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Math  /  Trigonometry

Question33. cos⁡6x−sin⁡6x=cos⁡2x(1−14sin⁡22x)\cos ^{6} x-\sin ^{6} x=\cos 2 x\left(1-\frac{1}{4} \sin ^{2} 2 x\right)

Studdy Solution

STEP 1

What is this asking? We need to prove that this trigonometric identity is true! Watch out! Remember your difference of squares and difference of cubes formulas, and keep track of those exponents!

STEP 2

1. Rewrite the left side using the difference of squares.
2. Apply difference of squares again.
3. Rewrite using the difference of cubes.
4. Simplify using trigonometric identities.
5. Factor and simplify further.

STEP 3

Let's **begin** by writing down the left-hand side of our equation: cos⁡6x−sin⁡6x\cos^6 x - \sin^6 x.
We can think of this as a **difference of squares**: a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b), where a=cos⁡3xa = \cos^3 x and b=sin⁡3xb = \sin^3 x.

STEP 4

Applying the **difference of squares** formula, we get: (cos⁡3x)2−(sin⁡3x)2=(cos⁡3x−sin⁡3x)(cos⁡3x+sin⁡3x)(\cos^3 x)^2 - (\sin^3 x)^2 = (\cos^3 x - \sin^3 x)(\cos^3 x + \sin^3 x)

STEP 5

Now, notice that cos⁡2x+sin⁡2x=1\cos^2 x + \sin^2 x = 1.
We can **sneak this in** by multiplying and dividing by one, which doesn't change anything!

STEP 6

Let's **rewrite** our expression as: (cos⁡3x−sin⁡3x)⋅(cos⁡x+sin⁡x)(cos⁡2x−cos⁡xsin⁡x+sin⁡2x)(cos⁡x+sin⁡x)⋅(cos⁡x−sin⁡x)(cos⁡2x+cos⁡xsin⁡x+sin⁡2x)(cos⁡x−sin⁡x)(\cos^3 x - \sin^3 x) \cdot \frac{(\cos x + \sin x)(\cos^2 x - \cos x \sin x + \sin^2 x)}{(\cos x + \sin x)} \cdot \frac{(\cos x - \sin x)(\cos^2 x + \cos x \sin x + \sin^2 x)}{(\cos x - \sin x)} This looks complicated, but it's just a clever way to **introduce** our helpful identity!

STEP 7

We can **apply** the difference of cubes formula a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a-b)(a^2 + ab + b^2) to cos⁡3x−sin⁡3x\cos^3 x - \sin^3 x.

STEP 8

This gives us (cos⁡x−sin⁡x)(cos⁡2x+cos⁡xsin⁡x+sin⁡2x)(\cos x - \sin x)(\cos^2 x + \cos x \sin x + \sin^2 x).
Similarly, the sum of cubes formula a3+b3=(a+b)(a2−ab+b2)a^3 + b^3 = (a+b)(a^2 - ab + b^2) applied to cos⁡3x+sin⁡3x\cos^3 x + \sin^3 x gives us (cos⁡x+sin⁡x)(cos⁡2x−cos⁡xsin⁡x+sin⁡2x)(\cos x + \sin x)(\cos^2 x - \cos x \sin x + \sin^2 x).

STEP 9

Substituting these back into our expression from the previous step, and remembering that cos⁡2x+sin⁡2x=1\cos^2 x + \sin^2 x = 1, we get: (cos⁡x−sin⁡x)(cos⁡x+sin⁡x)(1−cos⁡xsin⁡x)(1+cos⁡xsin⁡x)(\cos x - \sin x)(\cos x + \sin x)(1 - \cos x \sin x)(1 + \cos x \sin x)

STEP 10

Recall the **double angle identities**: cos⁡2x=cos⁡2x−sin⁡2x=(cos⁡x−sin⁡x)(cos⁡x+sin⁡x)\cos 2x = \cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x) and sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x \cos x.

STEP 11

We can **rewrite** our expression as: (cos⁡2x)(1−12sin⁡2x)(1+12sin⁡2x)(\cos 2x)(1 - \frac{1}{2}\sin 2x)(1 + \frac{1}{2}\sin 2x) This uses the fact that cos⁡xsin⁡x=12sin⁡2x\cos x \sin x = \frac{1}{2} \sin 2x.

STEP 12

Using the **difference of squares** one more time, we get: cos⁡2x(1−14sin⁡22x)\cos 2x (1 - \frac{1}{4}\sin^2 2x)

STEP 13

Look at that!
We've **successfully transformed** the left-hand side of the original identity into the right-hand side.

STEP 14

We have shown that cos⁡6x−sin⁡6x=cos⁡2x(1−14sin⁡22x)\cos^6 x - \sin^6 x = \cos 2x (1 - \frac{1}{4}\sin^2 2x).

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