Manipulating Trig Expressions

[Jakotheshadows]

Use the given substitution to express the given radical expression as a trigonometric function without radicals. Assume that a > 0 and 0 < theta < Pi/2. Then find expressions for the indicated trigonometric functions.

I'm just going to use T for theta instead of spelling it out repeatedly.

Here is the problem:
Let x = 3secT in ?(x²-9). Then find sinT and cosT
My work -
3secT=3/cosT
?[(3/cosT)²-9]
?[(3/cosT + 3)(3/cosT - 3)]
3?[(1/cosT + 1)(1/cosT - 1)]
3?(1/cos²T-1)
3?(1/cos²T - cos²T/cos²T)
3?[(1-cos²T)/cos²T]
3?(sin²T/cos²T) = 3?(tan²T) and 3tanT = ?(x²-9) so tanT = ?(x²-9) / 3

since tanT = sinT/cosT , I took the numerator from my tanT and made that my sinT and my denominator became the cosT
so sinT=?(x²-9) and cosT=3
This seems fine to me, however my textbook gives more abstract answers that I can't figure out for certain whether they mean my answer is correct or incorrect. Did I pwn my textbook, or am I being an arrogant fool?
According to the textbook:
SinT=?(x²-9)/x and cosT=3/x...
I'm thinking that my tangent is wrong, because if I'm right that makes x = 1? If that isn't the case I'm being a fool.
If I'm right, I guess I am just more thorough than the grad students making the answers. (or just an arrogant fool)

(edit: 3secT=3/cosT is part of my work, not part of the problem's wording)

 

[royhaas]

Since \(\displaystyle x = 3\sec(T)\), by definition we have that \(\displaystyle \cos(T)=3/x\). But \(\displaystyle sin^2(T)=1-cos^2(T)=1-9/x^2\), from which the result follows.

 

[Jakotheshadows]

Since \(\displaystyle x = 3\sec(T)\), by definition we have that \(\displaystyle \cos(T)=3/x\). But \(\displaystyle sin^2(T)=1-cos^2(T)=1-9/x^2\), from which the result follows.

Thanks for the reply.

sin²T=1-9/x²
sinT=?(1-9/x²)
?(x²/x²-9/x²)= ?(x²-9)/?(x²)= ?(x²-9)/x gotcha.

1-cos²T=1-9/x²
cos²T=9/x²
cosT=3/x I see what you mean here.

That was a productive post, but I still don't see the error in my original work. If anyone can please explain to me why my original solutions were incorrect I'd greatly appreciate it. I don't want to make the same mistake on the exam.

 

[Deleted member 4993]

Use the given substitution to express the given radical expression as a trigonometric function without radicals. Assume that a > 0 and 0 < theta < Pi/2. Then find expressions for the indicated trigonometric functions.

I'm just going to use T for theta instead of spelling it out repeatedly.

Here is the problem:
Let x = 3secT in ?(x²-9). Then find sinT and cosT
My work -
3secT=3/cosT
?[(3/cosT)²-9]
?[(3/cosT + 3)(3/cosT - 3)]
3?[(1/cosT + 1)(1/cosT - 1)]
3?(1/cos²T-1)
3?(1/cos²T - cos²T/cos²T)
3?[(1-cos²T)/cos²T]
3?(sin²T/cos²T) = 3?(tan²T) and 3tanT = ?(x²-9) so tanT = ?(x²-9) / 3

since tanT = sinT/cosT , I took the numerator from my tanT and made that my sinT and my denominator became the cosT
so sinT=?(x²-9)

and cosT=3 <<<< This is not fine, because the value of cosT is NEVER greater than 1.


This seems fine to me, however my textbook gives more abstract answers that I can't figure out for certain whether they mean my answer is correct or incorrect. Did I pwn my textbook, or am I being an arrogant fool?
According to the textbook:
SinT=?(x²-9)/x and cosT=3/x...
I'm thinking that my tangent is wrong, because if I'm right that makes x = 1? If that isn't the case I'm being a fool.
If I'm right, I guess I am just more thorough than the grad students making the answers. (or just an arrogant fool)

(edit: 3secT=3/cosT is part of my work, not part of the problem's wording)

 

[Jakotheshadows]

Thanks! I totally forgot the graph of the sine and cosine functions. If I had kept them in mind I would have known that cosT couldn't be 3. The absolute value of any output from the sine or cosine functions is never greater than one.

 

[Deleted member 4993]

Incedentally, if;

\(\displaystyle \tan(\theta) \, = \, \frac{a}{b}\)

then

\(\displaystyle \sin(\theta) \, = \, \frac{a}{\sqrt {a^2+b^2}}\)

and

\(\displaystyle \cos(\theta) \, = \, \frac{b}{\sqrt {a^2+b^2}}\)

 

[Jakotheshadows]

Incedentally, if;

\(\displaystyle \tan(\theta) \, = \, \frac{a}{b}\)

then

\(\displaystyle \sin(\theta) \, = \, \frac{a}{\sqrt {a^2+b^2}}\)

and

\(\displaystyle \cos(\theta) \, = \, \frac{b}{\sqrt {a^2+b^2}}\)

I have a feeling those will prove to be very useful. I'd even say they give me a nice advantage over the other classmates if they aren't in my textbook. Thanks