# Question #e8ab5

Aug 21, 2016

$\cos \left(x + y\right) = \frac{{a}^{2} + {b}^{2}}{2} - 1$

#### Explanation:

First, recall what $\cos \left(x + y\right)$ is:
$\cos \left(x + y\right) = \cos x \cos y + \sin x \sin y$

Note that:
${\left(\sin x + \sin y\right)}^{2} = {a}^{2}$
$\to {\sin}^{2} x + 2 \sin x \sin y + {\sin}^{2} y = {a}^{2}$

And:
${\left(\cos x + \cos y\right)}^{2} = {b}^{2}$
$\to {\cos}^{2} x + 2 \cos x \cos y + {\cos}^{2} y = {b}^{2}$

Now we have these two equations:
${\sin}^{2} x + 2 \sin x \sin y + {\sin}^{2} y = {a}^{2}$
${\cos}^{2} x + 2 \cos x \cos y + {\cos}^{2} y = {b}^{2}$

If we add them together, we have:
${\sin}^{2} x + 2 \sin x \sin y + {\sin}^{2} y + {\cos}^{2} x + 2 \cos x \cos y + {\cos}^{2} y = {a}^{2} + {b}^{2}$

Don't let the size of this equation throw you off. Look for identities and simplifications:
$\left({\sin}^{2} x + {\cos}^{2} x\right) + \left(2 \sin x \sin y + 2 \cos x \cos y\right) + \left({\cos}^{2} y + {\sin}^{2} y\right) = {a}^{2} + {b}^{2}$

Since ${\sin}^{2} x + {\cos}^{2} x = 1$ (Pythagorean Identity) and ${\cos}^{2} y + {\sin}^{2} y = 1$ (Pythagorean Identity), we can simplify the equation to:
$1 + \left(2 \sin x \sin y + 2 \cos x \cos y\right) + 1 = {a}^{2} + {b}^{2}$
$\to \left(2 \sin x \sin y + 2 \cos x \cos y\right) + 2 = {a}^{2} + {b}^{2}$

We can factor out a $2$ twice:
$2 \left(\sin x \sin y + \cos x \cos y\right) + 2 = {a}^{2} + {b}^{2}$
$\to 2 \left(\left(\sin x \sin y + \cos x \cos y\right) + 1\right) = {a}^{2} + {b}^{2}$

And divide:
$\left(\sin x \sin y + \cos x \cos y\right) + 1 = \frac{{a}^{2} + {b}^{2}}{2}$

And subtract:
$\sin x \sin y + \cos x \cos y = \frac{{a}^{2} + {b}^{2}}{2} - 1$

Finally, since $\cos \left(x + y\right) = \cos x \cos y + \sin x \sin y$, we have:
$\cos \left(x + y\right) = \frac{{a}^{2} + {b}^{2}}{2} - 1$

Aug 21, 2016

Given

$\sin x + \sin y = a \ldots \ldots . \left(1\right)$

$\cos x + \cos y = b \ldots \ldots . \left(2\right)$

Squaring and adding (1) & (2)

${\left(\cos x + \cos y\right)}^{2} + {\left(\sin x + \sin y\right)}^{2} = {a}^{2} + {b}^{2}$

$\implies 2 \left(\cos x \cos y + \sin x \sin y\right) + 2 = {a}^{2} + {b}^{2}$

$\implies 2 \cos \left(x - y\right) = {a}^{2} + {b}^{2} - 2. \ldots \left(3\right)$

Squaring and Subtracting (1) from(2)

${\left(\cos x + \cos y\right)}^{2} - {\left(\sin x + \sin y\right)}^{2} = {b}^{2} - {a}^{2}$

$\implies 2 \cos \left(x + y\right) + {\cos}^{2} x - {\sin}^{2} x + {\cos}^{2} y - {\sin}^{2} y = {b}^{2} - {a}^{2}$

$\implies 2 \cos \left(x + y\right) + \cos 2 x + \cos 2 y = {b}^{2} - {a}^{2}$

$\implies 2 \cos \left(x + y\right) + 2 \cos \left(x + y\right) \cos \left(x - y\right) = {b}^{2} - {a}^{2}$

$\implies \cos \left(x + y\right) \left(2 + 2 \cos \left(x - y\right)\right) = {b}^{2} - {a}^{2}$

($\text{From (3) } 2 \cos \left(x - y\right) = {a}^{2} + {b}^{2} - 2$)

$\implies \cos \left(x + y\right) \left(2 + {b}^{2} + {a}^{2} - 2\right) = {b}^{2} - {a}^{2}$

$\implies \cos \left(x + y\right) \left({b}^{2} + {a}^{2}\right) = {b}^{2} - {a}^{2}$

$\implies \cos \left(x + y\right) = \frac{{b}^{2} - {a}^{2}}{{b}^{2} + {a}^{2}}$

Aug 21, 2016

$\cos \left(x + y\right) = \frac{{b}^{2} - {a}^{2}}{{b}^{2} + {a}^{2}}$.

#### Explanation:

$\sin x + \sin y = a \Rightarrow 2 \sin \left(\frac{x + y}{2}\right) \cos \left(\frac{x - y}{2}\right) = a \ldots \ldots \ldots \left(1\right)$.

$\cos x + \cos y = b \Rightarrow 2 \cos \left(\frac{x + y}{2}\right) \cos \left(\frac{x - y}{2}\right) = b \ldots \ldots \ldots . \left(2\right)$.

Dividing $\left(1\right)$ by $\left(2\right)$, we have, $\tan \left(\frac{x + y}{2}\right) = \frac{a}{b}$.

Now, $\cos \left(x + y\right) = \frac{1 - {\tan}^{2} \left(\frac{x + y}{2}\right)}{1 + {\tan}^{2} \left(\frac{x + y}{2}\right)}$

$= \frac{1 - {a}^{2} / {b}^{2}}{1 + {a}^{2} / {b}^{2}} = \frac{{b}^{2} - {a}^{2}}{{b}^{2} + {a}^{2}}$.

Enjoy Maths.!