# Solve for $x,y$: $\sin (x-y)=3\sin x \cos y-1$ and $\sin (x+y)=-2\cos x \sin y$

I was trying to solve $$\sin (x-y)=3\sin x \cos y-1 \\ \sin (x+y)=-2\cos x \sin y$$ for $$x,y$$, so first I added both the equations and then subtracted both of them and solved these two equations the result I got was $$\frac{\tan y}{\tan x}=-\frac{1}{3}$$ but I am not able to further find the value of $$x$$ and $$y$$. Please can you help me solve this further?

• Welcome to MSE. A question should be written in such a way that it can be understood even by someone who did not read the title. Commented Jan 14, 2022 at 8:26
• Hint: you can use these two formulas: sin(x+y)=sin(x)cos(y)+cos(x)sin(y), sin(x-y)=sin(x)cos(y)-cos(x)sin(y). Commented Jan 14, 2022 at 8:31

Let $$a = x + y, b = x - y$$. Then using the product-to-sum identities:

$$\sin b = 3 \sin \left( \frac{a+b}{2} \right) \cos \left( \frac{a-b}{2} \right) - 1 = \frac{3}{2} (\sin a + \sin b) - 1$$ $$\sin a = -2 \cos \left( \frac{a+b}{2} \right) \sin \left( \frac{a-b}{2} \right) = -(\sin a - \sin b)$$

and by a further substitution $$p = \sin a, q = \sin b$$:

$$2q = 3p + 3q - 2, p = q - p$$

which gives $$0 = 3p + q - 2 = 3p + 2p - 2$$, thus $$p = \frac{2}{5}$$ and $$q = \frac{4}{5}$$ and you can work your way backwards.

• thankyou so much! Commented Jan 14, 2022 at 14:37

(See below a graphical representation of the two implicit curves, the first one in black, the second one in red).

Expanding the two LHS, you get a system:

$$\begin{cases}-2 \sin x \cos y-\sin y \cos x & =& -1 \\ \sin x \cos y+3\sin y \cos x & =& 0\end{cases}$$

which solved as the linear system $$\begin{cases}-2 a-b & =& -1 \\ a+3 b & =& 0\end{cases}$$

gives

$$a=\sin x \cos y=\frac35 \ \ \ and \ \ \ b=\sin y \cos x=-\frac15. \tag{1}$$

Now express that :

$$(\cos y)^2+(\sin y)^2=1 \tag{2}$$

giving:

$$\dfrac{a^2}{(\sin x)^2}+\dfrac{b^2}{(\cos x)^2}=1$$

involving only $$(\cos x)^2$$ and $$(\sin x)^2 = 1-(\cos x)^2$$ finally leaving you with a quadratic equation in $$(\cos x)^2$$.

Having solved it, you just have to replace the values of $$x$$ in equations (1) to get the values of $$y$$.

But you may obtain spurious solutions due to the squarings in (2). Therefore you must check all of them in the initial system.

• Good solution. But if I may suggest an improvement, in the final steps, you have $\sin x\cos y = 0.6, \cos x\sin y = - 0.2$. Add and subtract to get equations $\sin(x+y) = 0.4$ and $\sin(x-y) = 0.8$ respectively, following which you can find general solutions for $x+y$ and $x-y$ respectively, then add and subtract them to find all possible values of $x$ and $y$. A bit neater than squaring maybe? Commented Jan 14, 2022 at 10:07
• @Deepak Very good remark ! Commented Jan 14, 2022 at 10:36
• thankyou so much!! Commented Jan 14, 2022 at 14:38
• There is no ambiguity about the values of $\ x \$ and $\ y \$. Your results show that $\ \sin(x+y) \ > \ 0 \ \ ,$ but the given equation is $\ \sin (x+y) \ = \ -2\cos x \sin y \ \ .$ So $\ \cos x \$ is negative, placing $\ x \$ and $\ y \$ in the second and first quadrants, respectively.
– user882145
Commented Jan 21, 2022 at 1:03
• @boojum You are right. Besides, I just added a graphical representation. Thanks to you... and to Geogebra. Commented Jan 21, 2022 at 6:03

You were well on the way when you got to your tangent ratio; you just needed to continue in that vein. Presumably, you found from the "angle-addition" formulas that $$\sin(x+y) \ \ = \ \ \sin x · \cos y \ + \ \cos x · \sin y \ \ = \ \ -2· \cos x · \sin y$$ $$\Rightarrow \ \ \sin x · \cos y \ \ = \ \ -3· \cos x · \sin y \ \ . \quad \quad \quad \mathbf{[1]}$$ You will want to use this in the other equation at some point.

