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]