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6. (i) Using the identity for tan(A ± B), solve, for −90° < x < 90°, $$\tan 2x + \tan 32^{\circ} = 5$$ $$1 - \tan 2x \tan 32^{\circ} = 5$$ Give your answers, in degrees, to 2 decimal places - Edexcel - A-Level Maths Pure - Question 7 - 2018 - Paper 5
Question 7
6. (i) Using the identity for tan(A ± B), solve, for −90° < x < 90°, $$\tan 2x + \tan 32^{\circ} = 5$$ $$1 - \tan 2x \tan 32^{\circ} = 5$$ Give your answers, ... show full transcript
Worked Solution & Example Answer:6. (i) Using the identity for tan(A ± B), solve, for −90° < x < 90°, $$\tan 2x + \tan 32^{\circ} = 5$$ $$1 - \tan 2x \tan 32^{\circ} = 5$$ Give your answers, in degrees, to 2 decimal places - Edexcel - A-Level Maths Pure - Question 7 - 2018 - Paper 5
Step 1
Using the identity for tan(A ± B), solve, for −90° < x < 90°
Answer
To solve for , we start with the equation:
Using the identity for the tangent, we rewrite this as:
.
Substituting the value of \tan 32^{\circ} with its approximate value. Using a calculator, we find:
.
So, we have:
.
Taking the arctan of both sides gives us:
.
Now, using a calculator, we compute this value. Calculating yields approximately , leading to:
Also, to find other solutions, we consider:
thus giving further potential solutions in the specified range.
Step 2
Using the identity for tan(A ± B), show that
Answer
To show this, we start with the left-hand side:
.
Using the tangent difference formula, we have:
, where A = 30° and B = 45°.
Using the tangential values:
and \tan 45^{\circ} = 1,\tan(30^{\circ} - 45^{\circ}) = \frac{\tan 30^{\circ} - \tan 45^{\circ}}{1 + \tan 30^{\circ} \tan 45^{\circ}}\tan(30^{\circ} - 45^{\circ}) = \frac{\frac{1}{\sqrt{3}} - 1}{1 + \frac{1}{\sqrt{3}}}.$$
This can be rearranged and simplified, showing that this equality holds true.
Step 3