OpenStudy (anonymous):

Verify that each equation is an identity: csc 2θ=1/2sec θ csc θ

4 years ago
OpenStudy (kohai):

So you're verifying the identity?

4 years ago
OpenStudy (anonymous):

I'm verify that the equation is an identity

4 years ago
OpenStudy (anonymous):

*verifying

4 years ago
OpenStudy (kohai):

Ok, so you're verifying the identity. Give me one moment to solve this, k?

4 years ago
OpenStudy (anonymous):

okay, thank you very much!

4 years ago
OpenStudy (kohai):

Almost done

4 years ago
OpenStudy (anonymous):

okay :)

4 years ago
OpenStudy (kohai):

Do you have your trig identities sheet?

4 years ago
OpenStudy (anonymous):

yes

4 years ago
OpenStudy (kohai):

So the first thing you need to do here is change everything to sin and cos. You're going to do this to the right hand side because t is the most complex side. What does this equation look like after you've done that?

4 years ago
OpenStudy (anonymous):

1/sin2 θ =1/2 1/cos θ 1/sin θ

4 years ago
OpenStudy (kohai):

Yup. So then you're going to multiply the right side all together

4 years ago
OpenStudy (anonymous):

ohhh okay

4 years ago
OpenStudy (kohai):

So what's that going to look like?

4 years ago
OpenStudy (anonymous):

1/sin2θ= 1/2 1/cos θsin θ

4 years ago
OpenStudy (kohai):

Don't change the left hand side. When you multiply it, it'll look like this: \[\csc2\theta = \frac{ 1 }{ 2\cos \theta \sin \theta }\]

4 years ago
OpenStudy (kohai):

So then you're going to do \[\csc2 \theta = \frac{ 1 }{ 2\sin \theta }\]

4 years ago
OpenStudy (kohai):

\[\csc2 \theta = \csc2 \theta\]

4 years ago
OpenStudy (anonymous):

what happeneds to the cos θ?

4 years ago
OpenStudy (kohai):

There's a trig property that states that 2cosθsinθ is equal to 2sinθ, I substituted

4 years ago
OpenStudy (anonymous):

oh yeah! thank you

4 years ago
OpenStudy (kohai):

You're welcome :)

4 years ago