Mathematics
OpenStudy (anonymous):

mind blowing problem, HELP!

OpenStudy (anonymous):

OpenStudy (kinggeorge):

I'm going to assume you have a calculator, and can plug in the necessary numbers in to complete the chart. I will however, help you prove that $\sec(x)-\cos(x)=\sin(x)\tan(x)$

OpenStudy (anonymous):

i still don't know what to do.

OpenStudy (kinggeorge):

Do you have a calculator with the $$\cos$$ function?

OpenStudy (anonymous):

no :/

OpenStudy (saifoo.khan):

Which calc do you have? @sammy90210

OpenStudy (anonymous):

pre

OpenStudy (anonymous):

ohh haha, i thought you meant calculus,

OpenStudy (saifoo.khan):

Model # ?

OpenStudy (saifoo.khan):

lol.

OpenStudy (anonymous):

i don't have my scientific one with me, i have just a generic one atm

OpenStudy (kinggeorge):

Try using http://www.wolframalpha.com/ to do your calculations then. For example, to find $\sec(0.2)-\cos(0.2)$Type in "sec(.2)-cos(.2)"

OpenStudy (anonymous):
OpenStudy (kinggeorge):

@sammy90210 You forgot the "s" at the beginning.

OpenStudy (anonymous):
OpenStudy (anonymous):

so i plug in 0.04 for y1 under 0.2?

OpenStudy (kinggeorge):

Precisely.

OpenStudy (kinggeorge):

Now instead of using .2, use .4. Then type that number in y1 under 0.4, and so on.

OpenStudy (kinggeorge):

Once you're done with those, switch over to "sin(x)tan(x)" where instead of typing "x," type .2, .4, .6, ... and record those for y2 under their respective x-values.

OpenStudy (anonymous):

thank you :), and once the graph is filled, thats it? or do i have to write something else?

OpenStudy (kinggeorge):

After that, you need to analytically prove that $$y_1=y_2$$. I can help you through that as it's a little trickier.

OpenStudy (anonymous):

thanks :)

OpenStudy (anonymous):

btw, am i supposed to end up with the same answers for both columns? because i am

OpenStudy (kinggeorge):

Yes, you're supposed to have the same numbers.

OpenStudy (anonymous):

ohh ok :)

OpenStudy (kinggeorge):

First, rewrite $\sec(x)=\frac{1}{\cos(x)}$So you have$y_1=\frac{1}{\cos(x)}+\cos(x)$Now, we need to get a common denominator, so...$y_1=\frac{1}{\cos(x)}+\cos(x)\cdot\frac{\cos(x)}{\cos(x)}=\frac{1}{\cos(x)}+\frac{\cos^2(x)}{\cos(x)}$ Do all these steps make sense to you so far?

OpenStudy (anonymous):

yes

OpenStudy (kinggeorge):

There's actually a repeated typo in there. I should have minuses instead of pluses.

OpenStudy (kinggeorge):

Continuing however, we get $y_1=\frac{1-\cos^2(x)}{\cos(x)}$Can you tell me another way to write $1-\cos^2(x)?$

OpenStudy (anonymous):

hmm i dunno

OpenStudy (kinggeorge):

Do you remember the identity $\sin^2(x)+\cos^2(x)=1$

OpenStudy (anonymous):

yes

OpenStudy (kinggeorge):

Notice that if you rearrange that, you get $\cos^2(x)-1=\sin^2(x)$This means that we can rewrite $y_1=\frac{1-\cos^2(x)}{\cos(x)}=\frac{\sin^2(x)}{\cos(x)}$

OpenStudy (kinggeorge):

Can you finish it from here to show that $y_1=\frac{\sin^2(x)}{\cos(x)}=\sin(x)\tan(x)=y_2$

OpenStudy (anonymous):

x=0 y1=0 y2=0

OpenStudy (kinggeorge):

Can you rather do some algebraic manipulation to turn $\frac{\sin^2(x)}{\cos(x)}$into$\sin(x)\tan(x)$Without plugging in values?

OpenStudy (kinggeorge):

Perhaps this would help. $\frac{\sin^2(x)}{\cos(x)}=\sin(x)\cdot \frac{\sin(x)}{\cos(x)}$What is $\frac{\sin(x)}{\cos(x)}?$

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