# Thread: 1 Last question Guys

1. ## 1 Last question Guys

Guys I really appreciate all of your help today, as I am trying very, very hard to learn all of this. Thanks to Soroban I am actually starting to grasp this concept, but I have one last question.

Cos(2x)=3/5 and I know that 2x is in quadrant I.

My Unknown is: tan(x)=?

Since this is equal to 3/5 how would I go about appyling the double angle formula so than I can say cos(x)=adj/hyp, and so fourth. I like putting it into this form because it makes it way easier for me to solve the problem. I know that Cos(2x)=Cos(2+x) and that Cos(2+x)= Cos(2)*Cos(x)-Sin(2)*Sin(x) but I am just not really sure how to implement the 3/5 what would I substitue it for?

Again Thanks for all of your help!

I love this site!

2. Originally Posted by qbkr21
Guys I really appreciate all of your help today, as I am trying very, very hard to learn all of this. Thanks to Soroban I am actually starting to grasp this concept, but I have one last question.

Cos(2x)=3/5 and I know that 2x is in quadrant I.

My Unknown is: tan(x)=?

Since this is equal to 3/5 how would I go about appyling the double angle formula so than I can say cos(x)=adj/hyp, and so fourth. I like putting it into this form because it makes it way easier for me to solve the problem. I know that Cos(2x)=Cos(2+x) and that Cos(2+x)= Cos(2)*Cos(x)-Sin(2)*Sin(x) but I am just not really sure how to implement the 3/5 what would I substitue it for?

Again Thanks for all of your help!

I love this site!
it is
Originally Posted by qbkr21
I know that Cos(2x)=Cos(2+x) and that Cos(2+x)= Cos(2)*Cos(x)-Sin(2)*Sin(x)
2x =x+x
2x is not equal to 2+x
Correct it and you will easily get the answer.
While solving, remember that 2x is in first quadrant.
If you are not able to solve, I willpost it for you. But just try, you will enjoy doing it.

Keep Smiling
Malay

3. Originally Posted by qbkr21
Guys I really appreciate all of your help today, as I am trying very, very hard to learn all of this. Thanks to Soroban I am actually starting to grasp this concept, but I have one last question.

Cos(2x)=3/5 and I know that 2x is in quadrant I.

My Unknown is: tan(x)=?

Since this is equal to 3/5 how would I go about appyling the double angle formula so than I can say cos(x)=adj/hyp, and so fourth. I like putting it into this form because it makes it way easier for me to solve the problem. I know that Cos(2x)=Cos(2+x) and that Cos(2+x)= Cos(2)*Cos(x)-Sin(2)*Sin(x) but I am just not really sure how to implement the 3/5 what would I substitue it for?

Again Thanks for all of your help!

I love this site!
There may be a fancier and more direct way to do this, but since we know what quadrant the angle 2x is in, this will work.

$cos(2x) = 2cos^2(x) - 1$ will let you find cos(x):

$\frac{3}{5} = 2cos^2(x) - 1$

$2cos^2(x) = \frac{8}{5}$

$cos^2(x) = \frac{4}{5}$

$cos(x) = \frac{2}{\sqrt{5}}$ (which we know is positive, since if 2x is in QI, then x must also be.)

Again:
$cos(2x) = 1 - 2sin^2(x)$ will let you find sin(x):

$\frac{3}{5} = 1 - 2sin^2(x)$

$2sin^2(x) = \frac{2}{5}$

$sin^2(x) = \frac{1}{5}$

$sin(x) = \frac{1}{\sqrt{5}}$ (which we know is positive since x is in QI)

So
$tan(x) = \frac{sin(x)}{cos(x)} = \frac{ \frac{1}{\sqrt{5}} }{ \frac{2}{\sqrt{5}} }$

$tan(x) = \frac{1}{2}$

-Dan

4. ## Re:

Ok so when Cos(2x)=3/5

I break the problem about using the Double Angle Identity:

so...

Cos(2x)= Cos(x+x) so:

Cos(x+x)= Cos(x)*Cos(x)-Sin(x)*Sin(x)

but I am still trying to figure out out you implement the 3/5 where would I put this? Where each of the "x's" are?

Thanks

5. Originally Posted by qbkr21
Ok so when Cos(2x)=3/5

I break the problem about using the Double Angle Identity:

so...

Cos(2x)= Cos(x+x) so:

Cos(x+x)= Cos(x)*Cos(x)-Sin(x)*Sin(x)

but I am still trying to figure out out you implement the 3/5 where would I put this? Where each of the "x's" are?

Thanks
There are 3 forms of the cosine double angle formula.
$cos(2x) = cos^2(x) - sin^2(x)$
$cos(2x) = 2cos^2(x) - 1$
$cos(2x) = 1 - 2sin^2(x)$

If you know that $cos(2x) = 3/5$ you don't substitute x = 3/5, what you have is an equation that you need to solve for x. (Like $x^2 + 5 = 10$, you don't put x = 10.)

