# Thread: Making A Trig Identity

1. ## Making A Trig Identity

Ok so I have to make up a trig identity

The instructions are to start with a statement that I know is true, and than keep changing both sides to equivalent equations until it is complex

So far I have:

cos²x = cos²x

1-sin²x = cos2x + sin²x

I need 2 more lines of changing them until the final identity..

But I don't know how to keep going!

Any help?

Thanks

Exactly what are the directions?

Ok, so I have to make up a trig identity

The instructions are to start with a statement that I know is true,
and then keep changing both sides to equivalent equations until it is complex.

So far I have: . $\cos^2\!x \:=\:\cos^2\!x$

. . . . . . $1-\sin^2\!x \:=\: \cos2x + \sin^2\!x$

I need 2 more lines of changing . . . . Why two more lines?

How about: . $1-\sin^2\!x \;=\;\cos^2\!x$

. . . $(1-\sin x)(1 + \sin x) \;=\;\cos^2\!x$

. . . . . . . . . $\frac{1-\sin x}{\cos x} \;=\;\frac{\cos x}{1 + \sin x}$

Another: . $(\sin x + \cos x)^2 \;=\;(\sin x + \cos x)^2$

. . . . . . . $(\sin x + \cos x)^2 \;=\;\sin^2\!x + 2\sin x\cos x + \cos^2\!x$

. . . . . . . $(\sin x + \cos x)^2 \;=\;\underbrace{\sin^2\!x + \cos^2\!x}_{\text{This is 1}} + \underbrace{2\sin x\cos x}_{\text{This is }\sin2x}$

. . . . . . . $(\sin x + \cos x)^2 \;=\;1 + \sin2x$

3. The exact instructions are:

Start with a statement you know is true for all values of x; for example sin²x = sin²x

Progressively change each side into an equivalent, but more complex form; for example 1 - cos²x = -cos2x + cos²x

Continue replacing terms with equivalent expressions until you decide that your identity is sufficiently complex
( min. 3 lines of changing before the final identity )

So pretty much just have to keep changing them.. but my problem was running out of things to change into lol

is all i could go.. i didn't know how to change them any further

thanks so much for helping!