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Math  /  Trigonometry

QuestionIf cos⁡2xsin⁡4x=A+Bcos⁡2x+Ccos⁡4x+Dcos⁡2xcos⁡4x\cos ^{2} x \sin ^{4} x=A+B \cos 2 x+C \cos 4 x+D \cos 2 x \cos 4 x then A=A= B=B= C=C= D=D=

Studdy Solution

STEP 1

What is this asking? We need to find the constants *A*, *B*, *C*, and *D* that make this trigonometric equation true for all *x*. Watch out! Remember your double and quadruple angle formulas, and don't mix them up!
Also, be careful with those exponents – they can be tricky!

STEP 2

1. Rewrite the left side using double angle formulas.
2. Expand and simplify the expression.
3. Match coefficients to find *A*, *B*, *C*, and *D*.

STEP 3

Alright, let's **kick things off** by rewriting the left side of the equation.
We know that sin⁡2x=1−cos⁡2x2\sin^2 x = \frac{1 - \cos 2x}{2} and cos⁡2x=1+cos⁡2x2\cos^2 x = \frac{1 + \cos 2x}{2}.
This is a great start because we're trying to express everything in terms of cos(2*x*)!

STEP 4

Let's substitute these into our original equation: (1+cos⁡2x2)(1−cos⁡2x2)2=A+Bcos⁡2x+Ccos⁡4x+Dcos⁡2xcos⁡4x \left(\frac{1 + \cos 2x}{2}\right) \left(\frac{1 - \cos 2x}{2}\right)^2 = A + B \cos 2x + C \cos 4x + D \cos 2x \cos 4x

STEP 5

Time to **expand** that left side!
It might look a little scary, but we can totally handle it: 18(1+cos⁡2x)(1−2cos⁡2x+cos⁡22x) \frac{1}{8}(1 + \cos 2x)(1 - 2\cos 2x + \cos^2 2x) =18(1−2cos⁡2x+cos⁡22x+cos⁡2x−2cos⁡22x+cos⁡32x) = \frac{1}{8}(1 - 2\cos 2x + \cos^2 2x + \cos 2x - 2\cos^2 2x + \cos^3 2x) =18(1−cos⁡2x−cos⁡22x+cos⁡32x) = \frac{1}{8}(1 - \cos 2x - \cos^2 2x + \cos^3 2x)

STEP 6

Now, let's use that double angle formula again, but this time for cos⁡22x\cos^2 2x: cos⁡22x=1+cos⁡4x2\cos^2 2x = \frac{1 + \cos 4x}{2}.
Also, we can rewrite cos⁡32x\cos^3 2x as cos⁡2xcos⁡22x=cos⁡2x(1+cos⁡4x2)\cos 2x \cos^2 2x = \cos 2x \left(\frac{1 + \cos 4x}{2}\right).
Substituting these back in, we get: 18(1−cos⁡2x−1+cos⁡4x2+cos⁡2x+cos⁡2xcos⁡4x2) \frac{1}{8}\left(1 - \cos 2x - \frac{1 + \cos 4x}{2} + \frac{\cos 2x + \cos 2x \cos 4x}{2}\right)

STEP 7

Let's **simplify** this a bit more: 116(2−2cos⁡2x−1−cos⁡4x+cos⁡2x+cos⁡2xcos⁡4x) \frac{1}{16}(2 - 2\cos 2x - 1 - \cos 4x + \cos 2x + \cos 2x \cos 4x) =116(1−cos⁡2x−cos⁡4x+cos⁡2xcos⁡4x) = \frac{1}{16}(1 - \cos 2x - \cos 4x + \cos 2x \cos 4x)

STEP 8

Now, let's **compare** this with the right side of our original equation: 116(1−cos⁡2x−cos⁡4x+cos⁡2xcos⁡4x)=A+Bcos⁡2x+Ccos⁡4x+Dcos⁡2xcos⁡4x \frac{1}{16}(1 - \cos 2x - \cos 4x + \cos 2x \cos 4x) = A + B \cos 2x + C \cos 4x + D \cos 2x \cos 4x

STEP 9

By **matching** the coefficients of the corresponding terms, we can see that: A=116A = \frac{1}{16}, B=−116B = -\frac{1}{16}, C=−116C = -\frac{1}{16}, and D=116D = \frac{1}{16}.

STEP 10

A=116A = \frac{1}{16} B=−116B = -\frac{1}{16} C=−116C = -\frac{1}{16}D=116D = \frac{1}{16}

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