C3 2016 3 Edexcel

3

a) 2cos(\theta)-sin(\theta)=R(cos\theta cos\alpha -sin\theta sin\alpha)
Rcos\alpha = 2
Rsin\alpha=1
R=\sqrt{5}
\alpha=26.56505...\approx 26.57
\therefore 2cos(\theta)-sin(\theta)=Rcos(\theta + 26.57)

b) \frac{2}{\sqrt{5}cos(\theta +26.57)}=15
\frac{17}{15\sqrt{5}}=cos(\theta+26.57)
\theta+26.57=59.54629 and \theta+26.57=300.45371
\theta= 33.0 and 273.9 to 1dp

c) using the answer to a) we know 2cos\theta + sin\theta =\sqrt{5}cos(\theta-26.57)
Similarly as before we know that
59.55=\theta - 26.57 and \theta - 26.57=-300.45
\therefore \theta=86.12 and \theta=-273.88
So we can see that \theta=86.12\approx86.1 must be the smallest positive value of \theta

:cactus: :cactus: :cactus: PERFECTION :cactus: :cactus: :cactus: