Rewrite #sin^4(x)tan^2(x)# in terms of the first power of cosine?

#sin^4(x)tan^2(x)# .

3 Answers
Youssuf E.
Mar 7, 2018

#=> (1-3cos^2(x) +3cos^4(x) -cos^6(x))/cos^2(x)#

Explanation:

#sin^4(x)tan^2(x)#

#=> (1-cos^2(x))^2(sin^2(x))/cos^2(x)#

#=> (1-2cos^2(x) + cos^4(x))(sin^2(x))/cos^2(x)#

#=>( sin^2(x)-2sin^2(x)cos^2(x) + sin^2(x)cos^4(x))/cos^2(x)#

#=> ((1-cos^2(x)) -2(1-cos^2(x))cos^2(x)+(1-cos^2(x))cos^4(x))/cos^2(x)#

#=> (1-cos^2(x) -2cos^2(x)+2cos^4(x)+cos^4(x)-cos^6(x))/cos^2(x)#

#=> (1-3cos^2(x) +3cos^4(x) -cos^6(x))/cos^2(x)#

1s2s2p
Mar 7, 2018

#sin^4xtan^2x=-(cos(6x)-6cos(4x)+15cos(2x)-10)/(16cos(2x)+16)#

Explanation:

#sin^4xtan^2x=sin^6x/cos^2x#

#cos(2x)=cos^2x-sin^2x#
#color(white)(cos(2x))=cos^2x-(1-cos^2x)#
#color(white)(cos(2x))=2cos^2x-1#

#cos^2x=(cos(2x)+1)/2#

Using De Moivre's Theoreom, we can evaluate #sin^6x#:
#2isin(x)=z-1/z# (where #z=cosx+isinx#)
#(2isin(x))^6=(z-1/z)^6#
#-64sin^6(x)=z^6-6z^4+15z^2-20+15/z^2-6/z^4+1/z^6#
#-64sin^6(x)=-20+(z^6+1/z^6)-6(z^4-1/z^4)+15(z^2-1/z^2)#
#(z^n-1/z^n)=2cos(nx)#
#sin^6(x)=(-20+2cos(6x)-12cos(4x)+30cos(2x))/-64#

#((-20+2cos(6x)-12cos(4x)+30cos(2x))/-64)/((cos(2x)+1)/2)=-(2cos(6x)-12cos(4x)+30cos(2x)-20)/(32cos(2x)+32)#

#sin^4xtan^2x=sin^6x/cos^2x=-(cos(6x)-6cos(4x)+15cos(2x)-10)/(16cos(2x)+16)#

Abhishek K.
Mar 7, 2018

#sin^4x*tan^2x=1/16[(10-15cos2x+6cos4x-cos6x)/(1+cos2x)]#

Explanation:

We'll use,

#rarrsin^2x=(1-cos2x)/2#

#rarrcos^2x=(1+cos2x)/2#

#rarr4cos^3x=cos3x+3cosx#

Now, #rArrtan^2x*sin^4x#

#=sin^2x/cos^2x*sin^4x#

#=(sin^2x)^3/cos^2x#

#=((1-cos2x)/2)^3/((1+cos2x)/2)#

#=1/4[(1-cos2x)^3/(1+cos2x)]#

#=1/4[(1-3cos2x+3cos^2(2x)-cos^3(2x))/(1+cos2x)]#

#=4/(4*4)[(1-3cos2x+3cos^2(2x)-cos^3(2x))/(1+cos2x)]#

#=1/16[(4-3*4cos2x+3*2*{2cos^2(2x)}-4cos^3(2x))/(1+cos2x)]#

#=1/16[(4-12cos2x+3*2*{1+cos4x}-{cos6x+3cos2x})/(1+cos2x)]#

#=1/16[(4-12cos2x+6+6cos4x-cos6x-3cos2x)/(1+cos2x)]#

#=1/16[(10-15cos2x+6cos4x-cos6x)/(1+cos2x)]#

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