can anybody help?

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- Dec 9th 2012, 03:02 AMtautvydukssimplify and proof that answer is this
can anybody help?

Attachment 26152 - Dec 9th 2012, 03:09 AMtopsquarkRe: simplify and proof that answer is this
For the first problem, note that

$\displaystyle sin(A - B) = sin(A)~cos(B) - sin(B)~cos(A)$

so we have:

$\displaystyle sin \left ( \frac{\pi}{3} - \alpha \right ) = sin \left ( \frac{\pi}{3} \right ) ~cos( \alpha ) - sin( \alpha) ~cos \left ( \frac{\pi}{3} \right )$

See if that helps.

-Dan - Dec 9th 2012, 03:10 AMMarkFLRe: simplify and proof that answer is this
How do you begin with an expression and end with an equation?

- Dec 9th 2012, 03:11 AMtautvyduksRe: simplify and proof that answer is this
but what i should do with 1/2sin(alpha)

- Dec 9th 2012, 03:14 AMMarkFLRe: simplify and proof that answer is this
Subtract it from the expression

**topsquark**gave you after you have simplified it. - Dec 9th 2012, 03:16 AMtopsquarkRe: simplify and proof that answer is this
For the second problem, note that $\displaystyle sin(2x) = 2~sin(x)~cos(x)$, so we have

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

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

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

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

Note that there will be*four*solutions to this.

-Dan

Wow. Lots of posting going on in this thread! - Dec 9th 2012, 03:17 AMtautvyduksRe: simplify and proof that answer is this
x

- Dec 9th 2012, 03:27 AMtautvyduksRe: 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

- Dec 9th 2012, 03:30 AMtautvyduksRe: simplify and proof that answer is this
- Dec 9th 2012, 03:42 AMMarkFLRe: simplify and proof that answer is this
The second problem is:

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

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

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

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

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

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

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

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

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

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

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

You will find 4 solutions on the interval $\displaystyle 0\le x<2\pi$. 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. - Dec 9th 2012, 04:36 AMtautvyduksRe: simplify and proof that answer is this
i need to suplify left side of this and get the right side.. not to solve X...

- Dec 9th 2012, 04:41 AMMarkFLRe: simplify and proof that answer is this
It is not an identity. It is only true for certain values of

*x*. - Dec 9th 2012, 04:54 AMtautvyduksRe: simplify and proof that answer is this
ohhh it`s imposibble tu supplify it am i right?

- Dec 9th 2012, 05:19 AMMarkFLRe: simplify and proof that answer is this
It is impossible to take the left side and get the right, yes.

- Dec 9th 2012, 05:35 AMtautvyduksRe: simplify and proof that answer is this
oh thanks. and can you help with the 1st one?