System of equations

Geometry Level 4

Given that sin ⁡ x sin ⁡ y = 4 \dfrac{\sin x}{\sin y}=4 and cos ⁡ x cos ⁡ y = 1 3 \dfrac{\cos x}{\cos y}=\dfrac{1}{3} .

And if sin ⁡ 2 x sin ⁡ 2 y + cos ⁡ 2 x cos ⁡ 2 y = m n \dfrac{\sin 2x}{\sin 2y}+\dfrac{\cos 2x}{\cos 2y}=\dfrac{m}{n} , where m , n m,n are co-prime positive integers, find the value of m + n m+n .


The answer is 550.

View wiki
Khang Nguyen Thanh by Khang Nguyen Thanh , 27, Vietnam

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

From the given equations we have 9 sin ⁡ 2 x = 144 sin ⁡ 2 y 9{{\sin }^{2}}x=144{{\sin }^{2}}y and 9 cos ⁡ 2 x = cos ⁡ 2 y 9{{\cos }^{2}}x={{\cos }^{2}}y .

It follows that 9 = 9 sin ⁡ 2 x + 9 cos ⁡ 2 x = 144 sin ⁡ 2 y + cos ⁡ 2 y = 144 sin ⁡ 2 y + ( 1 − sin ⁡ 2 y ) 9=9{{\sin }^{2}}x+9{{\cos }^{2}}x=144{{\sin }^{2}}y+{{\cos }^{2}}y=144{{\sin }^{2}}y+\left( 1-{{\sin }^{2}}y \right)

And so sin ⁡ 2 y = 8 143 {{\sin }^{2}}y=\dfrac{8}{143} .

In the same way we get sin ⁡ 2 x = 128 143 , cos ⁡ 2 x = 15 143 {{\sin }^{2}}x=\dfrac{128}{143},\ {{\cos }^{2}}x=\dfrac{15}{143} and cos ⁡ 2 y = 135 143 {{\cos }^{2}}y=\dfrac{135}{143}

Hence, sin ⁡ 2 x sin ⁡ 2 y + cos ⁡ 2 x cos ⁡ 2 y = 2 sin ⁡ x cos ⁡ x 2 sin ⁡ y cos ⁡ y + cos ⁡ 2 x − sin ⁡ 2 x cos ⁡ 2 y − sin ⁡ 2 y = 4. 1 3 + 15 143 − 128 143 135 143 − 8 143 = 169 381 \dfrac{\sin 2x}{\sin 2y}+\dfrac{\cos 2x}{\cos 2y}=\dfrac{2\sin x\cos x}{2\sin y\cos y}+\dfrac{{{\cos }^{2}}x-{{\sin }^{2}}x}{{{\cos }^{2}}y-{{\sin }^{2}}y}=4.\dfrac{1}{3}+\dfrac{\dfrac{15}{143}-\dfrac{128}{143}}{\dfrac{135}{143}-\dfrac{8}{143}}=\dfrac{169}{381}

So, m + n = 169 + 381 = 550 m+n=169+381=\boxed{550}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

I did the whole same process 3 times!! I couldn't believe, I would get such an ugly answer!

Navneel Mandal - 5 years, 7 months ago
Chew-Seong Cheong
Nov 13, 2015

{ sin ⁡ x sin ⁡ y = 4 ⇒ sin ⁡ x = 4 sin ⁡ y ⇒ sin ⁡ 2 x = 16 sin ⁡ 2 y . . . ( 1 ) cos ⁡ x cos ⁡ y = 1 3 ⇒ 3 cos ⁡ x = cos ⁡ y ⇒ 9 cos ⁡ 2 x = cos ⁡ 2 y . . . ( 2 ) \begin{cases} \dfrac{\sin x}{\sin y} = 4 & \Rightarrow \sin x = 4 \sin y & \Rightarrow \sin^2 x = 16 \sin^2 y & ... (1) \\ \dfrac{\cos x}{\cos y} = \dfrac{1}{3} & \Rightarrow 3 \cos x = \cos y & \Rightarrow 9\cos^2 x = \cos^2 y & ... (2) \end{cases}

