C3 trig help Urgent

Solve for 0< θ<360

Cos(θ-30) = 2Sinθ

Im stuck please help, i tried expanding the brackets and solving but it seems wrong

Original post by Rockyboy

Solve for 0< θ<360

Cos(θ-30) = 2Sinθ

Im stuck please help, i tried expanding the brackets and solving but it seems wrong

How so?

\cos(\theta - 30)=\cos(\theta)\cos(30)+\sin( \theta)\sin(30)

Then you can get

\tan(\theta)

out of it.

Reply 2

black1blade

19

Dividing by an expression with just sin and cos to get tan is a handy trick like knowing lne=1. I forgot it was best to divide by cos when I was doing a past paper and did the whole Rsin(x+a) thing when you didn't have to.

Reply 3

Rockyboy

OP

11

Original post by RDKGames

How so?

\cos(\theta - 30)=\cos(\theta)\cos(30)+\sin( \theta)\sin(30)

Then you can get

\tan(\theta)

out of it.

Yes but what to i do after. Cuz i got

1+tan(θ)tan(30) =2sin( θ)

Original post by Rockyboy

Yes but what to i do after. Cuz i got

1+tan(θ)tan(30) =2sin( θ)

That's wrong.

\cos(\theta)\cos(30)+\sin( \theta)\sin(30) = 2\sin(\theta)

So

\cos(\theta)\cos(30) = \sin( \theta)[2-\sin(30)]

Thus

\tan(\theta)=\frac{\cos(30)}{2-\sin(30) }

Reply 5

ForestShadow

22

Original post by black1blade

Dividing by an expression with just sin and cos to get tan is a handy trick like knowing lne=1. I forgot it was best to divide by cos when I was doing a past paper and did the whole Rsin(x+a) thing when you didn't have to.

dont you lose a solution if you divide by sin/cos or does that not apply to this question?

Reply 6

Rockyboy

OP

11

Original post by RDKGames

That's wrong.

\cos(\theta)\cos(30)+\sin( \theta)\sin(30) = 2\sin(\theta)

So

\cos(\theta)\cos(30) = \sin( \theta)[2-\sin(30)]

Thus

\tan(\theta)=\frac{\cos(30)}{2-\sin(30) }

I got it thx

(edited 6 years ago)

Reply 7

Rockyboy

OP

11

Never mind im being stupid. i get it now thank you very much

Reply 8

black1blade

19

Original post by ForestShadow

dont you lose a solution if you divide by sin/cos or does that not apply to this question?

Say if you had sinx+cosx=0 then the quickest approach is to divide by cosx. However yeah if possible you should always fractorise so if it was cos^2(x)+sinxcosx=0 you would factorise first to get cosx(cosx+sinx)=0 then say cosx=0 and cosx+sinx=0.

Reply 9

ForestShadow

22

Original post by black1blade

Say if you had sinx+cosx=0 then the quickest approach is to divide by cosx. However yeah if possible you should always fractorise so if it was cos^2(x)+sinxcosx=0 you would factorise first to get cosx(cosx+sinx)=0 then say cosx=0 and cosx+sinx=0.

ok thanks! :h:

C3 trig help Urgent

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