Simplify Pythagoras Theorem Homework Statement

[shanshan]

Homework Statement

I need to simplify vf = [(vcosx/2)^2+((2v-vsinx)/2)^2)]^1/2
as far as possible

Homework Equations

The Attempt at a Solution

vf = [(vcosx/2)^2+((2v-vsinx)/2)^2)]^1/2
= [(v^2cos^2(x)+v^2sin^2(x)+4v^2+2vsinx)/2]^1/2
= [(v^2(1)+4v^2-2vsinx)/2]^1/2 ---- using trig identity sin^2x+cos^x = 1
= [(5v^2-2vsinx)/2]^1/2

But this is not the answer - is anyone able to find any problems in my work, or can I simplify it further?

 

[Char. Limit]

Hint: you forgot to square the 2's in the denominator.

 

[shanshan]

tried this - ended up with:
[5v^2-2vsinx)^1/2]/2
and this still wasn't right.
Do you see any other problems?

 

[Char. Limit]

Yes, I do. You expanded (2v - v sin(x))² incorrectly. Check that again.

 

[mjjoga]

ok, your first step has an error in it. "v^2cos^2(x)" should be over 4 because it was over 2 in the original equation. You did the same thing with the second term of the original equation. Changing that will let you add the fractions more easily. Also when you squared 2v-vsinx you should get v^2sin^2(x)+4v^2-4v^2sin(x) all over 4 Now, you can put all the terms over 4 and pull out 1/4 and v^2. so,
v^2(1/4)[cos^2(x)+sin^2(x)+4-4sin(x)]^1/2=[v^2(1/4)[5-4sin(x)]]^1/2=
=v/2[5-4sin(x)]^1/2. That's as far as I could take it. I think that it was a good effort by you but you may have rushed through and made some algebra mistakes which took you way off course. Actually, using the identity is easy, it's manipulating with algebra that's hard! Good luck with the class
mjjoga