Rewrite sin^4(x)tan^2(x)sin4(x)tan2(x) in terms of the first power of cosine?

sin^4(x)tan^2(x)sin4(x)tan2(x) .

3 Answers
Mar 7, 2018

=> (1-3cos^2(x) +3cos^4(x) -cos^6(x))/cos^2(x)13cos2(x)+3cos4(x)cos6(x)cos2(x)

Explanation:

sin^4(x)tan^2(x)sin4(x)tan2(x)

=> (1-cos^2(x))^2(sin^2(x))/cos^2(x)(1cos2(x))2sin2(x)cos2(x)

=> (1-2cos^2(x) + cos^4(x))(sin^2(x))/cos^2(x)(12cos2(x)+cos4(x))sin2(x)cos2(x)

=>( sin^2(x)-2sin^2(x)cos^2(x) + sin^2(x)cos^4(x))/cos^2(x)sin2(x)2sin2(x)cos2(x)+sin2(x)cos4(x)cos2(x)

=> ((1-cos^2(x)) -2(1-cos^2(x))cos^2(x)+(1-cos^2(x))cos^4(x))/cos^2(x)(1cos2(x))2(1cos2(x))cos2(x)+(1cos2(x))cos4(x)cos2(x)

=> (1-cos^2(x) -2cos^2(x)+2cos^4(x)+cos^4(x)-cos^6(x))/cos^2(x)1cos2(x)2cos2(x)+2cos4(x)+cos4(x)cos6(x)cos2(x)

=> (1-3cos^2(x) +3cos^4(x) -cos^6(x))/cos^2(x)13cos2(x)+3cos4(x)cos6(x)cos2(x)

Mar 7, 2018

sin^4xtan^2x=-(cos(6x)-6cos(4x)+15cos(2x)-10)/(16cos(2x)+16)sin4xtan2x=cos(6x)6cos(4x)+15cos(2x)1016cos(2x)+16

Explanation:

sin^4xtan^2x=sin^6x/cos^2xsin4xtan2x=sin6xcos2x

cos(2x)=cos^2x-sin^2xcos(2x)=cos2xsin2x
color(white)(cos(2x))=cos^2x-(1-cos^2x)cos(2x)=cos2x(1cos2x)
color(white)(cos(2x))=2cos^2x-1cos(2x)=2cos2x1

cos^2x=(cos(2x)+1)/2cos2x=cos(2x)+12

Using De Moivre's Theoreom, we can evaluate sin^6xsin6x:
2isin(x)=z-1/z2isin(x)=z1z (where z=cosx+isinxz=cosx+isinx)
(2isin(x))^6=(z-1/z)^6(2isin(x))6=(z1z)6
-64sin^6(x)=z^6-6z^4+15z^2-20+15/z^2-6/z^4+1/z^664sin6(x)=z66z4+15z220+15z26z4+1z6
-64sin^6(x)=-20+(z^6+1/z^6)-6(z^4-1/z^4)+15(z^2-1/z^2)64sin6(x)=20+(z6+1z6)6(z41z4)+15(z21z2)
(z^n-1/z^n)=2cos(nx)(zn1zn)=2cos(nx)
sin^6(x)=(-20+2cos(6x)-12cos(4x)+30cos(2x))/-64sin6(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)20+2cos(6x)12cos(4x)+30cos(2x)64cos(2x)+12=2cos(6x)12cos(4x)+30cos(2x)2032cos(2x)+32

sin^4xtan^2x=sin^6x/cos^2x=-(cos(6x)-6cos(4x)+15cos(2x)-10)/(16cos(2x)+16)sin4xtan2x=sin6xcos2x=cos(6x)6cos(4x)+15cos(2x)1016cos(2x)+16

Mar 7, 2018

sin^4x*tan^2x=1/16[(10-15cos2x+6cos4x-cos6x)/(1+cos2x)]sin4xtan2x=116[1015cos2x+6cos4xcos6x1+cos2x]

Explanation:

We'll use,

rarrsin^2x=(1-cos2x)/2sin2x=1cos2x2

rarrcos^2x=(1+cos2x)/2cos2x=1+cos2x2

rarr4cos^3x=cos3x+3cosx4cos3x=cos3x+3cosx

Now, rArrtan^2x*sin^4xtan2xsin4x

=sin^2x/cos^2x*sin^4x=sin2xcos2xsin4x

=(sin^2x)^3/cos^2x=(sin2x)3cos2x

=((1-cos2x)/2)^3/((1+cos2x)/2)=(1cos2x2)31+cos2x2

=1/4[(1-cos2x)^3/(1+cos2x)]=14[(1cos2x)31+cos2x]

=1/4[(1-3cos2x+3cos^2(2x)-cos^3(2x))/(1+cos2x)]=14[13cos2x+3cos2(2x)cos3(2x)1+cos2x]

=4/(4*4)[(1-3cos2x+3cos^2(2x)-cos^3(2x))/(1+cos2x)]=444[13cos2x+3cos2(2x)cos3(2x)1+cos2x]

=1/16[(4-3*4cos2x+3*2*{2cos^2(2x)}-4cos^3(2x))/(1+cos2x)]=116[434cos2x+32{2cos2(2x)}4cos3(2x)1+cos2x]

=1/16[(4-12cos2x+3*2*{1+cos4x}-{cos6x+3cos2x})/(1+cos2x)]=116[412cos2x+32{1+cos4x}{cos6x+3cos2x}1+cos2x]

=1/16[(4-12cos2x+6+6cos4x-cos6x-3cos2x)/(1+cos2x)]=116[412cos2x+6+6cos4xcos6x3cos2x1+cos2x]

=1/16[(10-15cos2x+6cos4x-cos6x)/(1+cos2x)]=116[1015cos2x+6cos4xcos6x1+cos2x]