Question

If sin^8x+cos^8x-1=0 then what is the value of cos^2x(sin^2x)

Answer 1

Step-by-step explanation:

First you (Sin^2x+Cos^2x)^4=Sin^8x+Cos^8x+ 6 sin^4x.Cos^4x+4 Sin^6x.Cos ^2x+ 4 Cos^4x.Sin^2x

but sin8x+cos8x=1 LHS=1

so You can have this FINALLY

3 Sin^2x.Cos^2x+2 Sin^4x+2Cos^4x=0

2sin^2x(sin2^x+cos^2x)+Cos^2x(2sin^2x+Cos^2x)

then Subject that.you will have -1

Answer 2

The value of $cos^2xsin^2x$ are 0 and 2.

Step-by-step explanation:

$Cos^8x+sin^8x-1=0$

$sin^8x+cos^8x+2sin^4xcos^4x-2sin^4xcos^4x=1$

Adding and subtracting $2sin^4xcos^4x$

$(sin^4x)^2+(cos^4x)^2+2sin^4xcos^4x-2sin^4xcos^4x=1$

$(sin^4x+cos^4x)^2-2sin^4xcos^4x=1$

Using identity :$(a+b)^2=a^2+b^2+2ab$

$((sin^2x)^2+(cos^2x)^2+2sin^2xcos^2x-2sin^2xcos^2)^2-2sin^4xcos^4x=1$

$((sin^2x+cos^2x)^2-2sin^2xcos^2)^2-2sin^4xcos^4x=1$

$(1-2sin^2xcos^2x)^2-2sin^4xcos^4x=1$

$1+4sin^4xcos^4x-4sin^2xcos^2x-2sin^4xcos^4x=1$

$2sin^4xcos^4x-4sin^2xcos^2x=1-1=0$

$2(sin^2xcos^2x)^2-4sin^2xcos^2x=0$

It is quadratic equation in $sin^2xcos^2x$

Quadratic formula  for quadratic equation$ax^2+bx+c=0$

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Where a=Coefficient of x square

b=Coefficient of x

c=Constant term

Using quadratic formula  

$sin^2xcos^2x=\frac{4\pm\sqrt{(-4)^2-4(2)(0)}}{2(2)}$

$sin^2xcos^2=\frac{4\pm\sqrt{16}}{4}=\frac{4\pm 4}{4}$

$sin^2xcos^2x=\frac{4+4}{4}=\frac{8}{4}=2$

$sin^2xcos^2x=\frac{4-4}{4}=0$

Hence, the value of $cos^2xsin^2x$ are 0 and 2.

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