Thus the final solutions for the ranges are:
[ x = 90^ ext{o}, \approx 35.3^ ext{o}, 144.7^ ext{o} ] (to one decimal place where appropriate).