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# prove sin4x-sin2x/ cos4x+cos2x= ?

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prove sin4x-sin2x/ cos4x+cos2x= ?

Guest Apr 13, 2015

#1
+92221
+10

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!

$$\\=\frac{(2sin2xcos2x-sin2x)}{(cos^22x-sin^22x+cos2x)}\\\\ =\frac{sin2x(2cos2x-1)}{cos^22x-(1-cos^22x)+cos2x}\\\\ =\frac{sin2x(2cos2x-1)}{cos^22x-1+cos^22x+cos2x}\\\\ =\frac{sin2x(2cos2x-1)}{2cos^22x+cos2x-1}\\\\ =\frac{sin2x(2cos2x-1)}{2cos^22x \; +2cos2x-1cos2x \; -1}\\\\ =\frac{sin2x(2cos2x-1)}{2cos2x(cos2x+1)\;-1(cos2x +1)}\\\\ =\frac{sin2x(2cos2x-1)}{(2cos2x-1)(cos2x+1)}\\\\ =\frac{sin2x}{(cos2x+1)}\\\\$$

$$\\=\frac{2sinxcosx}{(cos^2x-sin^2x+1)}\\\\ =\frac{2sinxcosx}{(cos^2x-(1-cos^2x)+1)}\\\\ =\frac{2sinxcosx}{(cos^2x-1+cos^2x+1)}\\\\ =\frac{2sinxcosx}{(2cos^2x)}\\\\ =\frac{sinx}{(cosx)}\\\\ =tanx$$

Melody  Apr 13, 2015
Sort:

#1
+92221
+10

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!

$$\\=\frac{(2sin2xcos2x-sin2x)}{(cos^22x-sin^22x+cos2x)}\\\\ =\frac{sin2x(2cos2x-1)}{cos^22x-(1-cos^22x)+cos2x}\\\\ =\frac{sin2x(2cos2x-1)}{cos^22x-1+cos^22x+cos2x}\\\\ =\frac{sin2x(2cos2x-1)}{2cos^22x+cos2x-1}\\\\ =\frac{sin2x(2cos2x-1)}{2cos^22x \; +2cos2x-1cos2x \; -1}\\\\ =\frac{sin2x(2cos2x-1)}{2cos2x(cos2x+1)\;-1(cos2x +1)}\\\\ =\frac{sin2x(2cos2x-1)}{(2cos2x-1)(cos2x+1)}\\\\ =\frac{sin2x}{(cos2x+1)}\\\\$$

$$\\=\frac{2sinxcosx}{(cos^2x-sin^2x+1)}\\\\ =\frac{2sinxcosx}{(cos^2x-(1-cos^2x)+1)}\\\\ =\frac{2sinxcosx}{(cos^2x-1+cos^2x+1)}\\\\ =\frac{2sinxcosx}{(2cos^2x)}\\\\ =\frac{sinx}{(cosx)}\\\\ =tanx$$

Melody  Apr 13, 2015
#4
+92221
+1

Sagedragon has asked me how I factorised

$$2cos^2(2x)+cos(2x)-1$$

in the denominator.

His actual question was why/how I changed  cos(2x) into 2cos(2x)-1cos(2x)

It may be easier for you to see if I let     $$a=cos(2x)$$

then you have the expression

$$\textcolor{blue}{2} a^2+1a \textcolor{blue}{-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]

That would have to be +2 and -1

So I will now split the coefficient for a into  2-1

1a becomes  2a-1a

$$\textcolor{blue}{2} a^2+1a \textcolor{blue}{-1}\\ =\textcolor{blue}{2} a^2+2a-1a \textcolor{blue}{-1}\\ \text{Now I factorise in pairs}\\ =\textcolor{blue}{2a} (a+1) \textcolor{blue}{-1}(a+1)\\ \text{Now I have 2a lots of (a+1)and -1 lots of (a+1)}\\ \text{which makes (2a-1) lot of (a+1)}\\ =\textcolor{blue}{(2a-1)}(a+1)$$

so

$$2a^2+1a -1=(2a-1)(a+1)\\ but\;\;a=cos(2x)\\ so\\ 2cos^2(2x)+1cos(2x) -1=[2cos(2x)-1][cos(2x)+1]\\$$

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. :)

Melody  Jul 30, 2017
#2
+85759
0

Very nice, Melody  !!!!!

CPhill  Apr 13, 2015
#3
+92221
0

Thanks Chris :)

Melody  Apr 14, 2015

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