1. Proving

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

2. $2sin(x) cos^3(x) + 2sin^3(x)cos(x) = sin(2x)$

3. $(1/4)sin (4x) = sin (x) cos^3(x) - cos (x) sin^3(x)$

2. The first one is not bad at all. Just expand out, but first sub in $cos^{2}(x)=1-sin^{2}(x)$

$(1-2sin^{2}(x))^{2}+4sin^{2}(x)(1-sin^{2}(x))$

$1-4sin^{2}(x)+4sin^{4}(x)+4sin^{2}(x)-4sin^{4}(x)$

Now, see it?.

3. Originally Posted by Dragon
1. $(1-2sin^2(x))^2 + 4sin^2(x) cos^2(x) =1$

2. $2sin(x) cos^3(x) + 2sin^3(x)cos(x) = sin(2x)$

3. $(1/4)sin (4x) = sin (x) cos^3(x) - cos (x) sin^3(x)$
2. $2sin(x) cos^3(x) + 2sin^3(x)cos(x)$
$=2sin(x)cos(x)cos^2(x)+2sin(x)sin^2(x)cos(x)$
now, group $2sin(x)cos(x)$together,
and sub in $sin^2(x)+cos^2(x)=1$and $2sin(x)cos(x)=sin(2x)$
solved!

3. $(1/4)sin (4x) = sin (x) cos^3(x) - cos (x) sin^3(x)$
the method is similar with question 2. try it yourself. come back to me if you still have problems.

4. i got number 2 but not 3

5. Originally Posted by Godfather
i got number 2 but not 3
since right-hand side of question 3 is more complicated, so we start our proof from right-hand side.

RHS: $sin(x)cos^3(x)-cos(x)sin^3(x)$
$=sin(x)cos(x)cos^2(x)-cos(x)sin(x)sin^2(x)$
$=sin(x)cos(x)(cos^2(x)-sin^2(x))$
$=(1/2)sin(2x)cos(2x)$
$=(1/4)sin(4x)$

so, LHS=RHS, solved!