Alan

Like so:

what is the value of sin 15

Note that   \(\cos(30°)=\frac{\sqrt3}{2}\)

Also  \(\cos(30)=1-2\sin^2(15)\) so \(\sin(15)=\sqrt{\frac{1-\cos(30)}{2}} \rightarrow \sqrt{\frac{1-\sqrt3/2}{2}} \rightarrow \frac{1}{2}\sqrt{2-\sqrt{3}}\)

Can also write \(\cos(30)=2\cos^2(15)-1\) so can get cos(15) in a similar manner to that of sin(15), and then get tan(15) from sin(15)/cos(15).

You've gone a little wrong in the second line of your development of y₁ and y₂ Omi.

For y₁:  \(6-(3-\sqrt5)\rightarrow3+\sqrt5\) not \(9+\sqrt5\)

A similar comment holds for y₂.

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tan(45deg) = 1

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Probability p(x) = mu^xe^-mu/x! for Poisson distribution

Here:  p(3) = 11³e^-11/3! ≈ 0.0037

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Initial vertical velocity u_h = 22.5*sin(33deg) m/s

At the highest point the vertical velocity is zero, so using v = u + at we have:

0 = 22.5*sin(33deg) - 9.8t. where the acceleration of gravity is 9.81m/s² downwards.  From this you can get the time, t, in seconds.

The horizontal velocity is given by u_h = 22.5*cos(33deg) m/s so the distance (in metres) travelled in time t is given by multiplying this velocity by the value found for t.

97/12 = 8 r 1

i.e. It is 8 remainder 1

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\(-2\frac{1}{4}+\frac{4}{5}\rightarrow-\frac{9}{4}+\frac{4}{5}\rightarrow-\frac{45}{20}+\frac{16}{20}\rightarrow-\frac{29}{20}\rightarrow=-1\frac{9}{20}\)

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I don't know how to enter it in Wolfram Alpha either!  

The symbolic math in Mathcad gives 0 as the result.  Unfortunately, it doesn't show how it works it out.

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I should note that an assumption in the above is that a is greater than b.