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Mathematics 8 Online
OpenStudy (anonymous):

If sine of x equals square root of 2 over 2, what is cos(x) and tan(x)? Explain your steps in complete sentences.

OpenStudy (anonymous):

@jamesr @Xaze @MoonMoonWolf @miszzkeriee @mathway @Michele_Laino @mickey1513

OpenStudy (anonymous):

Ok, we have \( \huge \sin x = \frac{\sqrt{2}}{2}\) So we know that \( \huge \sin \theta = \frac{y}{r}\) \( \huge y = \sqrt{2}\) r = radius \( \huge r = 2\) \( \huge \cos x = \frac{x}{r} \) Since we already know what y equals and r, we can use the following formula to find x \( \huge x^2 + y^2 = r^2 \) \( \huge x^2 + \sqrt{2}^2 = 2^2 \) \( \huge x^2 + \sqrt{2}^2 = 2^2 - \sqrt{2}^2 \) \( \huge \sqrt{x^2} = \sqrt{2^2 - \sqrt{2}^2} \) Can you finish solving for x ?

OpenStudy (anonymous):

Note, we are working on cos right now. We will do tan after cos.

OpenStudy (anonymous):

I tried but couldn't get it.

OpenStudy (anonymous):

@Nixy

OpenStudy (anonymous):

Ok, what did you get for x?

OpenStudy (anonymous):

4

OpenStudy (anonymous):

Ok, one sec

OpenStudy (anonymous):

By the way, using complete sentences how would I explain the key features of the graph of the tangent function?

OpenStudy (anonymous):

\( \huge \sqrt{x^2} = \sqrt{2^2 - \sqrt{2}^2} \) \( \huge \sqrt{x^2} = x\) \( \huge x = \sqrt{2^2 - \sqrt{2}^2} \) \( \huge 2^2 = 4 \) \( \huge \sqrt{2}^2 = \sqrt{4} = 2 \) So now we have \( \huge x = \sqrt{4 - 2} \) What is the value of x ?

OpenStudy (anonymous):

x = pie2

OpenStudy (anonymous):

Once you know how to solve the problem you should be able to explain.

OpenStudy (anonymous):

\( \huge x = \sqrt{2} \)

OpenStudy (anonymous):

Yeah

OpenStudy (anonymous):

So, \( \huge \cos x = \frac{x}{r}\) since we know what x is and r, we have cos x = \( \huge \frac{\sqrt{2}}{2} \)

OpenStudy (anonymous):

Since we have found cos, what do you think tan x =? \( \huge \tan x = \frac{y}{x} \)

OpenStudy (anonymous):

tan x would be pie2/4?

OpenStudy (anonymous):

Tan x = \( \huge \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} \) but you need to divide this

OpenStudy (anonymous):

Can you divide that?

OpenStudy (anonymous):

Wouldn't it equal 1?

OpenStudy (anonymous):

You are correct!!!!

OpenStudy (anonymous):

tan x = 1

OpenStudy (anonymous):

AYYYY

OpenStudy (anonymous):

so for the final answer how would i explain

OpenStudy (anonymous):

Just go over the steps that we went through here

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

thank u

OpenStudy (anonymous):

YW

OpenStudy (anonymous):

but for this how would i explain

OpenStudy (anonymous):

Using complete sentences, explain the key features of the graph of the tangent function.

OpenStudy (anonymous):

Is that a separate question on your homework?

OpenStudy (anonymous):

nah separate

OpenStudy (anonymous):

It is separate ?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

Ok, you just need to explain the key feature of the tan when it comes to graphing.

OpenStudy (anonymous):

Look in your book. It should tell you step by step. For instance you will have asymptotes at odd multiples of \( \huge \frac{\pi}{2} \)

OpenStudy (anonymous):

Here are the properties of the tan The domain is the set of all real numbers except odd multiples of \( \huge \frac{\pi}{2} \) The range is the set of all real numbers The tangent function is an odd function, as the symmetry of the graph with respect to the origin indicates. The tangent function is periodic with period pi The x-intercepts are .....,-2pi, -pi, 0, pi, 2pi, 3pi, ..... Vertical asymptotes occur at x = odd multiples of \( \huge \frac{\pi}{2} \)

OpenStudy (anonymous):

Got it?

OpenStudy (anonymous):

Yes

OpenStudy (anonymous):

Good job. It is a lot to take in but keep practicing and you will get it.

OpenStudy (anonymous):

Thanks Nixy

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