why, how (1 Viewer)

ExtremelyBoredUser · Oct 22, 2021

yashbb said:
View attachment 32892

Not sure what you mean by why or how?

Just do the regular t formula method;

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

yashbb · Oct 22, 2021

ExtremelyBoredUser said:
Not sure what you mean by why or how?

Just do the regular t formula method;

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

View attachment 32894

we were not taught any method, teacher just did two examples and gave us work.

yashbb · Oct 22, 2021

tysm!

ExtremelyBoredUser · Oct 22, 2021

yashbb said:
we were not taught any method, teacher just did two examples and gave us work.

Oh. Its pretty straightforward.

Say you have a function;

cos(x) + 3sin(x) = k
you can find the value of theta through auxiliary angles but you can also use t formulae for this.

You would make t = tan(x/2) as to be able to substitute the trigonometric functions into one variable t.

Then you would solve for t.

This is one way of memorising the identities. cos(x) would be opposite over hypotenuse so 1-t^2/1+t^2.

There's not much to this, its just algebra. If you're algebra's pretty rusty and you're not confident, this is going to be brutal. If your algebraic manipulation is good, then this is pretty straightforward.

yashbb · Oct 22, 2021

ExtremelyBoredUser said:
Oh. Its pretty straightforward.

Say you have a function;

cos(x) + 3sin(x) = k
you can find the value of theta through auxiliary angles but you can also use t formulae for this.

You would make t = tan(x/2) as to be able to substitute the trigonometric functions into one variable t.

View attachment 32896

This is one way of memorising the identities. cos(x) would be opposite over hypotenuse so 1-t^2/1+t^2.

There's not much to this, its just algebra. If you're algebra's pretty rusty and you're not confident, this is going to be brutal. If your algebraic manipulation is good, then this is pretty straightforward.

my algebra is fine, its just that we are never really taught anything. Thank you. I can't express how grateful I am.

Drongoski · Oct 23, 2021

Here's a derivation; not necessarily the best approach:

$cos2\theta = cos^2\theta - sin^2\theta = \frac {cos^2\theta - sin^2\theta}{cos^2\theta + sin^2\theta} = \frac{\frac{cos^2\theta}{cos^2\theta} - \frac{sin^2\theta}{cos^2\theta}}{\frac{cos^2\theta}{cos^2\theta} + \frac {sin^2\theta}{cos^2\theta}}\\ \\ = \frac{1 - tan^2\theta}{1 + tan^2\theta} = \frac {1 - t^2}{1 + t^2}$

The question of course did not ask for a derivation!

5uckerberg · Oct 23, 2021

yashbb said:
View attachment 32892

If you do not want to use the t-formula as ExtremelyBoredUser said another way around it is this $t^{2}=\tan^{2}{\theta}$ now $t^{2}+1=\sec^{2}{\theta}$ . Now $\cos^{2}{\theta}=\frac{1}{t^{2}+1}$ and $\cos{2\theta}=\cos^{2}\{\theta}-\sin^{2}\{\theta}=\frac{1-t^{2}}{1+t^{2}}$

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