Current in circuit (trig problem)

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Hey this is my first post... been really stuck on this particular question from a paper:

In a particular circuit the current, I amperes, is given by:


I=4sinθ - 3cosθ θ>0


where θ is an angle related to the voltage.


Given that I=R x sin(θ-a), where R>0 and 0<=a<360degrees,


a) find the value of R, and the value of a to 1 decimal place.
b) Hence solve the equation 4sinθ - 3cosθ = 3 to find the values of θ between 0 and 360degrees.
c) Write down the greatest value for I.
d) Find the value of θ between 0 and 360degrees at which the greatest value of I occurs.

I think if someone could show me the steps to answering part a) I could be able to do the rest myself.
Thanks
 
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Rsin(\theta- a)= Rsin(\theta)cos(a)- Rcos(\theta)sin(a)

Compare that with I.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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