(i) Show that the equation
2 tan^2 θ sin^2 θ = 1 can be written in the form
2 sin^4 θ + sin^2 θ − 1 = 0.

Question

(i) Show that the equation 
2 tan^2 θ sin^2 θ = 1 can be written in the form 
2 sin^4 θ + sin^2 θ − 1 = 0.

lgbasallote · Answer

uhhh there's no over cos^2 on the bottom equation?

anonymous · Answer

Tan= Sin/Cos
Try messing with that.

ash2326 · Answer

We have
$$2 \tan^2 \theta\ \sin^2 \theta=1$$
We know that
$$\tan \theta= \frac{\sin \theta}{\cos \theta}$$
so
$$2 \sin ^2 \theta \sin^2 \theta= \cos ^2 \theta$$
Write
$$\cos^2 \theta= 1- \sin^2 \theta$$
I think you can do it from here:)

anonymous · Answer

@ash2326 Good job! So it the last part call the cos identity?

anonymous · Answer

Yes, I have it.

ash2326 · Answer

@Romero Thanks but I don't know what's it's called it's a standard identity!!

anonymous · Answer

I was just asking because I always encounter the half angle identity and I thought his had a specific name to it.

anonymous · Answer

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

anonymous · Answer

$$2\sin^4 \theta = 1-\cos^2 \theta$$

anonymous · Answer

$$2\sin^4 \theta = 1 -\sin^2 \theta$$

anonymous · Answer

which brings it back to it's origin

anonymous · Answer

I mean = cos^ theta. which gives 1- sin^2 theta

lgbasallote · Answer

uhmm why is it 1 - cos^2 theta? i think what you did was cross multiplication...i dunno

anonymous · Answer

I made a mistake there... I wrote cos^2 theta which gives 1-sin^2 instead again,. Hope that makes sense.

lgbasallote · Answer

oh i see good job :)