Question

sin4x-cos4x=2sin2x-1 ..prove the following

Answer 1

LHS

= sin^4(x)-cos^4(x)

= (sin^2(x)+cos^2(x))(Sin^2(x)-cos^2(x))

=sin^2 + cos^2 = 1......

= sin^2(x)-cos^2(x)

= sin^2(x) - (1-sin^2(x))

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

= 2sin^2(x) - 1

= RHS

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Answer 2

To prove that: $\sin^4 x$ - $\cos^4 x$ = 2$\sin^2 x$ - 1.

Solution:

L.H.S. = $\sin^4 x$ - $\cos^4 x$

= $(\sin^2 x)^2$ - $(\cos^2 x)^2$

Using the algebraic identity:

$a^{2} -b^2$ = (a + b)(a - b)

= ($\sin^2 x$ + $\cos^2 x$)($\sin^2 x$ - $\cos^2 x$)

Using the algebraic identity:

$\sin^2 A$ + $\cos^2 A$ = 1

= (1)($\sin^2 x$ - $\cos^2 x$)

= $\sin^2 x$ - $\cos^2 x$

Using the algebraic identity:

$\cos^2 A$ = 1 - $\sin^2 A$

= $\sin^2 x$ - (1 - $\sin^2 x$ )

= $\sin^2 x$ - 1 + $\sin^2 x$

= 2$\sin^2 x$ - 1

= R.H.S., proved.

Thus, $\sin^4 x$ - $\cos^4 x$ = 2$\sin^2 x$ - 1, proved.