Question

If sin x-cos x =0 then find the value of sin^4x+cos^4x =?

Answer 1

sinx-cosx = 0
sinx = cosx
it is only when
sinx= cosx = ±1/√2
=sin^4x+cos^4x
= (±1/√2)^4+(±1/√2)^4
= 1/4+1/4= 1/2

Answer 2

Answer:

Sinx-Cosx=0 (given)

Sinx=Cosx

Hence sin^2x=Cos^2x...(i)

We know,

Sin^2x+Cos^2x=1

Squaring both sides

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

Putting sin^2x=Cos^2x {From (i)}

(2Cos^2x)^2=1

Cos^2x=1/2

Cos^4x=1/4 (Squaring both sides)

Similarly Sin^4x=1/4

Sin^4x+Cos^4x

1/4+1/4=1/2