+118703 prove sin4x-sin2x/ cos4x+cos2x=
There is nothing to prove - do you just want it simplified?
simplify (sin4x-sin2x)/( cos4x+cos2x) is this what you mean? you need to use brackets!
=(2sin2xcos2x−sin2x)(cos22x−sin22x+cos2x)=sin2x(2cos2x−1)cos22x−(1−cos22x)+cos2x=sin2x(2cos2x−1)cos22x−1+cos22x+cos2x=sin2x(2cos2x−1)2cos22x+cos2x−1=sin2x(2cos2x−1)2cos22x+2cos2x−1cos2x−1=sin2x(2cos2x−1)2cos2x(cos2x+1)−1(cos2x+1)=sin2x(2cos2x−1)(2cos2x−1)(cos2x+1)=sin2x(cos2x+1)
=2sinxcosx(cos2x−sin2x+1)=2sinxcosx(cos2x−(1−cos2x)+1)=2sinxcosx(cos2x−1+cos2x+1)=2sinxcosx(2cos2x)=sinx(cosx)=tanx
+118703 prove sin4x-sin2x/ cos4x+cos2x=
There is nothing to prove - do you just want it simplified?
simplify (sin4x-sin2x)/( cos4x+cos2x) is this what you mean? you need to use brackets!
=(2sin2xcos2x−sin2x)(cos22x−sin22x+cos2x)=sin2x(2cos2x−1)cos22x−(1−cos22x)+cos2x=sin2x(2cos2x−1)cos22x−1+cos22x+cos2x=sin2x(2cos2x−1)2cos22x+cos2x−1=sin2x(2cos2x−1)2cos22x+2cos2x−1cos2x−1=sin2x(2cos2x−1)2cos2x(cos2x+1)−1(cos2x+1)=sin2x(2cos2x−1)(2cos2x−1)(cos2x+1)=sin2x(cos2x+1)
=2sinxcosx(cos2x−sin2x+1)=2sinxcosx(cos2x−(1−cos2x)+1)=2sinxcosx(cos2x−1+cos2x+1)=2sinxcosx(2cos2x)=sinx(cosx)=tanx
+118703 Sagedragon has asked me how I factorised
2cos2(2x)+cos(2x)−1
in the denominator.
His actual question was why/how I changed cos(2x) into 2cos(2x)-1cos(2x)
Hi Sagedragon I am really pleased that you have asked :)
It may be easier for you to see if I let a=cos(2x)
then you have the expression
2a2+1a−1
To factorize this I looked for two numbers
that would multiply to 2*-1=-2
[Since they multiply to a negative I know straight off that one is negative and one is positive]
and add to +1
That would have to be +2 and -1
So I will now split the coefficient for a into 2-1
1a becomes 2a-1a
2a2+1a−1=2a2+2a−1a−1Now I factorise in pairs=2a(a+1)−1(a+1)Now I have 2a lots of (a+1)and -1 lots of (a+1)which makes (2a-1) lot of (a+1)=(2a−1)(a+1)
so
2a2+1a−1=(2a−1)(a+1)buta=cos(2x)so2cos2(2x)+1cos(2x)−1=[2cos(2x)−1][cos(2x)+1]
Sagedragon,
Think about it and if you still have queries do not hesitate to ask.
You can ask on this thread if it is not locked but always send the link by private message as well becasue otherwise I may not see it. :)
