Trig help?

Question

If Theta is 5pi/6 why is its sine -1/2, cosine -sqrt3 /2, and tan sqrt3 /3?

What I don't understand is why I am not supposed to use the rules of 30 deg's instead of 60 deg's. Why not 60?

If I take 90 from 150 I get 60. Why does my math work tell me I should be using 30 deg???

Answer 1

sine of 5pi/6 should be -root 3/2 and cosine should be positive 1/2 and
tan should be neg root 3. You should use 60 as your reference angle.

Answer 2

If theta=5/6 = 60 degrees, then you certainly can use "60" side of the 30-60-90 triangle. The values you get should be:

sin(5/6) = sqrt(3)/2
cos(5/6) = 1/2
tan(5/6) = sqrt(3)

To be honest, I'm not sure where you're getting your answers of -1/2, -sqrt(3)/2 and sqrt(3)/3 from. The last one (sqrt(3)/3) never occurs on the standard angles (i.e. when the denominator for the radian measure is a power of 2 or three times a power of 2). The values of sine and cosine you give only occur when theta = 13/6 + 2k for some integer k.

Answer 3

Ok, let me show you a little thrick :

When I studied that, I used to transform rads to degrees, for example :

5pi / 6 = 5*180 / 6 = 150º

that's easy right, that angle is on the second cuadrant.

then :

you can do this : 150 = 180 - 30 = pi - 30

that's the thrick, because : sin180 = 0 and cos180 = -1

then :

sin(180 - 30)= sin180*cos30 - sin30*cos180 = 0 - (1/2)(-1)

tha result of sin(150) = 1/2, It's not -1/2, be careful, sinx is positive when x E [ 0 ; 180 ]

cos(180 - 30) = cos180*cos30 + sen30*sen180

-1*sqrt3/2 + 1/2*0 = -sqrt3/2

Yes, that's correct, because cosx is negative when x E [ 90 ; 180 ]

tan( 180 -30 ) = sin ( 180 -30 ) / cos (180 - 30 )

sin(150) = 1/2

cos(150) = -sqrt3/2

replacing them :

(1/2) / -sqrt3/2 = -sqrt3 / 3

Tanx is negative when x E [ 90 ; 180 ] BE CAREFUL WITH SIGNS.

that's why you need to use rules of 30 degrees, but you can also use 60 degrees, like this :

sin(150) = sin( 90 +60 )

sin90 = 1

cos90 = 0

sin( 90 + 60) = sin90*cos60 + sin60*cos90 = 1*1/2 + 0

1/2, then same result !!!!

cos( 90 + 60 ) = cos90*cos60 - sen90*sen60 = 0 - sqrt3/2

the same result !!.

Well, hope that helped you, remember :

sin( a - b) = sina*cosb - sinb*cosa

cos(a+b) = cosa*cosb - sin b*sina

sin( a + b) = sina*cosb + sinb*cosa

cos(a-b) = cosa*cosb + sinb*sina

Answer 4

5pi/6 is 150 degrees but instead of subtracting 90 you should subtract 150 from 180. this will give you a triangle in the second quadrant with 30 degrees as the base angle. the trig functions will work fine under that condition.

Answer 5

Well....... I thought I understood what you were asking. So I guess I'll just hit the whole thing.

The angle is measured from a 'zero reference' which is the positive side of the x-axis. If you subtract 90 from that, now you have a zero reference that's along the y-axis.

If you're wondering why subtract it from π, it's because the angle between the radius and the negative side of the x-axis is what you want to be using.

HTH ☺

Doug

Answer 6

Here's my answer from the first time you asked it:

your angle is 150deg. therefore, 180 - 150 = 30deg. However, the sin should be pos, cos neg and tan neg.

Answer 7

Good luck with that problem. I have no idea. Math is not by strong suit. I really hope that you find it out.

Answer 8

you sound a little confused, i think u need to revise radians my friend