Question

Precalculus

Answer 1

@Hero

Answer 2

\[\csc (\frac{ 5\pi }{ 6 }) = 2\]

Verify that this is true through cofunction identities.

Answer 3

Hey Shadow, how far did you get with this one?

Answer 4

I don't know this concept yet. Need to learn it.

Answer 5

I have the cofunction identities, I just don't know how to use them.

Answer 6

I see. Well basically, Cofunction identities involve the use of the the complement of the angle of the original trig function. For example,

\(\sin(60)^{\circ} = \cos(30^\circ)\)

Based on the example above, the general case is \(\sin(x) = \cos(90-x)\)

Answer 7

The problem you gave above works very similar except you're dealing with secants and cosecants so you would find the equivalent secant expression to the given secant expression, then go from there. Basically what you want to do is convert the expression to one of the trig functions, sine, cosine or tangent because the angles of those functions should be memorized.

Answer 8

\[\sec ( \frac{ \pi }{ 2 } - \frac{ 5\pi }{ 6 }) = 2\]

?

Answer 9

That's a good start. And you continue from there.

Answer 10

BTW, \(\pi = 180\) in this case.

Answer 11

\[\sec( \frac{ -\pi }{ 3 }) = 2\]

\[\frac{ 1 }{ \cos(\frac{ -\pi }{ 3 }) } = 2\]

\[\frac{ 1 }{ \frac{ 1 }{ 2 } } = 2\]

\[2 = 2\]

Answer 12

Boom! You got it.

Answer 13

Nice, thanks Hero.

Answer 14

You did a great job as well.

Answer 15

Im Confused.

Answer 16

Precalculus is pretty easy. It's just memorizing the Unit Circle, trigonometric/cofunction identities, then applying them then just doing basic pre-alg math. I learned this concept in literally five minutes, lol.

Answer 17

o, xD

Answer 18

Don't be discouraged, haha

Answer 19

What Grade are u in?

Answer 20

Senior

Answer 21

o

Answer 22

Im Far Behind, Lol