3. A question on trigonometry (Eton College 01C_MT1 Q15)

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This question comes from a test paper for A-level maths for Michaelmas Term at Eton College.

Q15. (a) Express $\cos(2\theta)+\sin(2\theta)$ in the form $R\sin(2\theta+\alpha)$ where $R>0$ and $0 < \alpha < \frac{\pi}{2}$. [3 marks]

(b) Hence show that $$ \begin{align} \frac{1-\sqrt{2}}{2}\le \cos\theta(\cos\theta+\sin\theta) \le\frac{1+\sqrt{2}}{2} \end{align} $$ for all real $\theta$. [3 marks]

 

Solution. (a) This is to find the harmonic identity. $$ \begin{align} \cos(2\theta)+\sin(2\theta) &=R\sin(2\theta+\alpha) \\ &=R(\sin2\theta\cos\alpha+\cos2\theta\sin\alpha) \\ &=(R\sin\alpha)cos2\theta+(R\cos\alpha)\sin2\theta \end{align} $$ Equating the cofficients yields the following simultaneous equations for $R$ and $\alpha$, $$ \left\{\begin{array}{l} R\sin\alpha=1 \\ R\cos\alpha=1 \end{array}\right. $$ which yields $$ \begin{align} {\rm (i)}&\qquad R^2\underbrace{\left(\sin^2\alpha+\cos^2\alpha\right)}_{=1}=R^2=2 & &\Rightarrow & R&=\sqrt{2} \\ {\rm (ii)}&\qquad \frac{R\sin\alpha}{R\cos\alpha}=\tan\alpha=1 & &\Rightarrow & \alpha&=\tan^{-1}(1)=\frac{\pi}{4} \end{align} $$ Thus we obtain $$ \begin{align} \cos(2\theta)+\sin(2\theta)=\sqrt{2}\sin\left(2\theta+\frac{\pi}{4}\right) \end{align} $$

 

(b) Using the double angle formulae, we can re-write the LHS of (a) as follows. $$ \begin{align} \cos(2\theta)+\sin(2\theta) &=\left(2\cos^2\theta-1\right)+2\sin\theta\cos\theta \\ &=2\cos\theta(\cos\theta+\sin\theta)-1 \end{align} $$ So the result of (a) now reads $$ \begin{align} 2\cos\theta(\cos\theta+\sin\theta)-1=\sqrt{2}\sin\left(2\theta+\frac{\pi}{4}\right) \end{align} $$ For the RHS, we also note that $$ \begin{align} -1\le \sin\left(2\theta+\frac{\pi}{4}\right)&\le 1 \\ \Rightarrow\quad -\sqrt{2}\le \sqrt{2} \sin\left(2\theta+\frac{\pi}{4}\right)&\le \sqrt{2} \end{align} $$ and find $$ \begin{align} -\sqrt{2}\le 2\cos\theta(\cos\theta+\sin\theta)-1 &\le \sqrt{2} \\ \Rightarrow\quad 1-\sqrt{2}\le 2\cos\theta(\cos\theta+\sin\theta) &\le 1+\sqrt{2} \\ \Rightarrow\quad \frac{1-\sqrt{2}}{2}\le \cos\theta(\cos\theta+\sin\theta) &\le \frac{1+\sqrt{2}}{2} \quad\checkmark \end{align} $$