Let $x$ and $y$ be real numbers such that $\frac{\sin x}{\sin y} = 3$ and $\frac{\cos x}{\cos y} = \frac{1}{2}$. The value of $\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}$ can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$.

My student left me with a list of questions just now (he also said that he wanted to be helped at school :P)

  • 1
    $\begingroup$ You can use the double angle formulae: $\sin 2x = 2 \sin x \cos x$ and $\cos 2x = 1 - 2 \sin^x $ to help you find $\sin 2x$, $\sin 2y$, $\cos 2x$ and $\cos 2y$. $\endgroup$ – Toby Mak Oct 26 '17 at 13:17
  • $\begingroup$ Okay. So, $\frac{3\sin y \cos x}{\sin y \times 2\cos y}$ is the same as $\frac{\sin 2x}{\sin 2y}$. $\endgroup$ – Bhavani Nevalkar Oct 26 '17 at 13:20
  • $\begingroup$ Can you show your working by posting an answer to your question? It would be easier for me to follow your steps this way. $\endgroup$ – Toby Mak Oct 26 '17 at 13:22

\begin{align} {\sin2x\over\sin2y}+{\cos2x\over\cos2y}= {}& {2\sin x\cos x\over2\sin y\cos y}+{\cos^2x-\sin^2x\over\cos^2y-\sin^2y} \\[10pt] = {} & {3\sin y\cos y\over2\sin y\cos y}+{(1/4)\cos^2y-9\sin^2y\over\cos^2y-\sin^2y} \\[10pt] = {} & {3\over2}+{(1/4)\cos^2y-9\sin^2y\over\cos^2y-\sin^2y}. \end{align}

On the other hand: $$ 1=\cos^2x+\sin^2x={1\over4}\cos^2y+9(1-\cos^2y), \quad\hbox{whence:}\quad \cos^2y={32\over35},\quad\sin^2y={3\over35}. $$ We finally have $$ {\sin2x\over\sin2y}+{\cos2x\over\cos2y}= {3\over2}+{{1\over4}{32\over35}-9{3\over35}\over{32\over35}-{3\over35}}={49\over58}. $$


\begin{align} \frac{\sin x}{\sin y} &= 3 \tag{1}\label{1} \\ \frac{\cos x}{\cos y} &= \frac{1}{2} \tag{2}\label{2} \end{align}

\begin{align} \frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y} &=\frac{p}{q} \tag{3}\label{3} \end{align}

Find $p+q$.

\begin{align} \frac{\sin 2x}{\sin 2y} &= \frac{2\sin x\cos x}{2\sin y\cos y} =\frac32 \tag{4}\label{4} \end{align}

\begin{align} \frac{\sin x}{\sin y} &= \frac{6\cos x}{\cos y} = 3 \tag{5}\label{5} \end{align}

\begin{align} \frac{\sin^2 x}{\sin^2 y} &= \frac{36\cos^2 x}{\cos^2 y} = 9 \tag{6}\label{6} \\ \frac{\sin^2 x+36\cos^2 x}{\sin^2 y+\cos^2 y} &= 9 \tag{7}\label{7} \\ 35\cos^2 x &= 8 \\ 2\cos^2 x &= \frac{16}{35} \tag{8}\label{8} \\ 2\cos^2 x-1 &= \frac{16}{35}-1 \tag{9}\label{9} \\ \cos 2x&=-\frac{19}{35} \tag{10}\label{10} \end{align}


\begin{align} \frac{\sin x}{6\sin y} &= \frac{\cos x}{\cos y} = \frac12 \tag{11}\label{11} \end{align}

\begin{align} \frac{\sin^2 x}{36\sin^2 y} &= \frac{\cos^2 x}{\cos^2 y} = \frac14 \tag{12}\label{12} \\ \frac{\sin^2 x+\cos^2 x}{36\sin^2 y+\cos^2 y} &= \frac14 \tag{13}\label{13} \end{align}

\begin{align} 35\sin^2 y+1&=4 \tag{14}\label{14} \\ \sin^2 y&=\frac{3}{35} \tag{15}\label{15} \\ 1-2\sin^2y&=1-\frac{6}{35} \tag{16}\label{16} \\ \cos2y&=\frac{29}{35} \tag{17}\label{17} . \end{align}


\begin{align} \frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y} &= \frac32-\frac{19}{29}=\frac{49}{58}=\frac{p}{q}. \end{align}


  • $\begingroup$ Thanks a lot! You are the only one who got to the answer fully. $\endgroup$ – Bhavani Nevalkar Oct 26 '17 at 17:38

The first two equations, squared, allow you to solve for $\sin^2$ and $\sin^2y$ as follows:


Then $\sin^2x=\dfrac{27}{35},\sin^2y=\dfrac3{35}$.

From this, by the double angle formula

$$\begin{cases}\cos 2x=-\dfrac{19}{35},\\\cos 2y=\dfrac{29}{35}.\end{cases}$$

The rational ratio is $$\dfrac pq=\frac32-\frac{19}{29}=\frac{49}{58}.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.