#### mathlover2358

##### New member

- Joined
- Feb 6, 2020

- Messages
- 2

The question is what value of

*a*gives a unique solution?

sin

*x*+ sin

*y*= 0.25

cos

*x*+ cos

*y*=

*a*

- Thread starter mathlover2358
- Start date

- Joined
- Feb 6, 2020

- Messages
- 2

The question is what value of

sin

cos

\(\displaystyle \sin(x) + \sin(y) = \dfrac 1 4\\

\sin^2(x) + \sin^2(y) + 2\sin(x)\sin(y) = \dfrac{1}{16}\\~\\

\cos(x) + \cos(y) = a\\

\cos^2(x) + \cos^2(y) + 2\cos(x)\cos(y) = a^2\\~\\

\text{now add them}\\

1 + 1 + 2(\sin(x)\sin(y) + \cos(x)\cos(y) = a^2 + \dfrac{1}{16}\\

2(1 + cos(x + y)) = a^2 + \dfrac{1}{16}\\

\cos(x+y) = \dfrac{a^2 + \frac{1}{16}}{2}-1\\

y = -x + \cos^{-1}\left(\dfrac{a^2 + \frac{1}{16}}{2}-1\right),~-x - \cos^{-1}\left(\dfrac{a^2 + \frac{1}{16}}{2}-1\right)\\

\text{In order for the solution to be unique we must have both arccosine terms evaluate to zero}\\

\text{This means that the argument of both must be 1}\\

\dfrac{a^2 + \frac{1}{16}}{2}-1 = 1\\

a = \pm \dfrac{3\sqrt{7}}{4}

\)

- Joined
- Feb 6, 2020

- Messages
- 2

Thank You so much