Pre-calculus help please
tanx-cotx/tanx+cotx = 1 - 2cos^2x
Verify that they are equal.
i googled this problem but i still dont understand it, despite their explanations
can somebody please answer this step my step, preferably with explanations on which identities were used etc?
any help would be appreciated, tyvm
May 7, 2012
9 Comments • Newest first
[quote=CrownedSorrow]Just take the derivative and find the tangent of the line with limit of infinity using the chain rule.[/quote]
lol, silly billy. You learn that in Calc.
Funny thing is that Kuta Software (The teacher part only- not the part you can find online) has this exact question on a test.
inbefore cheating on a test.
Also, I remember a similar problem on a quiz.
[quote=CrownedSorrow]Just take the derivative and find the tangent of the line with limit of infinity using the chain rule.[/quote]
I actually found this really funny. lmao Trolling at it's finest.
[i]wooooow[/i]
Getting basilmarket to do your trig identities for you? In no way will that help you get them right on the test.
Just take the derivative and find the tangent of the line with limit of infinity using the chain rule.
wow, even though I'm not the one asking the questions, thanks -the two posters- who gave the solutions... it helped me too!
What the? You learn about the cot (x) in pre Calc? I learned about it in Uni lol, but that's Queensland Education for you.
Yeah I read it over and edited it in hahaha. You caught me before I could change it lmaooo
EDIT: ohh @above you took a slightly different approach than me but that works too
Just got it . I'll try to explain it as best as I can but it'll be kinda hard on basil .. much easier seen when its written out.
Okay, so you only want to simplify one side and get it to equal the other. I chose the tanx-cotx/tanx + cotx side. No reason, but it just seemed much easier.
(tanx-cotx)/(tanx+cotx) - Now we simplify tanx and cotx to its components. It'll look a bit messy in text but its very simple.
[(sinx/cosx) - (cosx/sinx)]/[(sinx/cosx) + (cosx/sinx)] - Now we add together the fractions on both the bottom and top.
[(sin^2x-cos^2x)/(sinxcosx)]/[(sin^2x+cos^2x)/(sinxcosx)] - Since both the numerator and denominator have 1/sinxcosx, we cancel them out.
We get: (sin^2x-cos^2x)/(sin^2x+cos^2x) - now use the trig identity to set the bottom to 1.
sin^2x-cos^2x - This is where I got a little stumped since I haven't done this in a while. But you'll realize that if you take a negative out, you'll get an equivalent to cos2x using an identity.
- (cos^2x-sin^2x) - See it now? The inside is equivalent to cos2x.
Thus..
-(cos2x) - Now we substitute in with another identity for cos2x.
- (2cos^2x - 1).
And here's our answer: 1 - 2cos^2x
Hope that helped