What I did in the previous post gets around that. Rather than solving for x (which in this case is an inexact proceedure since x is (very likely) irrational and thus would need to be approximated) I solved for cos(x) and sin(x) and used them to find tan(x).

However if you like you can use your calculator to do the work for you:
$cos(2x) = 3/5$

Use the $cos^{-1}$ button to find
$2x = cos^{-1}(3/5) = 0.927295218$

Thus
$x = \frac{0.927295218}{2} = 0.463647609$

Thus
$tan(0.463647609) = 0.5$

-Dan

6. ## Re:

But tan(.927) doesn't = .5 it equals 1.3

in you spare time could you please show me how to use it in this formula

cos(2x)=cos(x)^2-sin2(x) or is this not possible.

She doesn't want decimals she wants straight up fractions

Thanks D

7. Originally Posted by qbkr21
But tan(.927) doesn't = .5 it equals 1.3

in you spare time could you please show me how to use it in this formula

cos(2x)=cos(x)^2-sin2(x) or is this not possible.

She doesn't want decimals she wants straight up fractions

Thanks D
First I copied and pasted the wrong number. I fixed it in the post. Thanks for catching that.

I couldn't figure out a way to use the formula for cos(2x) in that form, since it involves both sine and cosine functions at the same time. Is there something about the solution I gave you earlier that you are having trouble with?

-Dan

8. ## Re:

First off I have a crummy textbook that doesn't do a good job of expaining Trig., if I did have a good one I would read it before I posted so many questions, but needless to say I don't.

For tests and future references I want to know how to solve problems when you are given a Sin(2x), Cos(2x), Tan(2x), CSC(2x), Sec(2x), Cot(2x). The problems are really easy for me when there is just Sin(x), Cos(x), Tan(x), CSC(x), Sec(x), Cot(x). If you take a look at this post Soroban created he did a real nice job of of putting each in terms of (opp)^2+ (adj)^2= (Hyp)^2. I really had never though of using Path Theorm this way. Here is the post is there any you can corollate his reasoning with this problem?

http://www.mathhelpforum.com/math-he...-problems.html

9. Originally Posted by qbkr21
First off I have a crummy textbook that doesn't do a good job of expaining Trig., if I did have a good one I would read it before I posted so many questions, but needless to say I don't.

For tests and future references I want to know how to solve problems when you are given a Sin(2x), Cos(2x), Tan(2x), CSC(2x), Sec(2x), Cot(2x). The problems are really easy for me when there is just Sin(x), Cos(x), Tan(x), CSC(x), Sec(x), Cot(x). If you take a look at this post Soroban created he did a real nice job of of putting each in terms of (opp)^2+ (adj)^2= (Hyp)^2. I really had never though of using Path Theorm this way. Here is the post is there any you can corollate his reasoning with this problem?

http://www.mathhelpforum.com/math-he...-problems.html
The problem with the double angle formulas is that the triangle where you have an angle of 2x doesn't have a simple relation to the triangle where you have an angle x. There is possibly a way to do it this way, but it is beyond my ken. Sometimes you just have to go algebraic instead of visual, I'm afraid.

-Dan

10. ## Re:

Ok thanks Dan... so if you were me would you just suck it up and memorize those three formula's, and use the calculator. I mean to me this is the easiest way I have seen. Also I take it that you divided by 2 because you are solving for x? So if it was a 3x,4x, or 5x of cos or what not you would divide by 3,4, or 5 as well? Right now I am sitting a nearly a page of trig functions deviations all of which I memorized today. My head hurts. I use the TI voyage 200 Calculator and it has all of the trig fuctions so I guess I can just use it that way. Also if she wants all of this in degrees instead of radians, I probably could just set my calc to degrees instead of constantly mulitplying by (180/pi)? Thanks Dan and I look forward to hearing back from you...

11. Don't memorize all the formulas.

If you know $\sin^2(x)+\cos^2(x)=1$ and $\cos(2x)=\cos(x+x)=\cos^2(x)-\sin^2(x)$ you can derive them.

12. ## Re:

When using decimals how many significant figures would you use?

Also if I have sec(alpha/2)=-4 and I am looking for tan(alpha) is there a way on the calculator to find the inverse like we did earlier?

13. Sure. $\frac{1}{\cos(\frac{\alpha}{2})}=-4$

$\cos(\frac{\alpha}{2})=-\frac{1}{4}$

See it now?

14. ## Re:

So from here if I was look for the tan(x) what formula would I use?

15. ## Re:

For that problem in particular I got tan(x)= 3/5

I set the problem up like this:

alpha/2=arcsec(-4)

got the arcsec(-4) then multiplied it by 2

then took that answer and put it inside the tan() button on my calculator which then gave me= .6 which is about 3/5?

Is this right...Can I it this way?

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