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$$\frac{ \cos 6x + 6 \cos 4x + 15 \cos 2x + 10 }{ \cos 5x + 5 \cos 3x + 10 \cos x }$$ My approach so far : Tried to represent the denominator as a factor of numerator by manipulating numerator's $\cos 6x = \cos (5x+x)$ , $\cos 4x = \cos (3x+x)$ , so on .. but then $\sin x$ come up which make it more complex to solve .

The options for the answer are:

A) $\cos 2x$.
B) $2 \cos x$.
C) $\cos^2 x$.
D) $1 + \cos x$

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  • $\begingroup$ If you just need a quick answer, plugging in $x=0$ gived $32/16=2, so it's between B and D. Plugging $x=\pi/2$ gives $0$, so the answer must be B. $\endgroup$ – Lonidard Nov 25 '15 at 5:38
  • $\begingroup$ If you need to prove the identity, just keep using the angle sum formula that you mentioned with the goal to rewrite the entire expression in terms of the angle $x$. You'll have a bunch of $\sin x$ and $\cos x$ that ought to simplify. $\endgroup$ – zahbaz Nov 25 '15 at 5:39
  • $\begingroup$ @bharb that is cool . But I need to know how to solve them ' systematically ' . Thanks btw . Thats a good trick for entrance exams . $\endgroup$ – Ricky Nov 25 '15 at 5:41
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$$ 2^6 \cos^6 x = (e^{ix}+e^{-ix})^6=e^{6ix}+6e^{4ix}+15e^{2ix} +20+15e^{-2ix}+6e^{-4x}+e^{-6ix} \\ = 2(\cos 6x + 6 \cos 4x +15 \cos 2x +10) $$ and so forth

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  • $\begingroup$ thank you Ekaveera it was the binomial coefficients which gave me a clue, though at first the 10 in the numerator seemed anomalous. however that came out once i did the back-of-envelope calculation $\endgroup$ – David Holden Nov 25 '15 at 6:29
  • $\begingroup$ This solution makes the problem (and the solution) much easier to generalize. Beautiful! $\endgroup$ – Gyumin Roh Nov 25 '15 at 6:32
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Hint

Use multiple angles identities as they are given here and you will find a simple result (since $\cos(nx)$ can be expressed as a polynomial in $\cos(x)$).

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In a concept similar to the Chebyshev Polynomial,

For a fixed angle $x$, We let $T(n)=\cos nx$. Note the relation of $T(n+1)=2T(1)T(n)-T(n-1)$.

Now we have the numerator as $$T(6)+6T(4)+15T(2)+10=2T(1)T(5)-T(4)+6T(4)+15T(2)+10 = 2T(1)T(5)+5T(4)+15T(2)+10 = 2T(1)T(5)+5(2T(1)T(3)-T(2))+15T(2)+10=2T(1)T(5)+10T(1)T(3)+10T(2)+10=2T(1)T(5)+10T(1)T(3)+20T(1)T(1)-10T(0)+10=2T(1)T(5)+10T(1)T(3)+20T(1)T(1)=2T(1)(T(5)+5T(3)+10T(1))$$

Since the denominator is just $T(5)+5T(3)+10T(1)$, we have the answer as $2T(1)$, or $2 \cos x$.

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My answer is (B) because when I threw it into Wolfy and subtracted $2\cos(x)$ it came out zero.

Lazy in my old age.

(added later)

Well, at 0 it is 2, so only B and D are possible. and at $\pi$ it is $\frac{32}{-16} =-2 $ so it can't be D, so it must be B.

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  • $\begingroup$ This is high school maths , that needs to be done without a calculator . $\endgroup$ – Ricky Nov 25 '15 at 5:37
  • $\begingroup$ I added a high school way to solve it. $\endgroup$ – marty cohen Nov 25 '15 at 5:40
  • $\begingroup$ cool . But isnt there a ' systematic ' approach to solve this except simplifying it to cosx sinx components ? $\endgroup$ – Ricky Nov 25 '15 at 5:44
  • $\begingroup$ For these kind of questions, it pays to eliminate the wrong answers as quickly as possible, and then decide among the remaining answers. A god start is to always plug in zero, and them look at other special arguments. $\endgroup$ – marty cohen Nov 25 '15 at 5:52
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Use this identity and the work is done $cos(x)+cos(y)=2cos\frac{(x+y)}{2}.\frac{(x-y)}{2}$ and write $10=10cos0$ then create factors and nultiples to use this identity and after that multiple use of this identity you will get answer. As an example i give first step . $\frac{(cos(6x)+cos(4x)+5cos(4x)+5cos(2x)+10cos(2x)+10cos0)}{(cos(5x)+cos(3x)+4cos(3x)+4cos(x)+6cos(x))}=\frac{(2cos(5x).cos(x)+10cos(3x).cos(x)+20cos(x).cos(x))}{2cos(4x).cos(x)+8cos(2x).cos(x)+6cos(x)}$ now cancel $2,cos(x)$ then ........$=2cos(x)$ multiple times repeat the same process and its done. Bit long but thats necesarry. Hope its clear.

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  • $\begingroup$ @ricky Welcome ... $\endgroup$ – Archis Welankar Nov 25 '15 at 6:53

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