# How do I introduce a new variable in a trigonometric equation?

I have the trigonometric equation \begin{equation*} \sin^8 x + 2\cos^8 x -\dfrac{1}{2}\cos^2 2x + 4\sin^2 x= 0. \end{equation*} By putting $t = \cos 2x$, I have \begin{equation*} \dfrac{3}{16} t^4+ \dfrac{1}{4}t^3 + \dfrac{5}{8}t^2 -\dfrac{7}{4}t + \dfrac{35}{16} = 0. \end{equation*} How do I tell Mathematica to do that? Mathematica code is

Sin[x]^8 + 2 Cos[x]^8 - 1/2 Cos[2 x]^2 + 4 Sin[x]^2 == 0

• If you supply your terms in Mathematica syntax as well, that would make working with them much more comfortable. – Yves Klett Sep 27 '12 at 16:13
• With Maple, I used the following code restart; sort(simplify(algsubs(1-sin(x)^2=cos(x)^2, sin(x)^8+ 2*cos(x)^8 -1/2*cos(2*x)^2 + 4*sin(x)^2), {expand(cos(2*x))=t})); – minthao_2011 Sep 27 '12 at 16:17
• I meant: Please format your (LaTeX) expressions in Mathematica syntax so we can copy and paste easily. – Yves Klett Sep 27 '12 at 16:32

You can use TrigExpand to expand all trigonometric functions to fundamental forms and then Eliminate solves the rest

eq1 = Sin[x]^8 + 2 Cos[x]^8 - 1/2 Cos[2 x]^2 + 4 Sin[x]^2 == 0;
eq2 = t == Cos[2 x]

Eliminate[TrigExpand[{eq1, eq2}], x]


One way to do this is:

Sin[x]^8 + 2 Cos[x]^8 - 1/2 Cos[2 x]^2 + 4 Sin[x]^2 == 0 /.
Solve[t == Cos[2 x], x] //FullSimplify // Expand // Union // Column // TraditionalForm It gives exactly your answer if you get rid of your denominator 16 (multiply both sides of your equation by 16).

This will also work with more complex substitutions (for example t == Cos[x^2 - 1] ) when you can get multiple results:

Sin[x]^8 + 2 Cos[x]^8 - 1/2 Cos[2 x]^2 + 4 Sin[x]^2 == 0 /.
Solve[t == Cos[x^2 - 1], x]//FullSimplify//Expand//Union //Column // TraditionalForm A bit different approach :

Simplify @ TrigReduce[ Sin[x]^8 + 2 Cos[x]^8 - 1/2 Cos[2 x]^2 + 4 Sin[x]^2 == 0
/. Solve[ t == Cos[2 x], x, InverseFunctions -> True][]]

35 + 10 t^2 + 4 t^3 + 3 t^4 == 28 t


or using Eliminate :

Eliminate[ TrigToExp[{ Sin[x]^8 + 2 Cos[x]^8 - 1/2 Cos[2 x]^2 + 4 Sin[x]^2 == 0,
t == Cos[2 x]}], x, InverseFunctions -> True] //
PolynomialForm[#, TraditionalOrder -> True] &

3 t^4 + 4 t^3 + 10 t^2 - 28 t == -35


First plot the trigonometric equation:

Plot[Sin[x]^8 + 2 Cos[x]^8 - 1/2 Cos[2 x]^2 + 4 Sin[x]^2 == 0, {x, 0, 2 Pi}]


You will see that there are no (real) solutions. Is this what you expect?

• No. I only want as I asked. – minthao_2011 Oct 4 '12 at 2:43