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Maths Core 3 (C3)

Does Sin(theta/2) = sintheta,costheta?
-If not what does it equal?
(edited 6 years ago)
Original post by joyoustele
Does Sin(theta/2) = sintheta,costheta?


test using some value of theta to find out, for example when theta = pi.
How did you get this result btw?
(edited 6 years ago)
Original post by joyoustele
Does Sin(theta/2) = sintheta,costheta?
-If not what does it equal?


What do you mean by your latter sentence? Like what's a nice identity?

I guess you're using the compound angle formula sin(A+A)=2sinAcosA sin(A + A) = 2sinAcosA

Notice the angle of the argument on the RHS is half the angle on the argument on the LHS, so sin(a/2)=2sin(a/4)cos(a/4) sin(a/2) = 2sin(a/4)cos(a/4)

or if you want it in terms of another single trig function, start from
sin2(x/2)=1cos2(x/2)[br][br]sin(x/2)=1cos2(x/2)sin^2(x/2) = 1-cos^2(x/2) [br][br]sin(x/2) = \sqrt{1-cos^2(x/2) }

to get it in terms of tan, try starting from cos^2x + sin^2x =1 and divide through by sin^2x
Reply 3
Original post by NotNotBatman
What do you mean by your latter sentence? Like what's a nice identity?

I guess you're using the compound angle formula sin(A+A)=2sinAcosA sin(A + A) = 2sinAcosA

Notice the angle of the argument on the RHS is half the angle on the argument on the LHS, so sin(a/2)=2sin(a/4)cos(a/4) sin(a/2) = 2sin(a/4)cos(a/4)

or if you want it in terms of another single trig function, start from
sin2(x/2)=1cos2(x/2)[br][br]sin(x/2)=1cos2(x/2)sin^2(x/2) = 1-cos^2(x/2)[br][br]sin(x/2) = \sqrt{1-cos^2(x/2) }

to get it in terms of tan, try starting from cos^2x + sin^2x =1 and divide through by sin^2x


Thank you, I did workout that it was 2sin(a/4)cos(a/4) but thought it was wrong.Thanks now I know I was correct.
-I discovered these are called Half-angle formulas
Original post by joyoustele
Thank you, I did workout that it was 2sin(a/4)cos(a/4) but thought it was wrong.Thanks now I know I was correct.
-I discovered these are called Half-angle formulas


Careful, the half angle formula refers to sin a = 2sin(a/2)cos(a/2) and the cos(a/2) one.

but as long as you know the compound angle formula, the names are really unimportant.
Reply 5
Original post by NotNotBatman
Careful, the half angle formula refers to sin a = 2sin(a/2)cos(a/2) and the cos(a/2) one.

but as long as you know the compound angle formula, the names are really unimportant.


Oh:laugh:
Original post by NotNotBatman

or if you want it in terms of another single trig function, start from
sin2(x/2)=1cos2(x/2)[br][br]sin(x/2)=1cos2(x/2)sin^2(x/2) = 1-cos^2(x/2) [br] [br]sin(x/2) = \sqrt{1-cos^2(x/2) }


Few problems with this that I'm sure you can see?

Spoiler

(edited 6 years ago)
Original post by RDKGames
Few problems with this that I'm sure you can see?

Spoiler




oh yes, thanks.

SO op, the second one is not an identity.
Everything I said is C3 level...

Maybe not for the modulus definition but obviously square rooting both sides shouldn't seem as simple as it does at this level.
(edited 6 years ago)
Specifically, no. But if you were to use sin(x)1cos2(x)\sin(x) \equiv \sqrt{1-\cos^2(x)} as an identity in an exam question, most of the time you'd mess up. So its important to understand why it's not an identity on the natural domain.
I would say it is required. obviously not knowledge of 'if and only if' and the actual domain as he has defined it, but I think it's important to understand that the domain is restricted and it's not an identity as I mistakenly implied.

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