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How do you find the numerical value?

of 9sin(9)+cos(9)-6sin(6)-cos6

 May 23, 2014

Best Answer 

 #5
avatar+2354 
+5

Hey Melody,

 

No I didn't and I'm not having any trouble at the site either.

I do however block pop-up's so that might be the reason why.

 

To the one who posted the question,

could you please be more specific, so we can be more helpful?

 

Reinout 

 

p.s. right term is redirected

 May 23, 2014
 #1
avatar+2354 
+5

I'm not sure whether this is what you're looking for but...

 

$${\mathtt{9}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{9}}^\circ\right)}{\mathtt{\,\small\textbf+\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{9}}^\circ\right)}{\mathtt{\,-\,}}{\mathtt{6}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{6}}^\circ\right)}{\mathtt{\,-\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{6}}^\circ\right)} = {\mathtt{0.773\: \!905\: \!850\: \!979}}$$

 

If your question was in radians instead of degrees your answer is 3.514258807837

 

Reinout 

 May 23, 2014
 #2
avatar+118723 
+5

Hi Reinout, 

I think numerical value just means the number value. (not the absolute value)

That is what I would assume anyway.

 May 23, 2014
 #3
avatar+2354 
0

Yet, this is what he was looking for right? 

 

p.s. I edited the post

 May 23, 2014
 #4
avatar+118723 
+5

hi reinout,

I am really not sure. 

there are no units so perhaps it should be in radians.

I thought there mighe be some trick but I haven't seen it yet.

Did you send me a message or a pop-up?

got a pop-up but not a message. When I clicked on it I teleported here (I can't think of the right term)

There are spectres everywhere here.

 May 23, 2014
 #5
avatar+2354 
+5
Best Answer

Hey Melody,

 

No I didn't and I'm not having any trouble at the site either.

I do however block pop-up's so that might be the reason why.

 

To the one who posted the question,

could you please be more specific, so we can be more helpful?

 

Reinout 

 

p.s. right term is redirected

reinout-g May 23, 2014
 #6
avatar+118723 
0

Thanks - redirected - that's much better.  

Teleported sounds more fun though!

 May 23, 2014
 #7
avatar
0

no unit was given and the answer came from partial fraction.

 May 23, 2014

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