Using Power Reducing Formulas to rewrite Trig Expressions

[opus]

Homework Statement

Use the power reducing formulas to rewrite the expression that does not contain trigonometric functions of power greater than 1.

Given expression:
$4sin^2xcos^2x$

2. Homework Equations
Relevant Power-Reducing Formulas:
$sin^2x=\frac{1-cos2x}{2}$
$cos^2x=\frac{1+cos2x}{2}$

The Attempt at a Solution

$$4sin^2xcos^2x$$
$$=4\left(\frac{1-cos2x}{2}\right)\left(\frac{1+cos2x}{2}\right)$$
$$=4\left(\frac{1+cos2x-cos2x-cos^22x}{2}\right)$$
$$=4\left(\frac{1-cos^22x}{2}\right)$$
$$=2-cos^22x$$

The answer given is:
$\frac{1-cos4x}{2}$

Have I solved done a miscalculation somewhere, or is my entire approach to solving this wrong?
Thank you for any responses.

 

[Mark44]

Homework Statement

Use the power reducing formulas to rewrite the expression that does not contain trigonometric functions of power greater than 1.

Given expression:
$4sin^2xcos^2x$

2. Homework Equations
Relevant Power-Reducing Formulas:
$sin^2x=\frac{1-cos2x}{2}$
$cos^2x=\frac{1+cos2x}{2}$

The Attempt at a Solution

$$4sin^2xcos^2x$$
$$=4\left(\frac{1-cos2x}{2}\right)\left(\frac{1+cos2x}{2}\right)$$
$$=4\left(\frac{1+cos2x-cos2x-cos^22x}{2}\right)$$
$$=4\left(\frac{1-cos^22x}{2}\right)$$
$$=2-cos^22x$$

You have a mistake in the line above. Fix the mistake, and then use one of your formulas to reduce the power of the $\cos^2(2x)$ term.

The answer given is:
$\frac{1-cos4x}{2}$

Have I solved done a miscalculation somewhere, or is my entire approach to solving this wrong?
Thank you for any responses.

 

[SammyS]

Homework Statement

Use the power reducing formulas to rewrite the expression that does not contain trigonometric functions of power greater than 1.

Given expression:
$4sin^2xcos^2x$

2. Homework Equations
Relevant Power-Reducing Formulas:
$sin^2x=\frac{1-cos2x}{2}$
$cos^2x=\frac{1+cos2x}{2}$

The Attempt at a Solution

$$4sin^2xcos^2x$$ $$=4\left(\frac{1-cos2x}{2}\right)\left(\frac{1+cos2x}{2}\right)$$ $$=4\left(\frac{1+cos2x-cos2x-cos^22x}{2}\right)$$ $$=4\left(\frac{1-cos^22x}{2}\right)$$ $$=2-cos^22x$$
The answer given is:
$\frac{1-cos4x}{2}$

Have I solved done a miscalculation somewhere, or is my entire approach to solving this wrong?
Thank you for any responses.

$2-\cos^2 (2x)\$ has a cosine function with a power of 2 .

 

[opus]

Looking into your reply Mark! Give me a few moments to track down my mistake.
And thanks for that Sammy. So many twos floating around they all seem to blend into each other.

 

[opus]

Some progress I think. I've got it down to the power of 1, but can't seem to put my finger on this. Feels like untangling a spool of fishing line.
$2-cos^22x$
$=2-\frac{1+cos4x}{2}$
$=\frac{4}{2}-\frac{1+cos4x}{2}$=$\frac{3+cos4x}{2}$

Mark which like were you talking about where I made the error? Was it the last line in my post? The $2-cos^22x$?

 

[Mark44]

Some progress I think. I've got it down to the power of 1, but can't seem to put my finger on this. Feels like untangling a spool of fishing line.
$2-cos^22x$
$=2-\frac{1+cos4x}{2}$
$=\frac{4}{2}-\frac{1+cos4x}{2}$=$\frac{3+cos4x}{2}$

Mark which like were you talking about where I made the error? Was it the last line in my post? The $2-cos^22x$?

Yes, this one.
The preceding line has $4(\frac{1-cos^22x}{2})$. This is equal to $2(1 - \cos^2(2x)) \ne 2 - \cos^2(2x)$

 

[tnich]

$$=4\left(\frac{1-cos2x}{2}\right)\left(\frac{1+cos2x}{2}\right)$$
$$=4\left(\frac{1+cos2x-cos2x-cos^22x}{2}\right)$$

You also have an error between these two lines.

 

[epenguin]

There is also a well known formula about $sin~x.cos~x$ which it would have been my first step to use.