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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)

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    $\begingroup$ 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
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\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 $$ {\sin2x\over\sin2y}+{\cos2x\over\cos2y}= {3\over2}+{{1\over4}{32\over35}-9{3\over35}\over{32\over35}-{3\over35}}={49\over58}. $$

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\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}

Similarly,

\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}

So,

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

$p+q=49+58=107$.

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  • $\begingroup$ Thanks a lot! You are the only one who got to the answer fully. $\endgroup$ – Bhavani Nevalkar Oct 26 '17 at 17:38
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The first two equations, squared, allow you to solve for $\sin^2$ and $\sin^2y$ as follows:

$$\begin{cases}\sin^2x&=9\sin^2y,\\4(1-\sin^2x)&=1-\sin^2y.\end{cases}$$

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 $$\dfrac pq=\frac32-\frac{19}{29}=\frac{49}{58}.$$

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