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# Core 3 Trig Question

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1. Hi, I'm not really sure where to go with this Proof question.

Cos(θ+π/3)+√3Sinθ = (θ+π/6) I would really appreciate any help.
2. (Original post by mathshelp1956)
Hi, I'm not really sure where to go with this Proof question.

Cos(θ+π/3)+√3Sinθ = (θ+π/6) I would really appreciate any help.
Use the addition formula to expand the cos(θ+π/3). you should get cos(θ)cos(π/3) - sin(θ)sin(π/3). Put the sin(π/3) and cos(π/3) into your calculator and you should be able to go from there.
3. (Original post by OrionMusicNet)
Use the addition formula to expand the cos(θ+π/3). you should get cos()cos(π/3) - sin(θ)sin(π/3). Put the sin(π/3) and cos(π/3) into your calculator and you should be able to go from there.
Hi, the after the equal line in my first question should say =Sin(θ+π/6). When I expand the bracket I get
(Cosθ)(1/2)-(Sinθ)(√3/2)+ √3Sinθ. I don't know to to get from this to the Sin(θ+π/6).
4. (Original post by mathshelp1956)
Hi, I'm not really sure where to go with this Proof question.

Cos(θ+π/3)+√3Sinθ = (θ+π/6) I would really appreciate any help.
Im not convince this equation holds true.
Subbing in pi for theta gives
Cos(4pi/3) + root3 sin(pi) = 7pi/6

Sin pi is 0. Cos 4pi/3 is -1/2. the right hand side is clearly bigger than pi.

Incidentally sin (7pi/6) is -1/2
5. It should say =Sin(θ+π/6) not just (θ+π/6). Sorry

(Original post by tavtavtav)
Im not convince this equation holds true.
Subbing in pi for theta gives
Cos(4pi/3) + root3 sin(pi) = 7pi/6

Sin pi is 0. Cos 4pi/3 is -1/2. the right hand side is clearly bigger than pi.

Incidentally sin (7pi/6) is -1/2
6. (Original post by mathshelp1956)
Hi, the after the equal line in my first question should say =Sin(θ+π/6). When I expand the bracket I get
(Cosθ)(1/2)-(Sinθ)(√3/2)+ √3Sinθ. I don't know to to get from this to the Sin(θ+π/6).
√3Sinθ -(Sinθ)(√3/2) as a fraction should end you up with (√3Sinθ)/2 if you know how to deal with algebraic fractions. You will want to write 1/2 in terms of sin. sin(pi/6) is equivalent to that so you write that. √3/2 can also be written in terms of cos as cos(pi/6). Can you see that with that you will have sin(pi/6)(Cosθ) + Sin(θ)cos(pi/6) which if you flip around is the same as Sin(θ+π/6)?
7. Thank you very much, by the way would this be considered a more difficult trig question?

(Original post by OrionMusicNet)
√3Sinθ -(Sinθ)(√3/2) as a fraction should end you up with (√3Sinθ)/2 if you know how to deal with algebraic fractions. You will want to write 1/2 in terms of sin. sin(pi/6) is equivalent to that so you write that. √3/2 can also be written in terms of cos as cos(pi/6). Can you see that with that you will have sin(pi/6)(Cosθ) + Sin(θ)cos(pi/6) which if you flip around is the same as Sin(θ+π/6)?
8. (Original post by mathshelp1956)
Thank you very much, by the way would this be considered a more difficult trig question?
Probably. I haven't seen a proof like that come up in a paper but it's worth knowing how to do in case something of a similar format comes up. Do bear in mind though that questions expecting you to use the addition formula and plugging exact trig values (e.g. sin(pi/6)) into your calculator to prove stuff. Just make sure you keep going and don't get spooked and stop.

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