6. (i) Using the identity for tan(A ± B), solve, for −90° < x < 90°, \[ \tan 2x + \tan 32° = 5 \] \[ 1 - \tan 2x \tan 32° = 5 \] Give your answers, in degrees, to 2 decimal places. (ii) (a) Using the identity for tan(A ± B), show that \[ \tan(30° - 45°) = \frac{\tan 30° - 1}{1 + \tan 30°} \] \[ \theta \neq (60n + 45)°, n \in \mathbb{Z} \] (b) Hence solve, for 0 < θ < 180°, \[ (1 + \tan 30°)(\tan(θ + 28°) = \tan 30° - 1 \]

A-Level Edexcel Maths Pure Differentiation: 6. (i) Using the identity for ta

6. (i) Using the identity for tan(A ± B), solve, for −90° < x < 90°, \[ \tan 2x + \tan 32° = 5 \] \[ 1 - \tan 2x \tan 32° = 5 \] Give your answers, in degrees, to 2 decimal places - Edexcel - A-Level Maths Pure - Question 8 - 2018 - Paper 5

Question 8

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6. (i) Using the identity for tan(A ± B), solve, for −90° < x < 90°, \[ \tan 2x + \tan 32° = 5 \] \[ 1 - \tan 2x \tan 32° = 5 \] Give your answers, in degree... show full transcript

Worked Solution & Example Answer:6. (i) Using the identity for tan(A ± B), solve, for −90° < x < 90°, \[ \tan 2x + \tan 32° = 5 \] \[ 1 - \tan 2x \tan 32° = 5 \] Give your answers, in degrees, to 2 decimal places - Edexcel - A-Level Maths Pure - Question 8 - 2018 - Paper 5

Step 1

Using the identity for tan(A ± B), solve, for −90° < x < 90°

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Answer

To solve [ \tan 2x + \tan 32° = 5 \text{ and } 1 - \tan 2x \tan 32° = 5 ], we can rearrange the first equation:

[ \tan 2x = 5 - \tan 32° ]

Next, we substitute this into the second equation:

[ 1 - (5 - \tan 32°) \tan 32° = 5 ]

Expanding this gives us:

[ 1 - 5 \tan 32° + \tan^2 32° = 5 ]

Rearranging gives a quadratic equation in terms of ( \tan 2x ):

[ \tan^2 32° - 5 \tan 32° - 4 = 0 ]

Using the quadratic formula:

[ \tan 2x = \frac{5 \pm \sqrt{(5)^2 - 4(1)(-4)}}{2(1)} ]

This provides solutions for ( x ) that can then be calculated. Bringing these values into [-90° < x < 90°] will yield the final values for ( x ).

Step 2

Using the identity for tan(A ± B), show that

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Answer

We start with:

[ \tan(30° - 45°) = \frac{\tan 30° - \tan 45°}{1 + \tan 30° \tan 45°} ]

Substituting known values:

[ \tan 30° = \frac{1}{\sqrt{3}} \text{ and } \tan 45° = 1 ]

This leads to:

[ \tan(30° - 45°) = \frac{\frac{1}{\sqrt{3}} - 1}{1 + \frac{1}{\sqrt{3}}} = \frac{\frac{1 - \sqrt{3}}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} = \frac{1 - \sqrt{3}}{\sqrt{3} + 1} ]

Thus, we can confirm:

[ \tan(30° - 45°) = \frac{\tan 30° - 1}{1 + \tan 30°} ]

Step 3

Hence solve, for 0 < θ < 180°

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Answer

From subpart (a) we have:

[ (1 + \tan 30°)(\tan(θ + 28°) = \tan 30° - 1 ]

Substituting ( \tan 30° ):

[ (1 + \frac{1}{\sqrt{3}})(\tan(θ + 28°) = \frac{1}{\sqrt{3}} - 1 ]

This can be simplified to find ( θ ) by solving:

[ \tan(θ + 28°) = \frac{\frac{1 - \sqrt{3}}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} ]

Finding the angles that satisfy this equation would yield the required answers for ( θ ) within the domain specified.

6. (i) Using the identity for tan(A ± B), solve, for −90° < x < 90°, \[ \tan 2x + \tan 32° = 5 \] \[ 1 - \tan 2x \tan 32° = 5 \] Give your answers, in degrees, to 2 decimal places - Edexcel - A-Level Maths Pure - Question 8 - 2018 - Paper 5

Worked Solution & Example Answer:6. (i) Using the identity for tan(A ± B), solve, for −90° < x < 90°, \[ \tan 2x + \tan 32° = 5 \] \[ 1 - \tan 2x \tan 32° = 5 \] Give your answers, in degrees, to 2 decimal places - Edexcel - A-Level Maths Pure - Question 8 - 2018 - Paper 5

Using the identity for tan(A ± B), solve, for −90° < x < 90°

Using the identity for tan(A ± B), show that

Hence solve, for 0 < θ < 180°

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