Trig Identity Questins

[radharcnacille]

I've for part (i) and (ii) but can't get my head around part (iii)
The solutions give:


Sin2θ = 2 - 2Cos2θ


Then


Sin2θ = 2Sin2θTanθ


Can't figure this step out


Any help would be greatly appreciated

 

[Deleted member 4993]

I've for part (i) and (ii) but can't get my head around part (iii)
The solutions give:


Sin2θ = 2 - 2Cos2θ


Then


Sin2θ = 2Sin2θTanθ


Can't figure this step out


Any help would be greatly appreciated

Sin2θ = 2 - 2Cos2θ → factor out 2 from right-hand-side

Sin2θ = 2 (1- Cos2θ) ....................(1)

In step (i), you have proven that:

(1-Cos2θ)/Sin2θ = tanΘ

multiplying both sides by sin2Θ we get:

1-Cos2θ = Sin2θ * tanΘ

Now continue......

Then

Sin2θ = 2Sin2θTanθ


Can't figure this step out

 

[radharcnacille]

That's excellent!

Thank you!