STEP II 1994 question 3 solution

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STEP II 1994 question 3 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 1994 question 3 solution

$f(0) = 1 \newline f(x-y) = f(x)f(y) - f(a-x)f(a+y) \newline \newline \textb{Let;} x = a, y = 0\newline f(a) = f(a)f(0) - f(0)f(a) = 0$

$f(x-y) = f(x)f(y) - f(a-x)f(a+y) \newline \textb{So:} f(a-x) = f(a)f(x) - f(0)f(a+x)\newline f(a-x) = -f(a+x)$

By symmetry f(a+y) = -f(a-y)

$f(y-x) = f(x)f(y) - f(a-y)f(a+x)\newline = f(x)f(y) - [-f(a+y)][-f(a-x)]\newline f(y-x) = f(x-y)\newline f(t) = f(-t)$

ii)

$f(2a) = f(a - (-a)) f(a)f(-a) - [f(0)]^2\newline = - (1)^2\newline = -1$

iii)

$f(2a - t) = f(2a)f(t) - f(-a)f(a+t)\newline = -f(t) - f(a)f(a+t)\newline = -f(t)$

iv)

$f(4a) = f(2a - (-2a)) = f(2a)f(-2a) - f(-a)f(a)\newline = 1 - f(a)f(a)\newline = 1$

$f(3a) = f(2a - (-a)) = f(2a)f(-a) - f(-a)f(0)\newline = 0$

$f(4a + t) = f(4a - (-t)) = f(4a)f(-t) - f(-3a)f(a-t)\newline = f(t) - f(3a)\newline = f(t)$

$f(y-x) = f(x)f(y) - f(a-x)f(a+y)\newline \newline f = \cos(t) \newline \cos(y-x) = \cosx \cosy + \sinx \siny$

But $\sinx = \cos(\frac{\pi}{2} - x)$ and $\siny = -\cos(\frac{\pi}{2} + x)$ so $\cos(y-x)=-\sin(\frac{\pi}{2}-x)\sin(\frac{\pi}{2}+x)$

f = cos(Bt + C)

When t = -2, f = 0

cos(C - 2B) = 0

Take for example $C - 2B = \frac{\pi}{2}$

When t = 0, f = 1 $cos(C) = 1 \newline C = 0\newline B = -\frac{\pi}{4}\newline f = \cos(-\frac{\pi}{4}t) = \cos(\frac{\pi}{4}t)$

$f(x-y) = \cos(x\frac{\pi}{4} - y\frac{\pi}{4}) = \cos(x\frac{\pi}{4})\cos(y\frac{\pi}{4}) + \sin(x\frac{\pi}{4})\sin(y\frac{\pi}{4})$

But $\sin(y\frac{\pi}{4}) = \cos(\frac{\pi}{2} - y\frac{\pi}{4}) = \cos[(\frac{\pi}{4})(2 - y)] = \cos[(\frac{\pi}{4})(-2 + y)] = f(a+y)\newline \sin(x\frac{\pi}{4}) = -\cos(\frac{\pi}{2} + x\frac{\pi}{4}) = - \cos[(\frac{\pi}{4})(2 + x)] = - \cos[(\frac{\pi}{4})(-2 - x)] = -f(a-x)\newline f(x-y) = f(x)f(y) - f(a-x)f(a+y)$ as required.