Which Trigonemetric Identity do I need to use?

**Srengam** · August 31st 2011, 11:22 AM

Hi all,

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

Basically, how do I get from

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

to

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

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.

**Siron** · August 31st 2011, 11:28 AM

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

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

**Srengam** · August 31st 2011, 11:40 AM

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

how to get from

$(2sinx)(-sinx)$

to

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

this is the only missing step now,

many thanks again.

**Siron** · August 31st 2011, 11:45 AM

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

Use $\sin^2(x)=1-\cos^2(x)$ so you get:
$=\frac{1}{2}-2(1-\cos^2(x))$
$=...$

**Srengam** · August 31st 2011, 11:48 AM

Thank you so much,

I really appreciate it.

**Plato** · August 31st 2011, 11:55 AM

Originally Posted by Srengam

$(1+2sinx)(\frac{1}{2}-sinx)$
to
$2cos(x)^{2}-\frac{3}{2}$

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

**Siron** · August 31st 2011, 12:07 PM

Originally Posted by Srengam

Thank you so much,

I really appreciate it.

You're welcome!

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