From the "angle-difference" equation, we obtain $$\sin (x-y) \ \ = \ \ \sin x · \cos y \ - \ \cos x · \sin y \ \ = \ \ 3·\sin x \cos y \ - \ 1$$ $$= \ \ \sin x · \cos y \ - \ (1 \ - \ 2·\sin x · \cos y)$$ $$\Rightarrow \ \ \cos x · \sin y \ \ = \ \ 1 \ - \ 2·\sin x · \cos y \ \ ; \quad \quad \quad \mathbf{[2]}$$ but also $$\sin (x-y) \ \ = \ \ \sin x · \cos y \ - \ \cos x · \sin y \ \ = \ \ -4· \cos x · \sin y \ \ ,$$ inserting equation $$\ \mathbf{[1]} \ \ .$$ Consequently, $$\ \sin(x-y) \ = \ 2 · \sin (x+y) \ \ . \quad \quad \quad \mathbf{[3]}$$

We may then insert equation $$\ \mathbf{[2]} \$$ into the "angle-sum" , yielding $$\sin(x+y) \ \ = \ \ \sin x · \cos y \ + \ (1 \ - \ 2·\sin x · \cos y) \ \ = \ \ 1 \ - \ \sin x · \cos y \ \ ;$$ it follows from the given "angle-difference" equation, this most recent result, and $$\ \mathbf{[3]} \$$ that $$3·\sin x \cos y \ - \ 1 \ \ = \ \ 2 · (1 \ - \ \sin x · \cos y) \ \ \Rightarrow \ \ \sin x · \cos y \ \ = \ \ \frac35$$ $$\Rightarrow \ \ \sin(x+y) \ \ = \ \ 1 \ - \ \frac35 \ \ = \ \ \frac25 \ \ \Rightarrow \ \ \sin(x-y) \ \ = \ \ \frac45 \ \ .$$

It seems reasonable to conclude that $$\ 0 \ < \ (x + y) \ , \ (x - y) \ < \ \pi \ \ ,$$ since their sines are both positive. But as we were given that $$\sin(x+y) \ = \ -2· \cos x · \sin y \ ,$$ which is positive, we may then deduce that $$\ \cos x \ < \ 0 \ \ ,$$ indicating that $$\ x \$$ and $$\ y \$$ are in adjacent quadrants (that your tangent ratio is negative also tells us this; the result that $$\ \sin(x+y) \ < \ \sin(x-y) \$$ hints at it).

We may now complete our calculations by solving the system $$\begin{array}{c} x \ + \ y \ \ = \ \ \arcsin \left(\frac25 \right) \ \ \approx \ \ 0.4115 \ \ \text{or} \ \ \mathbf{\pi \ - \ 0.4115 \ = \ 2.7301} \\ x \ - \ y \ \ = \ \ \arcsin \left(\frac45 \right) \ \ \approx \ \ 0.9273 \end{array} \ \ \ ,$$ $$\Rightarrow \ \ x \ \ \approx \ \ 1.8287 \ \ [ \approx \ 104.8º ] \ \ \ , \ \ \ y \ \ \approx \ \ 0.9014 \ \ [ \approx \ 51.6º ] \ \ .$$

Testing these values in the original system of equations confirms them (and indeed, $$\ \frac{\tan 0.9014}{\tan 1.8287} \ \approx \ \frac{1.2638}{-3.7911} \ = \ -\frac13 \ ) \ \ .$$

$$2\sin x\cos y-\cos x\sin y+1=0$$

$$\sin x\cos y+3\cos x\sin y=0$$

Using cross multiplication,

$$\dfrac{\sin x}{-3\sin y}=\dfrac{\cos x}{\cos y}=\dfrac1{7\cos y\sin y}$$

$$\implies7\sin x=-3\sec y,7\cos x=\csc y$$

$$7^2=(-3\sec y)^2+(\csc y)^2=3(1+t)+\dfrac{1+t}t$$ where $$t=\tan^2y$$

So, we have a quadratic equation in $$t$$

Can you take it home from here?

• Sure ...thankyou!1 Commented Jan 14, 2022 at 14:38