Question #a6fe2

2 Answers
Feb 6, 2018

sin^4(x)+ cos^4(x) = 7/8

Explanation:

We know that:

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

Square both sides:

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

Expand the squares and mark as equation [1]:

sin^4(x)+ 2sin^2(x)cos^2(x)+cos^4(x) = 1" [1]"

Given sin(x)cos(x) = 1/4

Square both sides:

sin^2(x)cos^2(x) = 1/16

Multiply both sides by 2:

2sin^2(x)cos^2(x) = 2/16

Simplify and mark as equation [2]:

2sin^2(x)cos^2(x) = 1/8" [2]"

Subtract equation [2] from equation [1]:

sin^4(x)+ 2sin^2(x)cos^2(x)-2sin^2(x)cos^2(x)+cos^4(x) = 1-1/8

Simplify:

sin^4(x)+ cos^4(x) = 7/8

Feb 6, 2018

7/8

Explanation:

Squaring both sides and using the trig identity: sin^2x+cos^2x=1 we can write the expression in terms of sin^2x:

sin^2xtimescos^2x=1/16
sin^2xtimes(1-sin^2x)=1/16
sin^2x-sin^4x-1/16=0

Let y=sin^2x

y^2-y+1/16=0
y=(-(-1)+-sqrt((-1)^2-4times1times1/16))/(2times1)
y=1/2+-1/2sqrt(3/4)
y=1/2+-sqrt3/4

Substitute back:

sin^2x=1/2+sqrt3/4vvsin^2x=1/2-sqrt3/4

For each possible value of sin^2x we should have a value for the expression sin^4x+cos^4x.

if sin^2x=1/2+sqrt3/4:

  • cos^2x=1-(1/2+sqrt3/4)=>cos^4x=(1/2-sqrt3/4)^2=7/16-sqrt3/4
  • sin^4x=(1/2+sqrt3/4)^2=7/16+sqrt3/4

sin^4x+cos^4x=7/16+sqrt3/4+7/16-sqrt3/4=7/8

if sin^2x=1/2-sqrt3/4:

  • cos^2x=1-(1/2-sqrt3/4)=>cos^4x=(1/2+sqrt3/4)^2=7/16+sqrt3/4
  • sin^4x=(1/2-sqrt3/4)^2=7/16-sqrt3/4

sin^4x+cos^4x=7/16-sqrt3/4+7/16+sqrt3/4=7/8

In both cases, we get the same result, so we can conclude:

sin^4x+cos^4x=7/8