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Mathematics 8 Online
OpenStudy (anonymous):

find the exact value of the expression Cos(sin^-1 square root(3) divided by 2)

Directrix (directrix):

Where is the sine function equal to (square root(3) divided by 2) in Quadrant One?

Directrix (directrix):

By where, I mean what angle? Are we working in degrees or radians or does it matter?

OpenStudy (anonymous):

radians

Directrix (directrix):

Take a look at the attached table of selected trig values.

Directrix (directrix):

@NorthGeorgiaBoy17

OpenStudy (anonymous):

im lost now, so do i plug in one of those values for cos and sin?

Directrix (directrix):

@NorthGeorgiaBoy17 Look at the attached table. I wrote on it. Look at the section I highlighted.

OpenStudy (anonymous):

so i do cos(60)?

Directrix (directrix):

From the table, the sine is (square root(3) divided by 2) at pi/3 radians or 60 degrees.

Directrix (directrix):

Yes, @NorthGeorgiaBoy17

Directrix (directrix):

And, that value is ? @NorthGeorgiaBoy17

OpenStudy (anonymous):

\[\cos(\sin^{-1}(\frac{ \sqrt3 }{ 2 }))=\cos \left( \frac{ \pi }{ 3 } \right)\]Do you know how I got that? I got that because in order to solve the whole thing, we must first evaluate the expression arcsin(sqrt(3)/2) in the brackets. We know that sin^-1(sqrt(3)/2) is the same as saying for what value does sin = sqrt(3)/2? So we write it more formally like this:\[y=\sin^{-1}x \implies \sin y = x\]We know x = sqrt(3)/2, so we solve for y by finding for what value is the sine graph equal to sqrt(3)/2.\[\sin^{-1}x=y \implies \sin y=x \rightarrow \sin^{-1}\left( \frac{ \sqrt3 }{ 2 } \right) \implies \sin y = \frac{ \sqrt3 }{ 2 } \implies y=\frac{ \pi }{ 3 }\]So now can, can you evaluate cos(pi/3)? @NorthGeorgiaBoy17

Directrix (directrix):

And, that value is ? @NorthGeorgiaBoy17

OpenStudy (anonymous):

so the answer is 1/2 ?!

OpenStudy (anonymous):

Yes!

Directrix (directrix):

Yes.

OpenStudy (anonymous):

Good job.

OpenStudy (anonymous):

thanks friends

Directrix (directrix):

@NorthGeorgiaBoy17 I see that you may be new to OpenStudy. Welcome.

OpenStudy (anonymous):

yes i am, thank you!

OpenStudy (anonymous):

Well since you're new, I should also tell you that you may give a helpful answer a medal by clicking "Best Response" on the answer. @NorthGeorgiaBoy17

OpenStudy (anonymous):

@Directrix I'll take care of you lol.

Directrix (directrix):

That's the spirit. Thanks.

OpenStudy (anonymous):

And I just didn't give an answer. I thoroughly explained and left it on him to generate the answer =[

OpenStudy (anonymous):

And I wasn't indirectly asking for a medal, I swear lol...@Directrix

OpenStudy (anonymous):

directrix ill give you a medal too if you just type anything on the question i just poster

OpenStudy (anonymous):

posted

OpenStudy (anonymous):

becuase yall both helped and id like to be nice.

Directrix (directrix):

@genius12 Well, OS is somewhat like an online game. Earning medals or being awarded a medal for helping is tantamount to a "thank you." New users do not know that. If the Ambassadors would gently explain that, then the new users would understand how they can get quicker help. After all, we are human and like the medals. Just my opinon.

Directrix (directrix):

@NorthGeorgiaBoy17 @genius12 is a caring OpenStudier and solved the medal issue. So, no worries. It's all okay.

OpenStudy (anonymous):

okay! i didnt mean to offend anyone! hope we're all good.

Directrix (directrix):

We are all good. Nobody is offended.

OpenStudy (anonymous):

okay good!! thanks so much for you're help! i hope you help me again sometime:)

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