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A level Trigonometry Help

Express 4 sin x + 3 cos x in the form r sin (x + α). Hence find all the values of x in the range 0 x 360° for which cos 3x = cos 2x.

I‘ve done the first part, which I got 5 sin (x + 36.9°), but I have no idea how to solve the second part.

Can anyone help please, thanks!

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Reply 1
Can you upload a picture of the question?
Reply 2
B31495CE-6624-4B6D-A1B1-29EE22521F01.jpg.jpeg
Reply 3
I think they want you to.expand the multiple angles in terms of cos(x) and sin(x).
However, id just take acos directly and forget about the first part.
(edited 3 years ago)
Reply 4
Original post by mqb2766
I think they want you to.expand the multiple angles in terms of cos(x) and sin(x).
However, id just take acos directly and forget about the first part.


I‘ve tried to expand it in terms of cosx and sinx but I can’t link back to the first equation.
Cos(A+_B) = CosAcosB + SinASinB
Reply 6
Original post by charlenecsn
I‘ve tried to expand it in terms of cosx and sinx but I can’t link back to the first equation.

Can you upload?
Reply 7
Original post by CaptainDuckie
Cos(A+_B) = CosAcosB + SinASinB


I got 4cos^3(x)-3cos(x)=2cos^2x-1, but I can't relate this to the first equation.
Upload pic bro
Reply 9
Original post by mqb2766
Can you upload?


E76D687B-45FD-460A-947D-4091C5E29E96.jpg.jpeg
Cannot see that at all😂😂
Original post by CaptainDuckie
Cannot see that at all😂😂


you mean the pic cant be viewed?😂
Original post by charlenecsn
you mean the pic cant be viewed?😂


Your writing
Original post by CaptainDuckie
Your writing


4E935300-A40E-40F6-B3AE-070EBD92F1C0.jpg.jpeg
Original post by charlenecsn
4E935300-A40E-40F6-B3AE-070EBD92F1C0.jpg.jpeg

Will have a look in the morning, but the simple thing to do is take acos
to give
3x = +/-2x + 2*pi*n
Original post by mqb2766
Will have a look in the morning, but the simple thing to do is take acos
to give
3x = +/-2x + 2*pi*n


Alright, thank you so much!!!
Reply 16
As a side note, another method is to use the trigonometric sum-product results, specifically:

cos(ab)cos(a+b)=2sin(a)sin(b) \cos (a-b) - \cos (a+b) = 2 \sin(a) \sin(b) .

For this case: cos(2x)cos(3x)=2sin((5/2)x)sin((3/2)x)=0 \cos (2x) - \cos (3x) = 2 \sin ( (5/2)x ) \sin( (3/2) x) = 0 .
Original post by simon0
As a side note, another method is to use the trigonometric sum-product results, specifically:

cos(ab)cos(a+b)=2sin(a)sin(b) \cos (a-b) - \cos (a+b) = 2 \sin(a) \sin(b) .

For this case: cos(2x)cos(3x)=2sin((5/2)x)sin((3/2)x)=0 \cos (2x) - \cos (3x) = 2 \sin ( (5/2)x ) \sin( (3/2) x) = 0 .


Thanks! ))but that formula isn't in the syllabus
Reply 18
Original post by CaptainDuckie
Cos(A+_B) = CosAcosB + SinASinB

Close but not quite.
Original post by simon0
Close but not quite.


It’s the Sin(A _B) = SinASinB - CosAcosB

( I realised )

But forgot to edit
(edited 3 years ago)

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