# Trig Help.

• Sep 21st 2006, 12:40 PM
TomCat
Trig Help.
Hi,

I have a couple of questions. My prof is having us do a chart. It's of sin, cos, tan, cot, etc functions. You're given one and have to get another.

One of the ones I am stuck on is as follows.

You're given Sin theta and need to get Tan theta. I think I need to use a Pythagorean Identity to prove it. Here's what I did...

Sin theta to Tan theta
Using the ident. Cot squared theta + 1 = Csc squared theta
Then I used the Reciprocal Idents. to get 1 / Tan^2 theta + 1 = 1 / Sin^2 theta.
Then 1 / Tan^2 theta = 1 / Sin^2 theta - 1

That's where I'm stuck... What do I do next?

Thanks,

-Tom
• Sep 21st 2006, 01:14 PM
topsquark
Quote:

Originally Posted by TomCat
Hi,

I have a couple of questions. My prof is having us do a chart. It's of sin, cos, tan, cot, etc functions. You're given one and have to get another.

One of the ones I am stuck on is as follows.

You're given Sin theta and need to get Tan theta. I think I need to use a Pythagorean Identity to prove it. Here's what I did...

Sin theta to Tan theta
Using the ident. Cot squared theta + 1 = Csc squared theta
Then I used the Reciprocal Idents. to get 1 / Tan^2 theta + 1 = 1 / Sin^2 theta.
Then 1 / Tan^2 theta = 1 / Sin^2 theta - 1

That's where I'm stuck... What do I do next?

Thanks,

-Tom

There's a much simpler way:
You've got sin(theta). cos(theta) = (+/-)sqrt(1 - [sin(theta)]^2) then tan(theta) = sin(theta) / cos(theta)

NOTE: You need to make sure cos(theta) has the appropriate sign! You can tell that by noting which quadrant theta is in. (ie. cos(theta) is positive in I and IV, cos(theta) is negative in II, III.)

-Dan
• Sep 21st 2006, 02:12 PM
TomCat
That's not the answer is it? That is not like what my Prof. was doing in class earlier today. He said that we should not involve a third function.

Here's an example that I can do:

Sin (theta) to Cos (theta).
I use the Pathag. Ident of Sin^2 (theta) + Cos^2 (theta) = 1.
I subtract the Sin^2 (theta) from both sides to get...
cos^2 (theda) = 1 - Sin^2 (theta).
Then I take the sqrt and get...
Cos (theta) = (+/-) the sqrt of 1 - Sin^2 (theta).

That's the answer he is looking for... Does this make sense? There are 36 different things I need to find. 12 of them are extremely easy because the answers are sin (theta) or 1 / sin (theta) or Cos (theta) or 1 / Cos (theta... Etc...
• Sep 21st 2006, 03:12 PM
topsquark
Quote:

Originally Posted by TomCat
That's not the answer is it? That is not like what my Prof. was doing in class earlier today. He said that we should not involve a third function.

Here's an example that I can do:

Sin (theta) to Cos (theta).
I use the Pathag. Ident of Sin^2 (theta) + Cos^2 (theta) = 1.
I subtract the Sin^2 (theta) from both sides to get...
cos^2 (theda) = 1 - Sin^2 (theta).
Then I take the sqrt and get...
Cos (theta) = (+/-) the sqrt of 1 - Sin^2 (theta).

That's the answer he is looking for... Does this make sense? There are 36 different things I need to find. 12 of them are extremely easy because the answers are sin (theta) or 1 / sin (theta) or Cos (theta) or 1 / Cos (theta... Etc...

No matter what you do you are going to have to have either a cos(theta) or a sec(theta) in there to get tan(theta). So don't worry about introducing the third trig. function, its unavoidable.

-Dan
• Sep 21st 2006, 03:32 PM
TomCat
Okay. Thanks for the help. I sent him an e-mail a little bit ago asking for some help to see what he says. Maybe when he was talking about not introducing another function he was being specific to a certain problem. I'm not sure since he never specified. Hopefully he replies to my e-mail... He seems to be the type that doesn't use computers and such, but he did provide an e-mail address on the syllabus so we'll see I guess.

Thanks,

-Tom
• Sep 21st 2006, 08:15 PM
Soroban
Hello, TomCat!

Quote:

You're given sin θ and need to get tan θ.

I assume this means: express tan θ in terms of sin θ (only).

