1. ## 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

2. 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

3. 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...

4. 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

5. 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

6. Hello, TomCat!

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 - sin²θ

7. 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.

8. 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)

9. 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