Show that $\sin \left( {\theta  + \frac{\pi }{2}} \right) = \cos \theta$.

Consider $f(x) = \sin (ax)$ where $a$ is a constant. Prove by mathematical induction that ${f^{(n)}}(x) = {a^n}\sin \left( {ax + \frac{{n\pi }}{2}} \right)$ where $n \in {\mathbb{Z}^ + }$ and ${f^{(n)}}(x)$ represents the ${{\text{n}}^{{\text{th}}}}$ derivative of $f(x)$.

$\sin \left( {\theta  + \frac{\pi }{2}} \right) = \sin \theta \cos \frac{\pi }{2} + \cos \theta \sin \frac{\pi }{2}$     M1

$= \cos \theta$     AG

Note:     Accept a transformation/graphical based approach.

[1 mark]

consider $n = 1,{\text{ }}f'(x) = a\cos (ax)$     M1

since $\sin \left( {ax + \frac{\pi }{2}} \right) = \cos ax$ then the proposition is true for $n = 1$     R1

assume that the proposition is true for $n = k$ so ${f^{(k)}}(x) = {a^k}\sin \left( {ax + \frac{{k\pi }}{2}} \right)$     M1

${f^{(k + 1)}}(x) = \frac{{{\text{d}}\left( {{f^{(k)}}(x)} \right)}}{{{\text{d}}x}}\;\;\;\left( { = a\left( {{a^k}\cos \left( {ax + \frac{{k\pi }}{2}} \right)} \right)} \right)$     M1

$= {a^{k + 1}}\sin \left( {ax + \frac{{k\pi }}{2} + \frac{\pi }{2}} \right)$ (using part (a))     A1

$= {a^{k + 1}}\sin \left( {ax + \frac{{(k + 1)\pi }}{2}} \right)$     A1

given that the proposition is true for $n = k$ then we have shown that the proposition is true for $n = k + 1$. Since we have shown that the proposition is true for $n = 1$ then the proposition is true for all $n \in {\mathbb{Z}^ + }$     R1

Note:     Award final R1 only if all prior M and R marks have been awarded.

[7 marks]

Total [8 marks]