simplify and proof that answer is this

**tautvyduks** · December 9th 2012, 03:02 AM

can anybody help?

simplify and proof that answer is this-matttte.jpg

**topsquark** · December 9th 2012, 03:09 AM

Originally Posted by tautvyduks

can anybody help?

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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

**MarkFL** · December 9th 2012, 03:10 AM

How do you begin with an expression and end with an equation?

**tautvyduks** · December 9th 2012, 03:11 AM

but what i should do with 1/2sin(alpha)

**MarkFL** · December 9th 2012, 03:14 AM

Subtract it from the expression topsquark gave you after you have simplified it.

**topsquark** · December 9th 2012, 03:16 AM

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!

**tautvyduks** · December 9th 2012, 03:17 AM

x

**tautvyduks** · December 9th 2012, 03:27 AM

ok, thanks i did that but what i have to do with -1/2sin(alpha)? im bad at maths sorry

**tautvyduks** · December 9th 2012, 03:30 AM

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

**MarkFL** · December 9th 2012, 03:42 AM

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.

**tautvyduks** · December 9th 2012, 04:36 AM

i need to suplify left side of this and get the right side.. not to solve X...

**MarkFL** · December 9th 2012, 04:41 AM

It is not an identity. It is only true for certain values of x.

**tautvyduks** · December 9th 2012, 04:54 AM

ohhh it`s imposibble tu supplify it am i right?

**MarkFL** · December 9th 2012, 05:19 AM

It is impossible to take the left side and get the right, yes.

**tautvyduks** · December 9th 2012, 05:35 AM

oh thanks. and can you help with the 1st one?

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