. . . . . . . sin θ . . . . . . sin θ
tan θ .= .------- . = .-------------
. . . . . . . cos θ] . . . √1 - sinēθ

• Sep 21st 2006, 09:53 PM
ticbol
Quote:

Originally Posted by TomCat
Hi,

I have a couple of questions. My prof is having us do a chart. It's of sin, cos, tan, cot, etc functions. You're given one and have to get another.

One of the ones I am stuck on is as follows.

You're given Sin theta and need to get Tan theta. I think I need to use a Pythagorean Identity to prove it. Here's what I did...

Sin theta to Tan theta
Using the ident. Cot squared theta + 1 = Csc squared theta
Then I used the Reciprocal Idents. to get 1 / Tan^2 theta + 1 = 1 / Sin^2 theta.
Then 1 / Tan^2 theta = 1 / Sin^2 theta - 1

That's where I'm stuck... What do I do next?

Thanks,

-Tom

cot^2(A) +1 = csc^2(A)
If you get and equate their reciprocals,
1/[cot^2(A) +1] = 1/[sin^2(A)]
The lefthand side is not 1/[tan^2(A)] +1, so you cannot transpose the "1" to the righthand side.

sinA in terms of tanA:
tanA = sinA/cosA ----identity
Multiply both sides by cosA,
cosA*tanA = sinA
Express the cosA in terms of sinA,
[sqrt(1 -sin^2(A)]tanA = sinA ------------(i)
There, only sinA and tanA are involved.
Solve for sinA in terms of tanA and you're done.

Here is one way of doing that,
Square both sides of (i),
(1 -sin^2(A)[tan^2(A)] = sin^2(A)
Expand,
tan^2(A) -[sin^2(A)][tan^2(A)] = sin^2(A)
Collect the sinA terms,
tan^2(A) = sin^2(A) +[sin^2(A)][tan^2(A)]
Factor out the sin^2(A),
tan^2(A) = [sin^2(A)][1 +tan^2(A)]
Isolate the sin^2(A),
[tan^2(A)] /[1 +tan^2(A)] = sin^2(A)
So, take the square roots of both sides,
sinA = (tanA) /[sqrt(1 +tan^2(A))] -------------answer.
• Sep 22nd 2006, 12:22 AM
CaptainBlack
Quote:

Originally Posted by ticbol
:
:
sinA in terms of tanA:
tanA = sinA/cosA ----identity
Multiply both sides by cosA,
cosA*tanA = sinA
Express the cosA in terms of sinA,
[sqrt(1 -sin^2(A)]tanA = sinA ------------(i)
There, only sinA and tanA are involved.
Solve for sinA in terms of tanA and you're done.

Here is one way of doing that,
Square both sides of (i),
(1 -sin^2(A)[tan^2(A)] = sin^2(A)
Expand,
tan^2(A) -[sin^2(A)][tan^2(A)] = sin^2(A)
Collect the sinA terms,
tan^2(A) = sin^2(A) +[sin^2(A)][tan^2(A)]
Factor out the sin^2(A),
tan^2(A) = [sin^2(A)][1 +tan^2(A)]
Isolate the sin^2(A),
[tan^2(A)] /[1 +tan^2(A)] = sin^2(A)
So, take the square roots of both sides,
sinA = (tanA) /[sqrt(1 +tan^2(A))] -------------answer.

This is OK in the first and third quadrants, but in the second and third
sin and tan have different signs; so in full this should be:

Code:

```sinA=(tanA) /[sqrt(1 +tan^2(A))]  for 0<=A<90 or 270<pi<=360     =-(tanA) /[sqrt(1 +tan^2(A))] for 90<A<270```
also as tan is undefined at 90, 270 there is strictly no expression for sin in
terms of tan at these points.

RonL

(Note: It's a bit late to comment on it as Jameson has locked this thread and
I will not unlock something another has locked, but Soroban's post also suffers
from this problem)
• Sep 22nd 2006, 09:53 PM
CaptainBlack
Quote:

Originally Posted by CaptainBlack
This is OK in the first and third quadrants, but in the second and third
sin and tan have different signs; so in full this should be:

Code:

```sinA=(tanA) /[sqrt(1 +tan^2(A))]  for 0<=A<90 or 270<pi<=360     =-(tanA) /[sqrt(1 +tan^2(A))] for 90<A<270```
also as tan is undefined at 90, 270 there is strictly no expression for sin in
terms of tan at these points.

RonL