Prove that sinh(x+y)=sinxcoshy+coshxsinhy

Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is 11 passengers per minute.

If ${\displaystyle \sin x+\sin y=a\text{and}\cos x+\cos y=b}$ then find ${\displaystyle \tan (x-\frac{y}{2})}$

Solve the equation $\sec x-\sec xsi{n}^{2}x=\cos x$

I have the following equation:
${\displaystyle \sin (x-\frac{\pi }{6})+\cos (x+\frac{\pi }{4})=0}$

If $\alpha +\beta +\gamma =\frac{\pi }{2},$, then prove that
$\frac{(1-\tan \frac{\alpha }{2})(1-\tan \frac{\beta }{2})(1-\tan \frac{\gamma }{2})}{(1+\tan \frac{\alpha }{2})(1+\tan \frac{\beta }{2})(1+\tan \frac{\gamma }{2})}=\frac{\sin \alpha +\sin \beta +\sin \gamma -1}{\cos \alpha +\cos \beta +\cos \gamma }$
$\frac{(1-\tan \frac{\alpha }{2})(1-\tan \frac{\beta }{2})(1-\tan \frac{\gamma }{2})}{(1+\tan \frac{\alpha }{2})(1+\tan \frac{\beta }{2})(1+\tan \frac{\gamma }{2})}$
$=\frac{1-\tan \frac{\alpha }{2}-\tan \frac{\beta }{2}-\tan \frac{\gamma }{2}+\tan \frac{\alpha }{2}\tan \frac{\beta }{2}+\tan \frac{\beta }{2}\tan \frac{\gamma }{2}+\tan \frac{\alpha }{2}\tan \frac{\gamma }{2}-\tan \frac{\alpha }{2}\tan \frac{\beta }{2}\tan \frac{\gamma }{2}}{1+\tan \frac{\alpha }{2}+\tan \frac{\beta }{2}+\tan \frac{\gamma }{2}+\tan \frac{\alpha }{2}\tan \frac{\beta }{2}+\tan \frac{\beta }{2}\tan \frac{\gamma }{2}+\tan \frac{\alpha }{2}\tan \frac{\gamma }{2}+\tan \frac{\alpha }{2}\tan \frac{\beta }{2}\tan \frac{\gamma }{2}}$
I am stuck here

Use a half-angel formula to find the exact value of ${\displaystyle \cos \frac{5\pi }{8}}$

P(t) models the distance of a swinging pendulum (In CM) from the place it has travelled t seconds after it starts to swing. Here, t is entered in radians.
${\displaystyle P\left(t\right)=-5\cos \left(2\pi t\right)+5}$
What is the first time the pendulum reaches 3.5 CM from the place it was released?
Round your final answer to the hundredth of a second.
Ok, so I did this to get the solution:
${\displaystyle 3.5=-5\cos \left(2\pi t\right)+5}$
${\displaystyle -1.5=-5\cos \left(2\pi t\right)}$
${\displaystyle 0.3=\cos \left(2\pi t\right)}$
${\displaystyle \cos -1\left(0.3\right)=2\pi t}$
${\displaystyle \frac{{\cos }^{-1}\left(0.3\right)}{2\pi }=t}$
The fact is this gives me 11.55 seconds to get to 3.5 CM, which does not sound right. Where did I go wrong and how can I fix it?