What's the t-formula for the trig squares? (1 Viewer)

sadpwner · Mar 13, 2015

Like for cos^2 x and the others

FrankXie · Mar 13, 2015

depends $t=\tan x$ or $t=\tan\frac{x}{2}$

braintic · Mar 14, 2015

FrankXie said:
depends $t=\tan x$ or $t=\tan\frac{x}{2}$

Why would you use t=tanx when the angle is x?

seanieg89 · Mar 14, 2015

braintic said:
Why would you use t=tanx when the angle is x?

Because we are dealing with squares.

I think

$\cos^2(x)=\frac{1}{1+t^2}$

is nicer to work with than

$\cos^2(x)=\left(\frac{1-t^2}{1+t^2}\right)^2.$

If there are any odd powers floating around though you will get square roots involved, which are why we normally use tan(x/2) I guess.

sadpwner · Mar 14, 2015

seanieg89 said:
Because we are dealing with squares.

I think

$\cos^2(x)=\frac{1}{1+t^2}$

is nicer to work with than

$\cos^2(x)=\left(\frac{1-t^2}{1+t^2}\right)^2.$

If there are any odd powers floating around though you will get square roots involved, which are why we normally use tan(x/2) I guess.

Meh, why not include both for reference sakes? btw is there one for sin^2x

seanieg89 · Mar 14, 2015

sadpwner said:
Meh, why not include both for reference sakes? btw is there one for sin^2x

$\sin(x)=\frac{2\tan(x/2)}{1+\tan^2(x/2)}=\frac{\tan(x)}{\sqrt{1+\tan^2(x)}}$

$\cos(x)=\frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}=\frac{1}{\sqrt{1+\tan^2(x)}}$

braintic · Mar 14, 2015

seanieg89 said:
Because we are dealing with squares.

I think

$\cos^2(x)=\frac{1}{1+t^2}$

is nicer to work with than

$\cos^2(x)=\left(\frac{1-t^2}{1+t^2}\right)^2.$

If there are any odd powers floating around though you will get square roots involved, which are why we normally use tan(x/2) I guess.

Two issues with that:

(I) That formula is not mentioned in the syllabus, so you would have to first justify its use

(II) If students believe they can use that formula whenever they see a cos^2 (x), they are going to get into a jam if they come across an exam question that has both a cos^2 AND a straight cos.

seanieg89 · Mar 14, 2015

braintic said:
Two issues with that:

(I) That formula is not mentioned in the syllabus, so you would have to first justify its use

(II) If students believe they can use that formula whenever they see a cos^2 (x), they are going to get into a jam if they come across an exam question that has both a cos^2 AND a straight cos.

I was fairly sure things like

$\sin(x)=\frac{\tan(x)}{\sqrt{1+\tan^2(x)}}$

were in syllabus? I remember seeing the corresponding triangles in HS books, but I suppose I could be mistaken. Isn't that how the t-formula themselves would be derived anyway? Using the expressions for sin(x) and cos(x) in terms of tan(x) and the double angle formulae?

As for (II): any student that learns formulae without understanding their scope is asking to lose marks. If all you want is a guaranteed way to reduce rational functions of the trig functions to rational functions, then the standard t- formulae will always suffice, at the expense of being occasionally tedious.

Better students should be able to spot more efficient substitutions where available though.

Drsoccerball · Mar 14, 2015

sadpwner said:
Like for cos^2 x and the others

Possible formula's:
$cos2\theta=cos^2\theta -sin^2\theta$

$sin2\theta=2cos\theta sin\theta$

$sin\theta= \frac{2t}{1+t^2}$

$cos\theta =\frac{1-t^2}{1+t^2}$

I think this is what you asked for?

braintic · Mar 14, 2015

seanieg89 said:
I was fairly sure things like

$\sin(x)=\frac{\tan(x)}{\sqrt{1+\tan^2(x)}}$

were in syllabus?

I'm afraid not. And the current syllabus was published in 1982, so unless you're more ancient than me ....

FrankXie · Mar 14, 2015

of course one CAN use $t=\tan x$ because $\cos^2x$ is essentially $\cos2x$ . don't tell me 3U students do not need to know $\cos^2x=\frac{1+\cos2x}{2}$

braintic · Mar 14, 2015

FrankXie said:
of course one CAN use $t=\tan x$ because $\cos^2x$ is essentially $\cos2x$ . don't tell me 3U students do not need to know $\cos^2x=\frac{1+\cos2x}{2}$

Provided they justify it that way. They can't just quote the formula.

sadpwner · Mar 17, 2015

In the solutions, I always see them just instantly convert (cosx)^2, (sinx)^2 and (tanx)^2 into the t-formula. I assume they just use double angle formula then convert that. Do I have to show this? Btw do they use t=tanx or t=tanx/2

InteGrand · Mar 17, 2015

sadpwner said:
In the solutions, I always see them just instantly convert (cosx)^2, (sinx)^2 and (tanx)^2 into the t-formula. I assume they just use double angle formula then convert that. Do I have to show this? Btw do they use t=tanx or t=tanx/2

I'm guessing they use $t=\tan x$ , so that

$1. \tan^2 x = t^2$
$2. \cos^2 x = \frac{1}{t^2 + 1}\left(\text{as} \cos^2 x = \frac{1}{\sec^2 x}=\frac{1}{\tan^2 x + 1}\right)$
$3. \sin^2 x =\frac{t^2}{t^2+1}\left(\text{as} \sin^2 x = 1-\cos^2 x =1-\frac{1}{t^2 +1}=\frac{t^2}{t^2 +1} \right )$

as seanig89 posted earlier.

As for having to show them, you shouldn't need to show #1, since that's obvious, but #2 and #3 you would probably show in the HSC (although #2 is pretty obvious too (cos² x = 1/sec² x, and we know sec² x = tan² x + 1), and then #3 follows quickly). They don't take long to prove though.

seanieg89 · Mar 17, 2015

braintic said:
I'm afraid not. And the current syllabus was published in 1982, so unless you're more ancient than me ....

Fair enough. Anyway, as integrand pointed out these are trivial consequences of the pythagorean identity.

I do find it silly that these aren't allowed to be stated though, given that they are natural steps in the derivation of the "standard t-formulae".

braintic · Mar 17, 2015

seanieg89 said:
Fair enough. Anyway, as integrand pointed out these are trivial consequences of the pythagorean identity.

I do find it silly that these aren't allowed to be stated though, given that they are natural steps in the derivation of the "standard t-formulae".

And I'm sure all one would need to do is draw a right-angled triangle with sides labelled appropriately to justify the use of those formulae.
Just something that indicates they know where the formulae come from.

What's the t-formula for the trig squares? (1 Viewer)

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