Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Session 20 - Examples: Velocity, Speed and Arc Length In #2 in this PDF how does 2a get reduced to 2a (sin^2(θ/2))^1/2? I assume we start by squaring sin^2(θ/2) and sin(θ/2)cos(θ/2) and then taking the square root of the sums, yes? If yes I can reduce sin^2(θ/2) down to itself but I cannot seem to get rid of sin(θ/2)cos(θ/2)? My steps are: 1. Take the square root of the sum of (sin^2(θ/2))^2 + ((sin(θ/2)cos(θ/2))^2 2. Square root of: (1/2[1-cos(θ)])^2 + (1/2sin(θ))^2 3. Square root of: (sin^2(θ/2))^2 + (1/2sin(θ))^2 I'm stuck here.

OCW Scholar - Multivariable Calculus
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Note: I'm leaving out 2a until the end, after I find a ways to solve the problem of taking the square root of the sum of the squares. In other words, I'm OK with the 2a.
[sin(θ/2)]^2 +[sin(θ/2)cos(θ/2)]^2 =sin(θ/2)^4 +sin(θ/2)^2+cos(θ/2)^2 =sin(θ/2)^2 * [sin(θ/2)^2 + cos(θ/2)^2] =sin(θ/2)^ * 1
The second line in the above answer should have sin^2 * cos^2, not '+', sorry.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Of course! Thanks.

Not the answer you are looking for?

Search for more explanations.

Ask your own question