# Solve using trig identities

• Oct 28th 2010, 06:15 PM
inspire
Solve using trig identities
$\displaystyle sin(3t)=\frac{\sqrt3}{2}$

First off, let me just say that I know how to solve it by taking the arcsin and then dividing by 3, that isn't what I need to do.

It needs to be like this:
$\displaystyle sin(2t+t)$

I know that:
$\displaystyle sin(2t)cos(t)+cos(2t)+sin(t)$
then:
$\displaystyle 2sin(t)cos(t)cos(t)+(1-2sin^2(t))+sin(t)$

$\displaystyle 2sin(t)cos^2(t)+1-2sin^3(t)$

Then I get stuck. I'm trying to solve for t.
• Oct 28th 2010, 06:24 PM
pickslides
Quote:

Originally Posted by inspire
[
It needs to be like this:
$\displaystyle sin(2t+t)$

I know that:
$\displaystyle sin(2t)cos(t)+cos(2t)+sin(t)$
then:
$\displaystyle 2sin(t)cos(t)cos(t)+(1-2sin^2(t))+sin(t)$

Do you mean? $\displaystyle \sin(2t)\cos(t)+\cos(2t)\sin(t)$

Then

$\displaystyle 2\sin(t)\cos^2(t)+(1-2\sin^2(t))\sin(t)$

$\displaystyle 2\sin(t)\cos^2(t)+\sin(t)-2\sin^3(t)$
• Oct 28th 2010, 06:27 PM
Prove It
Actually

$\displaystyle \sin{(2t + t)} = \sin{2t}\cos{t} + \cos{2t}\sin{t}$

$\displaystyle = 2\sin{t}\cos{t}\cos{t} + (1 - 2\sin^2{t})\sin{t}$

$\displaystyle = 2\sin{t}\cos^2{t} + \sin{t} - 2\sin^3{t}$

$\displaystyle = 2\sin{t}(1 - \sin^2{t}) + \sin{t} - 2\sin^3{t}$

$\displaystyle = 2\sin{t} - 2\sin^2{t} + \sin{t} - 2\sin^3{t}$

$\displaystyle = 3\sin{t} - 4\sin^3{t}$.

So $\displaystyle \displaystyle 3\sin{t} - 4\sin^3{t} = \frac{\sqrt{3}}{2}$

$\displaystyle \displaystyle 0 = 4\sin^3{t} - 3\sin{t} + \frac{\sqrt{3}}{2}$.

Now try to solve.
• Oct 28th 2010, 06:37 PM
inspire
Quote:

Originally Posted by pickslides
Do you mean? $\displaystyle \sin(2t)\cos(t)+\cos(2t)\sin(t)$

Then

$\displaystyle 2\sin(t)\cos^2(t)+(1-2\sin^2(t))\sin(t)$

$\displaystyle 2\sin(t)\cos^2(t)+\sin(t)-2\sin^3(t)$

Isn't the addition for $\displaystyle sin(\alpha+\beta)=sin\alpha cos\beta+cos\alpha sin\beta$?
• Oct 28th 2010, 06:45 PM
skeeter
yes, but that's not the form you used in your OP.

Quote:

Originally Posted by inspire
$\displaystyle sin(2t+t)$

I know that:
$\displaystyle sin(2t)cos(t)+cos(2t)+sin(t)$

• Oct 28th 2010, 06:59 PM
inspire
Quote:

Originally Posted by Prove It
Actually

$\displaystyle \sin{(2t + t)} = \sin{2t}\cos{t} + \cos{2t}\sin{t}$

$\displaystyle = 2\sin{t}\cos{t}\cos{t} + (1 - 2\sin^2{t})\sin{t}$

$\displaystyle = 2\sin{t}\cos^2{t} + \sin{t} - 2\sin^3{t}$

$\displaystyle = 2\sin{t}(1 - \sin^2{t}) + \sin{t} - 2\sin^3{t}$

$\displaystyle = 2\sin{t} - 2\sin^2{t} + \sin{t} - 2\sin^3{t}$

$\displaystyle = 3\sin{t} - 4\sin^3{t}$.

So $\displaystyle \displaystyle 3\sin{t} - 4\sin^3{t} = \frac{\sqrt{3}}{2}$

$\displaystyle \displaystyle 0 = 4\sin^3{t} - 3\sin{t} + \frac{\sqrt{3}}{2}$.

Now try to solve.

Am I missing something, or is this incredibly hard to factor?