Question

Question #a6fe2

Answer 1

`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`

Answer 2

`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`