#11 (trig identities)

[scarlet willow]

i dont understand what they mean. isnt the equation already using sin?

[ashen pine]

It says

Sin theta

and cos theta

The equation is in sin 2(theta)

Try to get that "2" out of there.

You get it?

[scarlet willow]

then for #11 would getting 4sin(θ)cos^2(θ) - 1?

or would i still need to simplfy more

[ashen pine]

Wait what.

How'd you get that

[scarlet willow]

uh sin (3θ) = sin (2θ + θ)

then the addition sin thing

so sin (2θ) cos (θ) + sin(θ) cos(2θ)

and the sin(2θ) can be turned into 2 sin(θ)cos(θ)

no?

[ashen pine]

Yep

Then

[scarlet willow]

i turned cos(2θ) to 2cos^2(θ) -1

[ashen pine]

Yeah so

Then what does it become

[scarlet willow]

so then it was 2 sin(θ)cos(θ)cos(θ) + (2cos^2(θ) -1)sin(θ)

[ashen pine]

Yep

and not what you told earlier

Thats why I made you do all of it

[scarlet willow]

but wouldnt the 2 sin(θ)cos(θ)cos(θ) be simplified to 2 sin(θ)cos^2(θ)

[ashen pine]

Yeah you can

[scarlet willow]

and cant u distribute (2cos^2(θ) -1)sin(θ) so that its 2cos^2(θ)sin(θ) - sin(θ)

[ashen pine]

Yep?

Alternatively up there, you could use cos(2θ) = 1 - 2 sin^2 (θ)

[scarlet willow]

theres two 2cos^2(θ)sin(θ) so wouldnt adding them together make 4cos^2(θ)sin(θ)

o wait i think i did something wrong in my work

would it be correct if i got 4sin(θ)cos^2(θ) - sin(θ)

[ashen pine]

$\sin{3\theta} = \sin{(2\theta +\theta)}\
\sin{2\theta}\cos{\theta} + \cos{2\theta}\sin{\theta}\
2\sin\theta}\cos^{2}{\theta} + \left(2\cos^{2}{\theta} - 1\right)\sin{\theta}\
2\sin\theta}\cos^{2}{\theta} + 2\cos^{2}{\theta}\sin{\theta} - \sin{\theta}$

[ornate shuttle]

Quasar
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[scarlet willow]

is sin(θ)cos^2(θ) the same as cos^2(θ)sin(θ)?

[ashen pine]

You can use 4 sin theta cos ^2 theta

So uh

Well

[ashen pine]

Here.

See what you said earlier

[scarlet willow]

yea

[ashen pine]

Yu

Yup

[scarlet willow]

so then 4sin(θ)cos^2(θ) - sin(θ) would be the correct answer?

[ashen pine]

Yep

You could alternatively use

$\cos{2\theta} = 1 - 2\sin^{2}{\theta}$

[ornate shuttle]

Quasar

[ashen pine]

And get a cleaner answer

[scarlet willow]

oh ok