# simplify and proof that answer is this

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• December 9th 2012, 04:02 AM
tautvyduks
simplify and proof that answer is this
can anybody help?

Attachment 26152
• December 9th 2012, 04:09 AM
topsquark
Re: simplify and proof that answer is this
Quote:

Originally Posted by tautvyduks
can anybody help?

Attachment 26152

For the first problem, note that
$sin(A - B) = sin(A)~cos(B) - sin(B)~cos(A)$

so we have:
$sin \left ( \frac{\pi}{3} - \alpha \right ) = sin \left ( \frac{\pi}{3} \right ) ~cos( \alpha ) - sin( \alpha) ~cos \left ( \frac{\pi}{3} \right )$

See if that helps.

-Dan
• December 9th 2012, 04:10 AM
MarkFL
Re: simplify and proof that answer is this
How do you begin with an expression and end with an equation?
• December 9th 2012, 04:11 AM
tautvyduks
Re: simplify and proof that answer is this
but what i should do with 1/2sin(alpha)
• December 9th 2012, 04:14 AM
MarkFL
Re: simplify and proof that answer is this
Subtract it from the expression topsquark gave you after you have simplified it.
• December 9th 2012, 04:16 AM
topsquark
Re: simplify and proof that answer is this
For the second problem, note that $sin(2x) = 2~sin(x)~cos(x)$, so we have
$1 - sin^4(x) - sin^2(x) = \frac{1}{4}sin^2(x)$

$1 - sin^4(x) - sin^2(x) = \frac{1}{4} \left ( 2~sin(x)~cos(x) \right ) ^2$

Multiply out the RHS and use $cos^2(x) = 1 - sin^2(x)$, then solve for sin(x).

You will get $sin^2(x) = \frac{1}{2}$

Note that there will be four solutions to this.

-Dan

Wow. Lots of posting going on in this thread!
• December 9th 2012, 04:17 AM
tautvyduks
Re: simplify and proof that answer is this
x
• December 9th 2012, 04:27 AM
tautvyduks
Re: simplify and proof that answer is this
ok, thanks i did that but what i have to do with -1/2sin(alpha)? im bad at maths sorry
• December 9th 2012, 04:30 AM
tautvyduks
Re: simplify and proof that answer is this
Quote:

Originally Posted by topsquark
For the second problem, note that $sin(2x) = 2~sin(x)~cos(x)$, so we have
$1 - sin^4(x) - sin^2(x) = \frac{1}{4}sin^2(x)$

$1 - sin^4(x) - sin^2(x) = \frac{1}{4} \left ( 2~sin(x)~cos(x) \right ) ^2$

Multiply out the RHS and use $cos^2(x) = 1 - sin^2(x)$, then solve for sin(x).

You will get $sin^2(x) = \frac{1}{2}$

Note that there will be four solutions to this.

-Dan

Wow. Lots of posting going on in this thread!

Thanks, but i need to solve this i need to proof that
$1 - sin^4(x) - sin^2(x) is equal to frac{1}{4}sin^2(x)$

i
• December 9th 2012, 04:42 AM
MarkFL
Re: simplify and proof that answer is this
The second problem is:

$1 - \sin^4(x) - \sin^2(x) = \frac{1}{4}\sin^2(2x)$

This is not an identity to prove, but an equation to solve.

Using the double-angle identity for sine provided by topsquark, we have:

$1 - \sin^4(x) - \sin^2(x) = \frac{1}{4}(2\sin(x)\cos(x))^2$

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

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

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

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

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

$\sin^2(x)=\frac{1}{2}$

$\sin(x)=\pm\frac{1}{\sqrt{2}}$

You will find 4 solutions on the interval $0\le x<2\pi$. If you are not restricted to an interval, then you will have an infinite number of solutions, which is easily expressed as the integral multiple of an angle.
• December 9th 2012, 05:36 AM
tautvyduks
Re: simplify and proof that answer is this
i need to suplify left side of this and get the right side.. not to solve X...
• December 9th 2012, 05:41 AM
MarkFL
Re: simplify and proof that answer is this
It is not an identity. It is only true for certain values of x.
• December 9th 2012, 05:54 AM
tautvyduks
Re: simplify and proof that answer is this
ohhh it`s imposibble tu supplify it am i right?
• December 9th 2012, 06:19 AM
MarkFL
Re: simplify and proof that answer is this
It is impossible to take the left side and get the right, yes.
• December 9th 2012, 06:35 AM
tautvyduks
Re: simplify and proof that answer is this
oh thanks. and can you help with the 1st one?
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