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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?

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    $\begingroup$ 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. $\endgroup$ Commented Jan 14, 2022 at 8:26
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    $\begingroup$ 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). $\endgroup$
    – khashayar
    Commented Jan 14, 2022 at 8:31

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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.

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    $\begingroup$ thankyou so much! $\endgroup$
    – Sanjana
    Commented Jan 14, 2022 at 14:37
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(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.

enter image description here

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  • $\begingroup$ 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? $\endgroup$
    – Deepak
    Commented Jan 14, 2022 at 10:07
  • $\begingroup$ @Deepak Very good remark ! $\endgroup$
    – Jean Marie
    Commented Jan 14, 2022 at 10:36
  • $\begingroup$ thankyou so much!! $\endgroup$
    – Sanjana
    Commented Jan 14, 2022 at 14:38
  • $\begingroup$ 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. $\endgroup$
    – user882145
    Commented Jan 21, 2022 at 1:03
  • $\begingroup$ @boojum You are right. Besides, I just added a graphical representation. Thanks to you... and to Geogebra. $\endgroup$
    – Jean Marie
    Commented Jan 21, 2022 at 6:03
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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 \ ) \ \ . $

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

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

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?

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  • $\begingroup$ Sure ...thankyou!1 $\endgroup$
    – Sanjana
    Commented Jan 14, 2022 at 14:38

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