⇒ 9 ( 1 ) + ( 2 ) : 9 sin ⁡ 2 x + 9 cos ⁡ 2 x = 144 sin ⁡ 2 y + cos ⁡ 2 y 9 = 143 sin ⁡ 2 y + 1 ⇒ sin ⁡ 2 y = 8 143 ⇒ 1 − 2 sin ⁡ 2 y = 1 − 2 ( 8 143 ) ⇒ cos ⁡ 2 y = 127 143 ⇒ sin ⁡ 2 x = 16 sin ⁡ 2 y = 128 143 ⇒ 1 − 2 sin ⁡ 2 x = 1 − 2 ( 128 143 ) ⇒ cos ⁡ 2 x = − 113 143 \begin{aligned} \Rightarrow 9(1)+(2): \quad 9\sin^2 x + 9 \cos^2 x & = 144 \sin^2 y + \cos^2 y \\ 9 & = 143 \sin^2 y + 1 \\ \Rightarrow \sin^2 y & = \frac{8}{143} \Rightarrow 1 - 2 \sin^2 y = 1- 2 \left(\frac{8}{143}\right) \Rightarrow \color{#3D99F6}{\cos 2y = \frac{127}{143}} \\ \Rightarrow \sin^2 x & = 16 \sin^2 y = \frac{128}{143} \Rightarrow 1 - 2 \sin^2 x = 1- 2 \left(\frac{128}{143}\right) \Rightarrow \color{#D61F06}{ \cos 2x = - \frac{113}{143}} \end{aligned}

Now, we have:

sin ⁡ 2 x sin ⁡ 2 y + cos ⁡ 2 x cos ⁡ 2 y = 2 sin ⁡ x cos ⁡ x 2 sin ⁡ y cos ⁡ y + − 113 143 127 143 = 4 ( 1 3 ) − 113 127 = 508 − 339 381 = 169 381 \begin{aligned} \frac{\sin 2x}{\sin 2y} + \frac{\color{#D61F06}{\cos 2x}}{\color{#3D99F6}{\cos 2y}} & = \frac{2\sin x \cos x}{2 \sin y \cos y} + \frac{\color{#D61F06}{- \frac{113}{143}}}{\color{#3D99F6}{\frac{127}{143}}} & = 4 \left(\frac{1}{3}\right) - \frac{113}{127} \\ & = \frac{508-339}{381} = \frac{169}{381} \end{aligned}

⇒ m + n = 169 + 381 = 550 \Rightarrow m + n = 169+381 = \boxed{550}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Missed in calculation mistake ! My solution:

sin ⁡ x sin ⁡ y = 4 ⟹ 1 − cos ⁡ 2 x 1 − cos ⁡ 2 y = 16 ( 2 cos ⁡ 2 x − 1 ) − 1 ( 2 cos ⁡ 2 y − 1 ) − 1 = 16 ⟹ 16 cos ⁡ 2 y − 16 = cos ⁡ 2 x − 1 cos ⁡ 2 x − 16 cos ⁡ 2 y = − 15 . . . ( 1 ) \begin{aligned} \dfrac{\sin x}{\sin y} & =4 \\ \implies \dfrac{1-\cos^2 x}{1-\cos^2 y} & =16 \\ \dfrac{(2 \cos^2 x-1)-1}{(2\cos^2 y-1)-1} & =16 \\ \implies 16\cos 2y - 16 & = \cos 2x-1 \\ \cos 2x - 16 \cos 2y & = -15 & ... (1) \end{aligned} \
\dfrac{Cosx}{Cosy}=\dfrac 1 3,~~~~~\implies~~\dfrac{(2Cos^2x-1)+1}{(2Cos^2y-1)+1}=\dfrac 1 9.\
\implies~~9Cos2x+9=Cos2y+1,~~\therefore~~9Cos2x - Cos2y = -8.........................(2) \
(2)-9(1), ~~gives~~{\color{blue}{Cos2y=\dfrac{127}{143} } }.~~Substituting~this~in~(2),\
9Cos2x= \dfrac{127}{143}-8=\dfrac{127}{143}- \frac {16+7 143]\dfrac{127}{143}{143}.~~~\implies~\color{blue}{Cos2x =\dfrac{16}{9 143} }.\
\dfrac{Sin2x}{Sin2y} + \dfrac{Cos2x}{Cos2y} =dfrac{2Sinx}{2Siny} \dfrac{Cosx}{Cosy}+\dfrac{\frac{16}{9 143}}{\frac{127}{143} }})



Niranjan Khanderia - 3 years, 9 months ago
Will Jain
Nov 14, 2015

written sinx/siny in terms of cos and plugged the given cosx/cosy in that equation and found the required values

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...