math trig Q

[113zami]

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when do you use the property set sin(-x)=-sinx, cos(-x)=cosx,...etc.and when do you not use them?

for example: I need to find the sin(-150)degrees and cos(-150)degrees

for cos(-150)my solution is:
1) since cos(-x)=cosx ,then cos(-150)=cos150
2) I used the refrence angle of -150 which is 30 and cos30 is sqrt3/2, but our angle is in the 3rd quad. where cos is (-) so it should be -sqrt3/2
3) so cos150= -sqrt3/2 which is the correct ansr

but when i do the exact same thing to sin(-150) i get the wrong ansr

my solution is:
1) since sin(-x)=-sinx, then sin(-150)=-sin150
2) used the refrence angle which is still the same, so sin30 is 1/2, but our angle is in the 3rd quad. where sin is (-) so it should be -1/2
3) so -sin150 becomes -(-1/2)= +1/2
but the ansr is -1/2 why???

thanks for clarifying

 

[CannonD]

when do you use the property set sin(-x)=-sinx, cos(-x)=cosx,...etc.and when do you not use them?

for example: I need to find the sin(-150)degrees and cos(-150)degrees

for cos(-150)my solution is:
1) since cos(-x)=cosx ,then cos(-150)=cos150
2) I used the refrence angle of -150 which is 30 and cos30 is sqrt3/2, but our angle is in the 3rd quad. where cos is (-) so it should be -sqrt3/2
3) so cos150= -sqrt3/2 which is the correct ansr

but when i do the exact same thing to sin(-150) i get the wrong ansr

my solution is:
1) since sin(-x)=-sinx, then sin(-150)=-sin150
2) used the refrence angle which is still the same, so sin30 is 1/2, but our angle is in the 3rd quad. where sin is (-) so it should be -1/2
3) so -sin150 becomes -(-1/2)= +1/2
but the ansr is -1/2 why???

thanks for clarifying

Hey,
Your property set is wrong ( sin(-x)=-sinX and cos(-x)=-cosX this depends where the angle is, 1st Quadran,2nd 3rd, or 4th)
The first question was cos(-150)=-Cos30 this is true because its relative angle is (30 degree), Cos(-150)=Cos(150) is true, I mean in this case because when both of its relative angles are 30 and also they are in II,III quadran that's why they both have (-) sign. and it's in its 3rd quar that's why u use (-) sign on front of Cos. Remember the Unit circle for Trig.
On the 2nd problem, so Sin(-150) =-Sin(30) NOT sin(-150)=-sin(150) this is NOt true. Well just remember that Sin=Y axis and Cos=X axis

well that's the best i could explain, i hope i won't confuse you, Your idea and Property you set are TOTALLY wrong, you should read and make sure you understand.

 

[Streetwolf]

Hey,
Your property set is wrong ( sin(-x)=-sinX and cos(-x)=-cosX this depends where the angle is, 1st Quadran,2nd 3rd, or 4th)
The first question was cos(-150)=-Cos30 this is true because its relative angle is (30 degree), Cos(-150)=Cos(150) is true, I mean in this case because when both of its relative angles are 30 and also they are in II,III quadran that's why they both have (-) sign. and it's in its 3rd quar that's why u use (-) sign on front of Cos. Remember the Unit circle for Trig.
On the 2nd problem, so Sin(-150) =-Sin(30) NOT sin(-150)=-sin(150) this is NOt true. Well just remember that Sin=Y axis and Cos=X axis

well that's the best i could explain, i hope i won't confuse you, Your idea and Property you set are TOTALLY wrong, you should read and make sure you understand.

No his property is correct. His mistake is that 150 is in the second quadrant. You got lucky on the x because it's negative in both the 2nd and 3rd quadrants. But y is positive in the 2nd quadrant. So you end up with +1/2 at first and then you negate it to get -1/2.

To answer your first question, Zami, the property always holds. Think of the coordinate graph and the unit circle. You know where 0pi ( = 0 degrees) is located, at point (1,0). So if you have sin(x) or sin(-x), the difference is that the first goes counter-clockwise and the second goes clockwise around the circle - by the same amount. So if you picture the x-axis being a giant mirror, it's as though you see the sin(x) as the object and the sin(-x) as the reflection.

So what's up with that property? Well for cos(x) and cos(-x), you are always on different sides of the x-axis. But they are always at the same x-coordinate (again, think of the giant mirror). So if cos(x) is located at some x-coordinate, so is cos(-x). Thus cos(x) = cos(-x).

For sin(x) and sin(-x), since they are on opposite sides of the x-axis, and since sin deals with the y-coordinate, they will have opposite signs. One will be in the positive y territory and the other in the negative. So sin(-x) = -sin(x).

 

[113zami]

Thank you VERY much 🙂