Photo AI
8. Starting from the formulae for sin(A + B) and cos(A + B), prove that $$ tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\n$$ (b) Deduce that $$ \tan\left(\theta + \frac{\pi}{6}\right) = \frac{1 + \sqrt{3} \tan \theta}{\sqrt{3} - \tan \theta}\n$$ (c) Hence, or otherwise, solve, for $0 \leq \theta < \pi$, $$ 1 + \sqrt{3} \tan \theta = (\sqrt{3} - \tan \theta) \tan(\pi - \theta)\n$$ Give your answers as multiples of $\pi$. - Edexcel - A-Level Maths Pure - Question 1 - 2011 - Paper 6
Question 1
8. Starting from the formulae for sin(A + B) and cos(A + B), prove that $$ tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\n$$ (b) Deduce that $$ \tan\left... show full transcript
Worked Solution & Example Answer:8. Starting from the formulae for sin(A + B) and cos(A + B), prove that $$ tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\n$$ (b) Deduce that $$ \tan\left(\theta + \frac{\pi}{6}\right) = \frac{1 + \sqrt{3} \tan \theta}{\sqrt{3} - \tan \theta}\n$$ (c) Hence, or otherwise, solve, for $0 \leq \theta < \pi$, $$ 1 + \sqrt{3} \tan \theta = (\sqrt{3} - \tan \theta) \tan(\pi - \theta)\n$$ Give your answers as multiples of $\pi$. - Edexcel - A-Level Maths Pure - Question 1 - 2011 - Paper 6
Step 1
Starting from the formulae for sin(A + B) and cos(A + B), prove that
Answer
To prove the formula for tangent addition, we start with the definitions of sine and cosine:
sin(A + B) = sin A \cos B + cos A \sin B\n$$cos(A + B) = cos A \cos B - sin A \sin B\n$$
The tangent can then be expressed as:
tan(A + B) = \frac{sin(A + B)}{cos(A + B)}\n$$ Substituting the sine and cosine identities leads to:tan(A + B) = \frac{sin A \cos B + cos A \sin B}{cos A \cos B - sin A \sin B}\n$$
Dividing numerator and denominator by gives:
tan(A + B) = \frac{\frac{sin A}{\cos A} + \frac{sin B}{\cos B}}{1 - \frac{sin A}{\cos A} \cdot \frac{sin B}{\cos B}} = \frac{\tan A + \tan B}{1 - \tan A \tan B}\n$$ Thus, the required formula is proved.Step 2
Deduce that
Answer
We need to deduce the tangent addition for the specific angle:
Starting from:
\tan\left(\theta + \frac{\pi}{6}\right) = \frac{\tan \theta + \tan \frac{\pi}{6}}{1 - \tan \theta \tan \frac{\pi}{6}}\n$$ Knowing that $\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$, we substitute: $$\tan\left(\theta + \frac{\pi}{6}\right) = \frac{\tan \theta + \frac{1}{\sqrt{3}}}{1 - \tan \theta \cdot \frac{1}{\sqrt{3}}}\n$$ Upon simplification, we multiply numerator and denominator by $\sqrt{3}$: $$\tan\left(\theta + \frac{\pi}{6}\right) = \frac{\sqrt{3}\tan \theta + 1}{\sqrt{3} - \tan \theta}\n$$ Thus, we have deduced the relationship as required.Step 3
Hence, or otherwise, solve, for 0 ≤ θ < π
Answer
To solve the equation:
1 + \sqrt{3} \tan \theta = (\sqrt{3} - \tan \theta) \tan(\pi - \theta)\n$$ Since $\tan(\pi - \theta) = -\tan \theta$, we substitute:1 + \sqrt{3} \tan \theta = (\sqrt{3} - \tan \theta)(-\tan \theta)\n$$
This leads to:
1 + \sqrt{3} \tan \theta = -\sqrt{3} \tan \theta + \tan^2 \theta\n$$ Rearranging the terms gives us: $$\tan^2 \theta + (\sqrt{3} + \sqrt{3}) \tan \theta - 1 = 0\n$$ This is a quadratic equation in terms of $x = \tan \theta$: $$x^2 + 2\sqrt{3} x - 1 = 0\n$$ Using the quadratic formula, we find: $$x = \frac{-2\sqrt{3} \pm \sqrt{(2\sqrt{3})^2 + 4}}{2}\n$$ Solving this gives solutions for $\tan \theta$ as: Given the angles for $0 \leq \theta < \pi$, we then provide the answers in terms of multiples of $\pi$.