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Can someone explain this trig formula

I was doing C4 solomon paper C Q6C and i have to integrate: sin^2 (2t)

And in the markscheme it says
sin^2 (2t) = 4sin^2 (t) cos^2 (t)

Can anyone explain where they got that from?
Original post by Economistician
I was doing C4 solomon paper C Q6C and i have to integrate: sin^2 (2t)

And in the markscheme it says
sin^2 (2t) = 4sin^2 (t) cos^2 (t)

Can anyone explain where they got that from?


Use the half angle formula for sin(2t).
Reply 2
Avatar for Zacken
Zacken
22
Original post by Economistician
I was doing C4 solomon paper C Q6C and i have to integrate: sin^2 (2t)

And in the markscheme it says
sin^2 (2t) = 4sin^2 (t) cos^2 (t)

Can anyone explain where they got that from?


You should know that sin⁡2t=2sin⁡tcos⁡t\sin 2t = 2 \sin t\cos t, so squaring both sides yields:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\sin^2 t = (2\sin t\cos t)^2 = 2^2 \sin^2 t \cos^2 t = \cdots \end{equation*}



Edit: ninja-ed.
Original post by Zacken
You should know that sin⁡2t=2sin⁡tcos⁡t\sin 2t = 2 \sin t\cos t, so squaring both sides yields:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\sin^2 t = (2\sin t\cos t)^2 = 2^2 \sin^2 t \cos^2 t = \cdots \end{equation*}



Edit: ninja-ed.


You've put more effort, more likely he'll understand what you've said.
Is it as simple as:
Sin2x = 2SinxCosx
Sin^2 (2x) = 4Sin^2(x)Cos^2(x) ?

Edit: Oh, ok thanks for your help. I dont know why that had me stumped for so long
(edited 7 years ago)
Reply 5
Avatar for Zacken
Zacken
22
Original post by Economistician
Is it as simple as:
Sin2x = 2SinxCosx
Sin^2 (2x) = 4Sin^2(x)Cos^2(x) ?

Edit: Oh, ok thanks for your help. I dont know why that had me stumped for so long


Yep.

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