Josh's trigonometric equations

Given the two equations $\sin x+\sin y=1$ and $\cos x+\cos y=\sqrt{2}$ Squaring both the equations, $(\sin x+\sin y)^{2}=1$ $\Rightarrow \sin^{2}x+2\sin x\sin y+\sin^{2}y=1------1$ and $(\cos x +\cos y)^{2}=(\sqrt{2})^{2}$ $\Rightarrow \cos^{2}x+2\cos x\cos y+\cos^{2}y=2------2$ Adding the two equations we get, $\sin^{2}x+\sin^{2}y+2\sin x\sin y+\cos^{2}x+2\cos x\cos y+\cos^{2}y=3$ $\Rightarrow (\sin^{2}x+\cos^{2}x)+(\sin^{2}y+\cos^{2}y)+2\sin x\sin y+2\cos x\cos y=3$ $\Rightarrow 1+1+2\sin x\sin y+2\cos x\cos y=3$ $\Rightarrow +2\sin x\sin y+2\cos x\cos y=1$ $\Rightarrow \cos (x-y)-\cos (x+y)+\cos (x+y)+\cos (x-y)=1$ $\Rightarrow 2\cos (x-y)=1$ $\Rightarrow \cos (x-y)=\frac{1}{2}$ $\Rightarrow (x-y)=60^{\circ}$

Rewrite the first equation as

$\displaystyle 2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)=1 \,\,\, (*)$

And similarly the second one,

$\displaystyle 2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)=\sqrt{2} \,\,\, (**)$

Now divide both the equations to get,

$\displaystyle \tan\left(\frac{x+y}{2}\right)=\frac{1}{\sqrt{2}}$

$\displaystyle\Rightarrow x+y=2\arctan\left(\frac{1}{\sqrt{2}}\right)=2\arcsin\left(\frac{1}{\sqrt{3}}\right)$

Substitute the above value of $x+y$ in $(*)$ ,

$\displaystyle \Rightarrow 2\cdot \frac{1}{\sqrt{3}}\cos\left(\frac{x-y}{2}\right)=1$

$\displaystyle \Rightarrow \cos\left(\frac{x-y}{2}\right)=\frac{\sqrt{3}}{2}$

$\displaystyle \Rightarrow x-y=60^{\circ}$

Hence, the answer is $\fbox{60}$ degrees.

By squaring the two equations, we get:

$\sin^2 x + \sin^2 y + 2 \sin x \sin y = 1,$

$\cos^2 x + \cos^2 y + 2 \cos x \cos y = 2,$

by summing them, we get:

$\sin^2 x + \cos^2 x + \sin^2 y + \cos^2 y + 2 \sin x \sin y + 2 \cos x \cos y = 3$

since: $\sin^2 \theta + \cos^2 \theta = 1$ for any angle $\theta$ , therefore:

$2 + 2 \sin x \sin y + 2 \cos x \cos y = 3,$

since:

$\sin u \sin v = \frac{1}{2}( \cos(u-v) - \cos(u+v) )$

$\cos u \cos v = \frac{1}{2}( \cos(u-v) + \cos(u+v) )$

for any angles $u$ and $v$ , therefore:

$2 + \cos(x-y) - \cos(x+y) + \cos(x-y) + \cos(x+y) = 3,$

$2 \cos(x-y) = 1,$

$\cos(x-y) = \frac{1}{2},$

and $\cos(60°) = \frac{1}{2},$ therefore the answer is $\boxed{60°}$

Squaring the first and second equation gives:

$\sin^2 x + 2\sin x \sin y + \sin^2 y =1$ $\cos^x + 2\cos x \cos y + \cos^2 y =2$

Adding both equation yields:

$\sin^2 x + \cos^2 x + 2\cos x \cos y + 2\sin x \sin y + \sin^2 y + \cos^2 y = 3$ $2 + 2\cos x \cos y + 2\sin x \sin y = 3$ $2(\cos x \cos y + \sin x \sin y) = 1$ $\cos x \cos y + \sin x \sin y = \frac{1}{2}$ $\cos (x-y) = \frac{1}{2}$

Since $0^{\circ} \leq x-y \leq 180^{\circ}$ , $x-y$ can only be equal to $60^{\circ}$ .

$\sin^2 x+\sin^2 y+2\sin x\sin y=1$ $\cos^2 x+cos^2 y+2\cos x\cos y=2$ $\sin^2 x+\sin^2 y+2\sin x\sin y+\cos^2 x+cos^2 y+2\cos x\cos y=3$ $2\sin x\sin y+2\cos x\cos y=\cos(x-y)=\frac{1}{2}$

Therefore, $x-y=60$

One way to look at it is how we should get x-y from trigonometric identities we know. Recall that cos(x-y)=cosxcosy+sinxsiny. So we need to find a way to end up with this term from the given equations.

Square the first equation --> (sinx)^2+2sinxsiny+(siny)^2=1

Square the 2nd equation --> (cosx)^2+2cosxcosy+(cosy)^2=2

then add the two equations:

(sinx)^2+ (cosx)^2+2sinxsiny+2cosxcosy+(cosy)^2+(siny)^2=3

1+2cos(x-y)+1=3

x-y=60

[sin(x) + sin(y)^2 = sin^2(x) + 2sin(x)sin(y) + sin^2(y) = 1. [cos^2(x) + 2cos(x)cos(y) + cos^2(y) = 2. Adding these equations, and simplifying, 2[cos(x - y) = 1, cos(x - y) = 1/2, x - y = 60 degrees.]

$\begin{cases} \sin x + \sin y = 1 & ...(1) \\ \cos x + \cos y = \sqrt 2 & ...(2) \end{cases}$

$\begin{cases} (1)^2: & \sin^2 x + 2\sin x \sin y + \sin^2 y = 1 & ...(1a) \\ (2)^2: & \cos^2 x + 2\cos x \cos y + \cos^2 y = 2 & ...(2a) \end{cases}$

$\begin{aligned} (1a)+(2a): \quad \sin^2 x + \cos^2 x + 2\sin x \sin y + 2\cos x \cos y + \sin^2 y + \cos^2 y & = 1+2 \\ 1 + 2\cos (x-y) + 1 & = 3 \\ \implies \cos (x-y) & = \frac 12 \\ x - y & = \boxed{60}^\circ \end{aligned}$

Square both the equation and add , we will get the following : cos(x) cos(y)+sin(x) sin(y)=1/2 =>cos(x-y)=1/2 =>x-y=60