Question

mind blowing problem, HELP!

Answer 1

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 \[\sec(x)-\cos(x)=\sin(x)\tan(x)\]

Answer 2

i still don't know what to do.

Answer 3

Do you have a calculator with the \(\cos\) function?

Answer 4

no :/

Answer 5

Which calc do you have? @sammy90210

Answer 6

pre

Answer 7

ohh haha, i thought you meant calculus,

Answer 8

Model # ?

Answer 9

lol.

Answer 10

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

Answer 11

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)"

Answer 12

http://www.wolframalpha.com/input/?i=ec%28.2%29-cos%28.2%29

Answer 13

https://www.google.com.pk/search?sugexp=chrome,mod=12&sourceid=chrome&ie=UTF-8&q=cos+30#hl=en&safe=off&sa=X&ei=-qTBT9C7CYeg8gOawOn3Cg&ved=0CAQQBSgA&q=sec+0.2-cos+0.2&spell=1&bav=on.2,or.r_gc.r_pw.r_qf.,cf.osb&fp=269ae74d88b6f866&biw=1280&bih=695

Answer 14

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

Answer 15

http://www.wolframalpha.com/input/?i=sec%28.2%29-cos%28.2%29

Answer 16

so i plug in 0.04 for y1 under 0.2?

Answer 17

Precisely.

Answer 18

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

Answer 19

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.

Answer 20

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

Answer 21

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

Answer 22

thanks :)

Answer 23

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

Answer 24

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

Answer 25

ohh ok :)

Answer 26

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?

Answer 27

yes

Answer 28

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

Answer 29

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

Answer 30

hmm i dunno

Answer 31

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

Answer 32

yes

Answer 33

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

Answer 34

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

Answer 35

x=0 y1=0 y2=0

Answer 36

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

Answer 37

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