# Math Help - Simultaneous equations?

1. ## Simultaneous equations?

Solve the simultaneous equations sin x. cos y = 1/4 & 3 tan x = tan y.

Ans: x = -π/6 + (k + l)π ; k,l belong to Z
y = 2π/3 + (k - l)π ; k, l belong to Z

The approach to the problem is very direct. There is no difficulty in solving it. However, there seems to be a conflict with the given answer and my answer. Solving, I got two equations: x + y = kπ + (-1)^k * π/2
x - y = lπ - (-1)^l * π/6
Proceeding further, my answers dont match with the given solutions.

2. would you mind if you write your complete solution?

3. Originally Posted by fardeen_gen
Solve the simultaneous equations sin x. cos y = 1/4 & 3 tan x = tan y.
I probably went a roundabout and tediously long way of finding this, but the idea here is to abuse substitution and take advantage of the identity $sin^2(x)+cos^2(x)=1$...

You have:

$sin(x)cos(y)=\frac{1}{4}$

$3tan(x)=tan(y)$

Notice that...

$3tan(x)=tan(y)$

$3\frac{sin(x)}{cos(x)}=\frac{sin(y)}{cos(y)}$

$3sin(x)cos(y)=sin(y)cos(x)$

$\frac{3}{4}=sin(y)cos(x)$

I guess that really didn't get us anywhere, did it?

Let's use the identity: $sin^2(y)+cos^2(y)=1$

$1-cos^2(y)=sin^2(y)$

From $sin(x)cos(y)=\frac{1}{4}$, we have $cos(y)=\frac{1}{4sin(x)}$, so $cos^2(y)=\frac{1}{16sin^2(x)}$, and then

$1-\frac{1}{16sin^2(x)}=sin^2(y)$

where

$sin(y)=\sqrt{1-\frac{1}{16sin^2(x)}}$

We had $\frac{3}{4}=sin(y)cos(x)$ which can now simplify to just

$\frac{3}{4}=cos(x)\sqrt{1-\frac{1}{16sin^2(x)}}$

$\frac{3}{4cos(x)}=\sqrt{1-\frac{1}{16sin^2(x)}}$

Square both sides

$\frac{9}{16cos^2(x)}=1-\frac{1}{16sin^2(x)}$

Multiply by $16sin^2(x)cos^2(x)$

$9sin^2(x)=16sin^2(x)cos^2(x)-cos^2(x)$

Substitute $cos^2(x)=1-sin^2(x)$

$9sin^2(x)=16sin^2(x)(1-sin^2(x))-(1-sin^2(x))$

$9sin^2(x)=16sin^2(x)-16sin^4(x)-1+sin^2(x)$

$0=8sin^2(x)-16sin^4(x)-1$

Let $u = sin^2(x)$

$-16u^2+8u-1=0$

$16u^2-8u+1=0$

$u = \frac{8 \pm \sqrt{64-4(16)}}{32}$

$sin^2(x) = \frac{8 \pm \sqrt{0}}{32}$

$sin^2(x) = \frac{1}{4}$

$sin(x)=\frac{1}{2}$

Solve for x.

4. x = (k + l)π/2 + (-1)^k * π/4 - (-1)^l * π/12
y = from x

5. the previous post was in response to kalagota's request.

6. i was expecting the one before you came with x+y and x-y..
anyways, try to follow what colby wrote.. but take note of the last two steps..

$\sin^2 x = \frac{1}{4} \Longrightarrow \sin x = \pm\frac{1}{2}$

7. Originally Posted by kalagota
i was expecting the one before you came with x+y and x-y..
anyways, try to follow what colby wrote.. but take note of the last two steps..

$\sin^2 x = \frac{1}{4} \Longrightarrow \sin x = \pm\frac{1}{2}$
+/-, forgot to put that in there, thanks kalagota

8. Originally Posted by colby2152
I probably went a roundabout and tediously long way of finding this, but the idea here is to abuse substitution and take advantage of the identity $sin^2(x)+cos^2(x)=1$...

You have:

$sin(x)cos(y)=\frac{1}{4}$

$3tan(x)=tan(y)$

Notice that...

$3tan(x)=tan(y)$

$3\frac{sin(x)}{cos(x)}=\frac{sin(y)}{cos(y)}$

$3sin(x)cos(y)=sin(y)cos(x)$

$\frac{3}{4}=sin(y)cos(x)$

I guess that really didn't get us anywhere, did it?
Actually it did

This is simpler, I think.

Add the newly obtained equation to the original.
That is:

$\sin y \cos x+\sin x\cos y= \frac{3}{4}+\frac14 = 1$

But $\sin y \cos x+\sin x\cos y = \sin(x+y)$.

Thus $\sin(x+y) = 1 \Rightarrow \sin x = \sin \left(\frac{\pi}2 - y \right) = \cos y$

Substituting back in the original equation:

$\sin x\cos y = (\sin x)^2 = \frac14$