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Trigonometric Equations

Can anyone help me solve this equation sin(2 θ+50°)=0.5
Moved to the Maths section :smile:
Reply 2
Is there a range for the solutions? because there're an infinite number of solutions.

Do you typically use the cast diagram or the sin graph?

What have you tried so far?
Original post by cosv
Can anyone help me solve this equation sin(2 θ+50°)=0.5


2θ+30o=arcsin0.5=? 2\theta+30^o=\arcsin{0.5}=?
so 2θ=?2\theta=?
hence θ=?\theta=?
Reply 4
Original post by cosv
Can anyone help me solve this equation sin(2 θ+50°)=0.5


What are the boundaries for theta e.g. 0<θ<360°
Reply 5
Original post by mc1996
What are the boundaries for theta e.g. 0<θ<360°


Yes, the boundaies for theta are 0 to 360
Reply 6
Original post by cosv
Yes, the boundaies for theta are 0 to 360

2θ+50° = 150 boundaries change to 50°<2θ+50°<770
= 390
= 510
= 750
θ = 50
θ =170
θ =230
θ =350
Original post by mc1996
x


Don't give answers? :confused:

Forum policy.
(edited 10 years ago)
Reply 8
Original post by mc1996
2θ+50° = 150 boundaries change to 50°<2θ+50°<770
= 390
= 510
= 750
θ = 50
θ =170
θ =230
θ =350


Whoever is this ‘mc1996’? He has written something that is erroneous extremely erroneous at that. A true mathematician would understand that you cannot merely pick up a solution without writing in a logical and methodical manner. One of your solutions, ‘50’ has appeared from nowhere. You are supposed to write out the full solutions of u before getting the values of theta. Let u =2θ+50. u =150, 390, 510, 750. Therefore x =50, 155, 230, 350. What you have written here strongly suggests that you lack the required knowledge to study even GCSE Mathematics, let alone AS/A2. I suggest you go back to the basics before correcting people incorrectly.
Reply 9
Original post by natnalie
Whoever is this ‘mc1996’? He has written something that is erroneous extremely erroneous at that. A true mathematician would understand that you cannot merely pick up a solution without writing in a logical and methodical manner. One of your solutions, ‘50’ has appeared from nowhere. You are supposed to write out the full solutions of u before getting the values of theta. Let u =2θ+50. u =150, 390, 510, 750. Therefore x =50, 155, 230, 350. What you have written here strongly suggests that you lack the required knowledge to study even GCSE Mathematics, let alone AS/A2. I suggest you go back to the basics before correcting people incorrectly.


the harshness is strong in this one :eek:
Reply 10
Original post by mc1996
2θ+50° = 150 boundaries change to 50°<2θ+50°<770
= 390
= 510
= 750
θ = 50
θ =170
θ =230
θ =350


Can you please explain how you reached this answer? I get similar numbers for θ but I think I am missing a step.
Reply 11
Original post by cosv
Can you please explain how you reached this answer? I get similar numbers for θ but I think I am missing a step.


So the original questions was Sin(2θ+50)=0.5
next we do acrsin(0.5) to find 2θ+50. arcsin(0.5)=30
2θ+50 = 30 however the boundaries have changed to 50<2θ+50<770
30 does not appear within these boundaries. This means that we cannot use 30 to find a value of θ
However from the unit circle or sine graph, sin30 is the same as sin150, sin390, sin510 and sin750. All these values appear within our boundaries
Hence, 2θ+50 = 150, 390, 510, 750
making each one equal to 2θ+50, we can find a value of θ
Original post by mc1996
...


As has already been said

Please do not give full solutions and it is against the forum policy
Reply 13
Original post by TenOfThem
As has already been said

Please do not give full solutions and it is against the forum policy


my apologies
Reply 14
Original post by mc1996
my apologies


'mc1996' since you seem like a person who is quite confident on their mathematical ability, would you please help me with this question.

The curve C has equation xy = 1/2. The tangents to C at the distinct points P(p, 1/2p)
and Q(q, 1/2q), where p and q are positive, intersect at T and the normals to C at these points intersect at N. Show that T is the point ((2pq)/(p + q),1/(p + q)).
In the case pq = 1/2 , find the coordinates of N. Show (in this case) that T and N lie on the
line y = x and are such that the product of their distances from the origin is constant.

I understand how to attempt the first part but I'm finding it hard to complete the question. Any help on this question is appreciated as long as it isn't erroneous in any way! :smile:
(edited 10 years ago)
Original post by natnalie
Whoever is this ‘mc1996’? He has written something that is erroneous extremely erroneous at that. A true mathematician would understand that you cannot merely pick up a solution without writing in a logical and methodical manner. One of your solutions, ‘50’ has appeared from nowhere. You are supposed to write out the full solutions of u before getting the values of theta. Let u =2θ+50. u =150, 390, 510, 750. Therefore x =50, 155, 230, 350. What you have written here strongly suggests that you lack the required knowledge to study even GCSE Mathematics, let alone AS/A2. I suggest you go back to the basics before correcting people incorrectly.



Original post by natnalie
'mc1996' since you seem like a person who is quite confident on their mathematical ability, would you please help me with this question.

The curve C has equation xy = 1/2. The tangents to C at the distinct points P(p, 1/2p)
and Q(q, 1/2q), where p and q are positive, intersect at T and the normals to C at these points intersect at N. Show that T is the point ((2pq)/p + q,1/(p + q)).
In the case pq = 1/2 , find the coordinates of N. Show (in this case) that T and N lie on the
line y = x and are such that the product of their distances from the origin is constant.

I understand how to attempt the first part but I'm finding it hard to complete the question. Any help on this question is appreciated as long as it isn't erroneous in any way! :smile:


Hmmmmmmm quite a change of heart here
Reply 16
Original post by TenOfThem
Hmmmmmmm quite a change of heart here


Necessity makes strange bedfellows, even on TSR :biggrin:
Original post by davros
Necessity makes strange bedfellows, even on TSR :biggrin:


True

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