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STEP II 1994 question 3 solutionTSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 1994 question 3 solution
By symmetry
But
When t = -2, f = 0 cos(C - 2B) = 0 Take for example When t = 0, f = 1
Unparseable or potentially dangerous latex formula. Error 5: Image dimensions are out of bounds: 1626x23 as required.
\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
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![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)](http://www.thestudentroom.co.uk/latexrender/pictures/56/569f856fe3b334cd4f89b5930190ab7c.png "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)")
![f(2a) = f(a - (-a)) f(a)f(-a) - [f(0)]^2\newline
= - (1)^2\newline
= -1](http://www.thestudentroom.co.uk/latexrender/pictures/fe/feb8b09d14ac774f99b50daae6e70f4b.png "f(2a) = f(a - (-a)) f(a)f(-a) - [f(0)]^2\newline
= - (1)^2\newline
= -1")
