Power Reducing Forumula & other simply problem

[Nano-Passion]

Homework Statement

16sin²xcos²x

Rewrite expression in terms of the first power of cosine.

Homework Equations

sin²u = 1-cos2u / 2
cos²u = 1+cos2u /2

The Attempt at a Solution

16sin²xcos²x
= 16 (1-cos2u / 2)(1+cos2u/2)
= 16/4 ( 1 + cos2u - cos2u - cos²4u)
= 16/4 (1 - (1 + cos8u) / 2 )

There is something obviously wrong because the answer is 2-2cos4x but I can't seem to get it.

I'm not too sure how to multiply angles such as cos2u(cos2u).

Thanks in advance, I have a final tomorrow and aiming for perfection. =D

Homework Statement

ln (x+5) - ln(x-3) = ln x

The Attempt at a Solution

ln (x+5) - ln(x-3) = ln x
= e^{ln x+5} - e^{ln x-3}= e^{ln x}
= (x+5) - (x - 3) = x
= x + 5 - x + 3 = x
= x = 8

Now the answer key from my professor says x = 5.

Is it the professor's mistake or do I have to go back to high school lol.

 

[Mark44]

16sin²x cos²x = 4 * (4sin²x cos²x)
= 4 * (2 * sinx * cosx)²
Can you do something with that?

 

[Mark44]

Homework Statement

16sin²xcos²x

Rewrite expression in terms of the first power of cosine.

Homework Equations

sin²u = 1-cos2u / 2
cos²u = 1+cos2u /2

The Attempt at a Solution

16sin²xcos²x
= 16 (1-cos2u / 2)(1+cos2u/2)
= 16/4 ( 1 + cos2u - cos2u - cos²4u)
= 16/4 (1 - (1 + cos8u) / 2 )

There is something obviously wrong because the answer is 2-2cos4x but I can't seem to get it.

I'm not too sure how to multiply angles such as cos2u(cos2u).

Thanks in advance, I have a final tomorrow and aiming for perfection. =D

Homework Statement

ln (x+5) - ln(x-3) = ln x

The Attempt at a Solution

ln (x+5) - ln(x-3) = ln x
= e^{ln x+5} - e^{ln x-3}= e^{ln x}

The step above is incorrect - you can't do that. What you should do instead is combine the two terms on the left using the properties of logs; namely, the one about ln (A/B) = ln A - ln B.

= (x+5) - (x - 3) = x
= x + 5 - x + 3 = x
= x = 8

Now the answer key from my professor says x = 5.

Is it the professor's mistake or do I have to go back to high school lol.

 

[Dick]

cos(2u)*cos(2u)=cos^2(2u), not cos^2(4u).

 

[Nano-Passion]

cos(2u)*cos(2u)=cos^2(2u), not cos^2(4u).

Thats the problem as I've stated in my post. I don't know how to multiply them. =/

I can guess that cos2u * cos2u - cos²2u but then I don't know the steps or reasoning into it. =/

 

[Nano-Passion]

16sin²x cos²x = 4 * (4sin²x cos²x)
= 4 * (2 * sinx * cosx)²
Can you do something with that?

According to Double-Angle Forumula

2sinucosu = 2sinu

so..

4 (2sinxcosx)²
= 4 (sin2u)²
= 4(sin²4u)
= 4 ( (1 - cos 4u) / 2)
= 2 (1 - cos4u)
= 2 - 2cos4x

But I don't see how you got 16sin²x cos ²x = 4 (4sin²x cos ²x) because when you distribute that then you get...

4(4sin^2x cos^2x)
= 16sin^2x 4 cos^2x) which doesn't equal our original expression of 16sin^2x cos^2x

 

[Nano-Passion]

The step above is incorrect - you can't do that. What you should do instead is combine the two terms on the left using the properties of logs; namely, the one about ln (A/B) = ln A - ln B.

Ugh, I have no idea on how to follow up without violating some basic arithmetic laws on...

ln (x+5) / ln (x-3) = ln x

 

[Char. Limit]

Ugh, I have no idea on how to follow up without violating some basic arithmetic laws on...

ln (x+5) / ln (x-3) = ln x

It's not ln(x+5)/ln(x-3). It's ln((x+5)/(x-3)), and that's one of the properties of logarithms.

 

[Dick]

Thats the problem as I've stated in my post. I don't know how to multiply them. =/

I can guess that cos2u * cos2u - cos²2u but then I don't know the steps or reasoning into it. =/

There aren't any steps. A thing times itself is the thing squared. x*x=x^2. cos(2x)*cos(2x)=cos^2(2x).

 

[Mark44]

According to Double-Angle Forumula

2sinucosu = 2sinu

This is wrong right out of the gate. The formula is
2sinu cosu = sin(2u) \neq 2 sin u.

so..

4 (2sinxcosx)²
= 4 (sin2u)²
= 4(sin²4u)
= 4 ( (1 - cos 4u) / 2)
= 2 (1 - cos4u)
= 2 - 2cos4x

But I don't see how you got 16sin²x cos ²x = 4 (4sin²x cos ²x) because when you distribute that then you get...

4(4sin^2x cos^2x)
= 16sin^2x 4 cos^2x) which doesn't equal our original expression of 16sin^2x cos^2x