Trigonometry

[digitalmaster287]

I am at the end of my trigonometry unit and (since our class average was pretty low) offered us a chance at bonus marks. He gave us a sheet with some trig identities to prove and I'm stuck on the last one. It is as follows:

1 - (sin^2 x/1+cot x) - (cos^2 x/1+tan x) = sin x cos x

Can anyone help me?

 

[Crimson King]

I forgot all my trig identities, but I remember 1 - (sin X) = SOMETHING, I forgot what though. from there, you get whatever over Cot X which is the starting point to your solving this. Just remember to work out both sides a little to find the similarities.

 

[EEvisu]

divide by zero

 

[Bibbed]

Thought something was wrong, but thats def. right now.

 

[Bibbed]

1 - (sin^2 x/1+cot x) - (cos^2 x/1+tan x) = sin x cos x

1 - (sin^2 x/(1+(cos x/sin x))) - (cos^2 x/(1+(sin x/cos x))) = sin x cos x

1 - (sin^2 x/((sin x/sin x)+(cos x/sin x))) - (cos^2 x/((cos x/cos x)+(sin x/cos x))) = sin x cos x

1 - (sin^2 x/((1/sin x)*(sin x+cos x))) - (cos^2 x/((1/cos x)*(cos x+sin x))) = sin x cos x

1 - (sin^3 x/(sin x+cos x)) - (cos^3 x/(cos x+sin x)) = sin x cos x

Factor out -1 and then combine the two fractions (which can now be added)...
1 - ((sin^3 x+cos^3 x)/(sin x+cos x)) = sin x cos x

Because sin^2 x + cos^2 x = 1...
(sin^2 x + cos^2 x) - ((sin^3 x+cos^3 x)/(sin x+cos x)) = sin x cos x

Multiply out the two terms...
((sin x + cos x)(sin^2 x + cos^2 x))/(sin x + cos x) - ((sin^3 x+cos^3 x)/(sin x+cos x)) = sin x cos x

(sin^3 x + sin x cos^2 x + sin^2 x cos x + cos^3 x)/(sin x + cos x) - ((sin^3 x+cos^3 x)/(sin x+cos x)) = sin x cos x

(sin^3 x + sin x cos^2 x + sin^2 x cos x + cos^3 x - sin^3 x - cos^3 x)/(sin x+cos x)) = sin x cos x

(sin x cos^2 x + sin^2 x cos x)/(sin x+cos x)) = sin x cos x

Factor out sin x
(sin x*(cos^2 x + sin x cos x))/(sin x+cos x)) = sin x cos x

Factor out cos x
((sin x cos x)*(cos x + sin x))/(sin x+cos x)) = sin x cos x

The numerator and denominator now cancel out, and thats it.
sin x cos x) = sin x cos x


That's all I can give man, its screwing with my brain to write all that math. Granted, I haven't taken trigonometry, or done a proof in probably 8 years, all of that could be wrong.

 

[digitalmaster287]

Dude, Thanks a LOT. I understand it all, except for this one step near the beginning

Factor out -1 and then combine the two fractions (which can now be added)...
1 - ((sin^3 x+cos^3 x)/(sin x+cos x)) = sin x cos x

If you factor out -1 from sin^3 x - cos^3 x, why is the answer -sin^3 x +cos^3 x instead of just sin^3 x + cos^3 x? My calculator says what you did is right, but I just wanna know why...

EDIT: nvm I did a really stupid mistake...thanks again man

 

[Bibbed]

Whoops, see the edit now. Whichever works best for you though, just make sure the sin^3 and cos^3 cancel later and you'll be all good.

 

[cF=)]

Bibbed beated me to it. Why do I see these topics a few hours after everybody T__T

 

[Maeotian]

Noooo! Not trig identities! You can't believe how much I HATE those things! But what I would usually do is just try to put everything in terms of sin x's and cos x's, and you should be good from there.

 

[pikachun00b7]

**** I hated trig. And now I have to see it again in Calc!
Trigonometry= loose.