Question

Evaluate https://wca.sooschools.com/media/g_alg02_ccss_2013/11/img_alg02u11c02q08d_12.gif for θ = 30°.

The value of this expression is

Answer 1

@Preetha

Answer 2

@e.mccormick

Answer 3

@amistre64

Answer 4

Do you have a list of identities you have learned so far?

Answer 5

by identities do you mean csc, cos, sin? The answer supposed to be constant

Answer 6

Like the Pythagorean Identity:

\(\sin ^2 \theta + \cos ^2 \theta =1\)

Answer 7

cos2θ + sin2θ = 1

Answer 8

the 0 with the line going through it doesnt show the same when i send a message

Answer 9

(cos 0withLine)^2 + (sin 0withLine)^2=1

Answer 10

Yah, I use \(\LaTeX\) to make it:

`\(\sin ^2 \theta + \cos ^2 \theta =1\)` makes \(\sin ^2 \theta + \cos ^2 \theta =1\)

Answer 11

\(\sin ^2 \theta + \cos ^2 \theta =1\) is the answer?

Answer 12

No. That is an identity. Have you gone over identities at all?

Answer 13

a tiny bit. This question is the hardest and I need to pass it

Answer 14

Well, I think you can use identities to simplify it. Then it should be easier to solve.

Answer 15

ok

Answer 16

How do I use this one?

Answer 17

Well, it might or might not be used. You can replace anything with an identity value because they mean the same thing.

Now, another thing to remember is the basic rules of algebra. You were given:

\([\sin^2 \theta - (\cos^2\theta)(\sin^2\theta) ]^{-1}\)

Do you see anything in there you could factor? That might make it easier just by doing that.

Putting everything in terms of sine and cosine can also help a lot of times, but this time it already is.

And you can do algebra to the identities as well.

\(\sin ^2 \theta + \cos ^2 \theta =1\) implies two things by subtracting off either the sine or the cosine:
\(\cos ^2 \theta =1-\sin ^2 \theta\)
\(\sin ^2 \theta=1-\cos ^2 \theta \)

Answer 18

fctor 30?

Answer 19

The 30 is theta. So yes, you could put in the \(\dfrac{\sqrt{3}}{2}\) and \(\dfrac{1}{2}\), square them and do all the math. That works too.

I was just saying that it might be easier if you simplify it a bit first.

Answer 20

I don't really understand trig that well so I'm kind of stuck with (this) last problem because I received very little information concerning the Pythagorean Identities.

Answer 21

Identities are all about replacing things.

Answer 22

Ok

Answer 23

Now, you can work it like it is, putting in values from the unit circle and squaring them and solve it, or:

you can do identity subsitutions right off, then simplify it, and then do unit circle stuff, or:

you can do algebra and then identity subsitutions, then simplify it, and then do unit circle stuff.

Any way you do it you should get the same answer.

Answer 24

what is the answer?

Answer 25

Have not solve it. I am trying to find out what you need to know so that you can solve it.

Answer 26

I am new to trig, I took it yesterday. I did the rest of my problems by myself but on this question I don't have a clue on what to do at all I'm just trying to pass this question

Answer 27

OK, so you know how to find the values of sine and cosine at 30 degrees?

Answer 28

No, my connection going in and out

Answer 29

Do you have a unit circle?

Answer 30

No I don't see one. I read somewhere that the circle deals with pi 180 degrees

Answer 31

Did they give you anythign that said what sine and cosine of 30 degrees is?

The Trig sheet here might help:
http://tutorial.math.lamar.edu/cheat_table.aspx

Answer 32

cos230° + sin230° = 1
https://wca.sooschools.com/media/g_alg02_ccss_2013/11/p1_2ppwr2.gif = 1
3/4 + 1/4=1
1=1

Answer 33

First part: cos^2 30degrees + sin^2 30degrees = 1

Answer 34

Second Part really looks like this: https://wca.sooschools.com/media/g_alg02_ccss_2013/11/p1_2ppwr2.gif

Answer 35

You would only use that to do a substitution. Your problem is not in that form, so it does not work that way.

\([\sin^2 \theta - (\cos^2\theta)(\sin^2\theta) ]^{-1}\)

Answer 36

what way would it work

Answer 37

the picture really looks like (sqrt 3^2/2) + ( 1^2/2)= 1

Answer 38

@e.mccormick

Answer 39

You are using the wrong calculation. Do not use the identity. Use your original.

Answer 40

I still don't know how to do the problem period

Answer 41

I showed you the identity ti see if you knew about them. You seem confused by them so I am saying that rather than try and use the identity you can just go back to the original:

\([\sin^2 \theta - (\cos^2\theta)(\sin^2\theta) ]^{-1}\)

Answer 42

Also, it is good to know that \(\sin^2 \theta\) means \((\sin \theta)^2\). So if you have \(\theta = 30^\circ\) you can put in what theta is, square the values, and do the multiplication and subtraction.

Answer 43

square 30?

Answer 44

I received some information from open study, Well, you can factor out a sin^2 to get:
(sin^2(1-cos^2))^-1
1-cos^2 = sin^2
(sin^2)(sin^2)^-1
(sin^4)^-1
sin^-4
Either that is your answer, or csc^4.
The answer must be a constant

Answer 45

@e.mccormick

Answer 46

Yes, that is one way to do it. That will get you the valid solution. I did it 4 ways by now and all 4 I got the same answer. But I want you to get an answer using what you know.

Did they give you a list of sine and cosine for special angles? The sine and cosine on a 30-60-90 triangle?

Answer 47

No they didn't

Answer 48

Without that information, it is impossible to solve.

OK. So here is the typical 60-60-60 traingle with sides of 1:

Answer 49

ok

Answer 50

Sorry about the delay, had to feal with a problem user making threats... anyhow:

Let me cut it in half:


I now have a 30/60/90 with to known sides, 1 and 1/2

Answer 51

OOOPS! I typed 15 where I meant 30.

Answer 52

haha

Answer 53

Anyhow... I can use the Pythagorean therom to find the other side, which is \(\dfrac{\sqrt{3}}{2}\)