2
$\begingroup$

Question: Given sinx + sin2x = 1, Find cos12x + 3cos10x + 3cos8x + cos6x

This is not a homework question, but one I encountered in a test recently, and its striking similarity with complex numbers made me curious to look for more methods.

As a multiple choice question, four options were provided as follows:

A. 1 B. 2 C. 3 D. 4 E. 0

It is easy to observe that x = π/2 is a solution to sinx + sin2x = 1, and plugging that value (x=π/2) in the concerned expression definitely yields the result. However, this method is not entirely correct, or convincing.

We need to show that all x satisfying sinx + sin2x = 1 (hereafter referred to as equation 1) give the same value of cos12x + 3cos10x + 3cos8x + cos6x (hereafter referred to as expression 2) and I'm sure (intuitively, but also checked it graphically) that solutions more than just x = π/2 exist for equation 1.

I shall first state my multiple approaches to the problem in short, and then elaborate on them.

  1. Solving graphically or algebraically, finding x and substituting
  2. Algebraic and trigonometric manipulations leading to the second expression
  3. Use of geometry or complex numbers

A possible method could be to actually solve for all values of x that satisfy equation 1 and plug those in expression 2 to get the desired result. However, I suppose this approach is cumbersome and time taking. An alternative could be to algebraically manipulate the first equation to obtain the value of the second expression without actually solving for x. (The kind of method I'm looking for, as it resolves all troubles) This would mean that the truth of equation 1 implies the existence of a definite value for expression 2, hence showing that all values of x satisfying sinx + sin2x = 1 do indeed yield the same and unique value for cos12x + 3cos10x + 3cos8x + cos6x.

Lastly, I couldn't help but notice that the required expression is Re((cis2x + cis4x)^3) which might hint at the involvement of a complex number or geometry centric approach for a shorter and rather elegant solution. To proceed, I tried to use sinx + sin2x = 1 as Im(cisx + cis2x) = 1, to no avail.

Note that: Re(z) denotes real part of complex number z And Im(z) denotes imaginary part of complex number z.

Any help is appreciated, and as many possible methods and approaches are welcome. Thanks in advance!

$\endgroup$
4
  • 1
    $\begingroup$ When $x = \frac{\pi}{2}$, the value of the second expression is 0. $\endgroup$
    – user348749
    Sep 11, 2017 at 7:45
  • $\begingroup$ Please take some time to learn proper math formatting with MathJax and edit your question to fit it. $\endgroup$
    – g.kov
    Sep 11, 2017 at 7:58
  • $\begingroup$ I'm sure it isn't formatting that you give more importance over mathematics, is it? $\endgroup$ Sep 11, 2017 at 8:22
  • $\begingroup$ @user28968: Mathematics means beauty, and a nice typesetting of equations is a part of it. $\endgroup$
    – g.kov
    Sep 11, 2017 at 9:21

1 Answer 1

0
$\begingroup$

\begin{align} \sin\,x+\sin\,2x&=1 \tag{1}\label{1} \end{align}

Equation \eqref{1} transforms into \begin{align} (\sin\,x-1)(4\,\sin^3x+4\sin^2\,x+\sin\,x-1)&=0 , \end{align}

which has two real roots,

\begin{align} \sin x&=1 ,\\ \sin x&=\tfrac16\,((28+3\,\sqrt{87})^{\tfrac16} - (28+3\,\sqrt{87})^{-\tfrac16})^2 \approx 0.34781038 , \end{align} and none of them gives the value of $x$ that provides any of given answers (A. 1, B. 2. C. 3, D. 4) for the expression \begin{align} \cos\,12x + 3\cos10x + 3\cos8x + \cos6x . \end{align}

There is probably a typo somewhere. For example, $x=\tfrac\pi2$ gives the B, if the equation is \begin{align} \cos\,12x + 3\cos10x + 3\cos8x - \cos6x . \end{align}

$\endgroup$
1
  • $\begingroup$ Great! Understood it now. Must've been a typo, you're right $\endgroup$ Sep 11, 2017 at 9:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .