Finding greatest value of a trigonometric expression involving two angles 
A particle is projected up a plane inclined at an angle $b$ to the
  horizontal, the angle of projection being $a$ above the horizontal. 
Part 1: If the initial speed of projection is $V$ m/s show that 
  range, $C$, on the inclined plane is given by: 
  $$C= 2 V^2 \frac{\sin(a-b) \cos(a)}{g\cos(b)^2}$$
Part 2: Determine the maximum range ($C$ max) on the plane and the 
  value of $b$ for which this occurs.
  maximising the trigonometric function is the main problem here

Here is a diagram:

 A: Here is a diagram of the problem with a few helpful elements added:

The figure illustrates the fact that if not for the influence of gravity,
the projectile would simply travel a distance $Vt$ along the line at an angle $a$
above the horizontal in a certain time $t,$
but because of gravity, the projectile is deflected downward a distance
$\frac 12 gt^2,$ 
so it ends up at the intersection of the blue parabola and the vertical red line.
This point is the landing point on the inclined surface if we choose $t$
to be the amount of time the projectile is in flight before striking that surface.
Now we do a little trigonometry. We draw a right triangle with the
segment $Vt$ as the hypotenuse and an extension of segment $C$ as one of the legs.
The vertical red line cuts a smaller right triangle from this triangle;
the smaller triangle has hypotenuse $\frac 12 gt^2$
and angle $b$ at the topmost vertex, hence the adjacent leg is $\frac 12 gt^2 \cos b.$
But that leg is also the leg opposite the angle $a-b$ of the larger right triangle,
so we have
$$Vt \sin (a-b) = \frac 12 gt^2 \cos b.$$
Now consider the horizontal segment in the figure, which is the leg
adjacent to angle $b$ in a right triangle whose hypotenuse is $C,$
so its length is $C \cos b,$
but it is also the leg adjacent to angle $a$ in a right triangle 
whose hypontenuse is $Vt.$
This gives us
$$C \cos b = Vt \cos a.$$
From these two equations we can conclude that
$$\left(\frac 12 gt^2 \cos b\right)(C \cos b) = (Vt \sin (a-b))(Vt \cos a).$$
We have already assumed that $t > 0,$ so we can cancel factors of $t.$
Assuming $b$ is less than a right angle, $\cos b > 0$ as well, 
and we can assume $g > 0.$
We can use these facts to isolate $C$ on one side of the equation.
The result is
$$C = \frac{2V^2 \sin (a-b) \cos a}{g \cos^2 b}.$$
The second part of the problem is to maximize $C.$
At least one version of the posted question allows us to vary $b$ in order to maximize $C.$
But we can make $C$ as large as we want merely by keeping $a$ constant and
letting $b$ approach $-\frac\pi2$ radians--that is, incline the plane as steeply
downwards as we like, but never quite vertical (because then there is no second
intersection of the trajectory with the plane, and $C$ is undefined).
We can even allow $a$ to vary and still $C$ is unbounded as long as $a$ 
is not too close to $b.$
So I will answer a more interesting question, which I suspect is what was
meant to be asked: find $a$ that maximizes $C$ 
for given constant values of $V,\ g,$ and $b.$
To solve this, observe the trigonometric identity
$$\sin x - \sin y = 2 \sin\left(\frac{x-y}{2}\right) \cos\left(\frac{x+y}{2}\right).$$
Conveniently, there is a product of a sine and cosine in the quantity we want
to maximize. So we would like to have
$$\begin{eqnarray}
a - b &=& \frac12(x-y), \\
a     &=& \frac12(x+y).
\end{eqnarray}$$
Solving for $x$ and $y$ in terms of $a$ and $b,$ we find that
$$\begin{eqnarray}
x &=& 2a - b, \\
y &=& b.
\end{eqnarray}$$
Therefore, we can use the identity to substitute for $\sin (a-b) \cos a$ as follows:
$$C = \frac{2V^2}{g \cos^2 b} (\sin (2a-b) - \sin b).$$
But according to the conditions of the problem, everything on the right-hand side
of this equation is a constant except for $a.$
We therefore maximize $C$ by maximizing $\sin (2a-b),$
which occurs when $2a-b$ is a right angle.
Measuring in radians, that gives us $$a = \frac12\left(\frac\pi2 + b\right).$$
Measured in degrees, $a = 45 + \frac12 b.$
A: Here's me having a go at it:
Now, let the particle hit the plane after time t. 
Horizontal Displacement = $C*\cos b$ 
Vertical Displacement = $C*\sin b$
Horizontal velocity = $V*\cos a$ (along positive x-axis) 
Vertical velocity = $V*\sin a$ (along positive y-axis)
After time t:
$V*\cos a*t = C*\cos b$ . . . (1) 
$V*\sin a *t - 0.5*g*t^2 = C*\sin b$ . .  . (2)
From (1) obtain the value of $t$ and plug it into (2) and rearrange to answer Qn 1
