# Proving

• Mar 26th 2008, 03:55 PM
Dragon
Proving
1.\$\displaystyle (1-2sin^2(x))^2 + 4sin^2(x) cos^2(x) =1\$

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

3.\$\displaystyle (1/4)sin (4x) = sin (x) cos^3(x) - cos (x) sin^3(x)\$
• Mar 26th 2008, 04:04 PM
galactus
The first one is not bad at all. Just expand out, but first sub in \$\displaystyle cos^{2}(x)=1-sin^{2}(x)\$

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

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

Now, see it?.
• Mar 26th 2008, 04:20 PM
deniselim17
Quote:

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

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

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

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

3.\$\displaystyle (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.
• Mar 27th 2008, 04:57 PM
Godfather
i got number 2 but not 3
• Mar 27th 2008, 08:27 PM
deniselim17
Quote:

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

so, LHS=RHS, solved!