Which Trigonemetric Identity do I need to use?

Hi all,

I'm stuck with just one step now of a long problem.

Basically, how do I get from

$\displaystyle (1+2sinx)(\frac{1}{2}-sinx)$

to

$\displaystyle 2cos(x)^{2}-\frac{3}{2}$

I've been told I need to use just one trigonometric identity, but I honestly can't think of any that will do it?

Many thanks.

Re: Which Trigonemetric Identity do I need to use?

Expand:

$\displaystyle (1+2\sin(x))\cdot \left(\frac{1}{2}-\sin(x)\right)$

Afterwards use the fact $\displaystyle \sin^2(x)=1-\cos^2(x)$

Re: Which Trigonemetric Identity do I need to use?

Thanks but now theirs just one part I can't get after expanding,

how to get from

$\displaystyle (2sinx)(-sinx)$

to

$\displaystyle 2cos(x)^{2}-2$

this is the only missing step now,

many thanks again.

Re: Which Trigonemetric Identity do I need to use?

If you expand:

$\displaystyle (1+2\sin(x))\cdot \left(\frac{1}{2}-\sin(x)\right)$ then you get:

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

Use $\displaystyle \sin^2(x)=1-\cos^2(x)$ so you get:

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

$\displaystyle =...$

Re: Which Trigonemetric Identity do I need to use?

Thank you so much,

I really appreciate it.

Re: Which Trigonemetric Identity do I need to use?

Quote:

Originally Posted by

**Srengam** $\displaystyle (1+2sinx)(\frac{1}{2}-sinx)$

to

$\displaystyle 2cos(x)^{2}-\frac{3}{2}$

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

Re: Which Trigonemetric Identity do I need to use?

Quote:

Originally Posted by

**Srengam** Thank you so much,

I really appreciate it.

You're welcome! :)