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# P6 Question watch

1. Can someone please help. I am stuck on the last part of 5b), I got up to the second to last step, but I don't understand what the markscheme's done for the very last step of part b).

Thanks.
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2. (Original post by darkenergy)
Can someone please help. I am stuck on the last part of 5b), I got up to the second to last step, but I don't understand what the markscheme's done for the very last step of part b).

Thanks.
The solutions to cos (5 theta) = 0 are pi/10, 3pi/10 etc. You have just worked out that cos 5theta is identically equal to cos(theta) {16cos^4(theta) - 20cos^2(theta) +5). Since cos (3pi/10) certainly isnt 0, this means 16cos^4(theta) - 20cos^2(theta) +5 must be 0. This equation is then solved using the quadratic formula which leaves you with (5 ± sqrt 5)/8. Now if you consider the graph of cos theta against theta, you know that cos pi/10 will be larger than cos 3pi/10 so therefore cos 3pi/10 must be the negative root of the solution, i.e. cos 3pi/10 = 5 - sqrt 5/8, since this gives a smaller value than the positive root of the solution.
3. (Original post by darkenergy)
Can someone please help. I am stuck on the last part of 5b), I got up to the second to last step, but I don't understand what the markscheme's done for the very last step of part b).

Thanks.
It's said cos^2 (pi/10) > cos^2 (3pi/10)
[You know that for the function cosx: as x increases cosx decreases in the interval (0,pi/2)]
This allows us to identify that cos^2 (pi/10)=(1/8)(5+rt5) and so deduce that cos^2 (3pi/10) [the smaller value] is (1/8)(5-rt5)
4. got it. thank you very much the_anomaly and Gaz031.

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