Can't use a calculator for these, not sure how you do it...
sin(16pi/3) = ?
sin^-1(sin(7pi/6)) = ?
Thanks!
sin(16pi/3) = ?
sin^-1(sin(7pi/6)) = ?
Thanks!
Some Trigonometry Questions
- Thread starter Justin218
- Start date
Let us see if I can reply before a million others jump on this.
Ever heard of a little thing called the Unit Circle? It is your friend in manually calculating trigonometric functions of common angles
. A full revolution is 2pi, or 6/3pi. Therefore, 16/3pi returns you to the angle with the same terminal side as 4/3pi. So if I remember correctly (it's just a memorization thing really), sin(4/3pi)= -(sqrt[3]/2)/1 which is about -.866. (Remember sin=opposite/hypotenuse).
As for the second question, sin^-1 and sin are inverse operations and therefore cancel each other out - the answer is 7pi/6.
Edit: Note that if you do the second operation on a calculator it will give you -pi/6, since the sin^-1 function only returns angles in Quadrants I and IV.
Ever heard of a little thing called the Unit Circle? It is your friend in manually calculating trigonometric functions of common angles
. A full revolution is 2pi, or 6/3pi. Therefore, 16/3pi returns you to the angle with the same terminal side as 4/3pi. So if I remember correctly (it's just a memorization thing really), sin(4/3pi)= -(sqrt[3]/2)/1 which is about -.866. (Remember sin=opposite/hypotenuse).As for the second question, sin^-1 and sin are inverse operations and therefore cancel each other out - the answer is 7pi/6
. Edit: Note that if you do the second operation on a calculator it will give you -pi/6, since the sin^-1 function only returns angles in Quadrants I and IV.
Memorize the 30-60-90 triangle.
sin(16pi/3) = sin(4pi + 4pi/3) = sin(4pi/3) = -sqrt(3)/2
arcsin( sin(7pi/6) ) = arcsin( -1/2 ) = -pi/6
The range of arcsin() is [-pi/2, pi/2]
sin(16pi/3) = sin(4pi + 4pi/3) = sin(4pi/3) = -sqrt(3)/2
arcsin( sin(7pi/6) ) = arcsin( -1/2 ) = -pi/6
The range of arcsin() is [-pi/2, pi/2]
Just do the Sin of 240 degrees and Sin^-1 of 210 degrees I did this last year in Trig, good stuff
- Kevin
- Kevin
I'm still pretty confused
"So if I remember correctly (it's just a memorization thing really), sin(4/3pi)= -(sqrt[3]/2)/1 which is about -.866. (Remember sin=opposite/hypotenuse)."
So, uh, how do I go about memorizing all these angles? I appreciate the help, thanks
"So if I remember correctly (it's just a memorization thing really), sin(4/3pi)= -(sqrt[3]/2)/1 which is about -.866. (Remember sin=opposite/hypotenuse)."
So, uh, how do I go about memorizing all these angles? I appreciate the help, thanks
re: memorizing the values of sine:
Most elementary trigonometry tests ultimately end up asking you for the sine of just five angles: 0, 30, 45, 60, and 90 degrees. The values of each the sine for each of these angles are the square roots of 0, 1, 2, 3 or 4 divided by 2.
sin(0) = sqrt(0)/2 = 0
sin(30) = sqrt(1)/2 = 1/2
sin(45) = sqrt(2)/2 = sqrt(2)/2
sin(60) = sqrt(3)/2 = sqrt(3)/2
sin(90) = sqrt(4)/2 = 1
Most elementary trigonometry tests ultimately end up asking you for the sine of just five angles: 0, 30, 45, 60, and 90 degrees. The values of each the sine for each of these angles are the square roots of 0, 1, 2, 3 or 4 divided by 2.
sin(0) = sqrt(0)/2 = 0
sin(30) = sqrt(1)/2 = 1/2
sin(45) = sqrt(2)/2 = sqrt(2)/2
sin(60) = sqrt(3)/2 = sqrt(3)/2
sin(90) = sqrt(4)/2 = 1
and...
0 = 0 pi
30 = 1/6 pi = pi/6
45 = 1/4 pi = pi/4
60 = 1/3 pi = pi/3
90 = 1/2 pi = pi/2
etc....
180 = pi, 360 = 2pi ... and you can keep that in mind to do all your conversions...
You can skip this message if it is more confusing than helpful -- it's hard to describe without a picture.
The way to think of 16pi/3 is that we're thinking of going around the unit circle that SynthDude2001 mentioned. Every 2*pi is one complete revolution around the circle, and we end up back at the starting point. So sin(16pi/3) = sin(2pi + 2pi + 4pi/3) = sin(4pi/3).
To relate 4pi/3 to one of the previously mentioned angles of 0, 30, 45, 60 and 90 degrees, think of pi as half way around the circle or 180 degrees. 4pi/3 = pi + pi/3, which means we go half way around the circle, and then 60 degrees more. The 60 degrees tells us that the answer is going to be some variation of sqrt(3)/2, but is it negative or positive?
To answer the question of if it is positive or negative, we need to think of sine as "opposite over hypotenuse", but what triangle are we talking about? If we go back to the unit circle, the hypotenuse is the line going from (0, 0) out to the point on the circle. The "opposite" side is the line parallel to the y axis, and the "adjacent" side is the x axis. When we go around the circle by 4pi/3, we end up in the lower half of the unit circle, which means that the "opposite" line goes down into the negative side of the y axis. This means that sin(4pi/3) will have a negative value. If our point on the unit circle were on the positive side of the y axis, our value for sine woudl be positive.
Combining the two pieces of information (that sine(60) = sqrt(3)/2 AND that sin(4pi/3) must be negative), we can finally respond that the answer is -sqrt(3)/2.
The way to think of 16pi/3 is that we're thinking of going around the unit circle that SynthDude2001 mentioned. Every 2*pi is one complete revolution around the circle, and we end up back at the starting point. So sin(16pi/3) = sin(2pi + 2pi + 4pi/3) = sin(4pi/3).
To relate 4pi/3 to one of the previously mentioned angles of 0, 30, 45, 60 and 90 degrees, think of pi as half way around the circle or 180 degrees. 4pi/3 = pi + pi/3, which means we go half way around the circle, and then 60 degrees more. The 60 degrees tells us that the answer is going to be some variation of sqrt(3)/2, but is it negative or positive?
To answer the question of if it is positive or negative, we need to think of sine as "opposite over hypotenuse", but what triangle are we talking about? If we go back to the unit circle, the hypotenuse is the line going from (0, 0) out to the point on the circle. The "opposite" side is the line parallel to the y axis, and the "adjacent" side is the x axis. When we go around the circle by 4pi/3, we end up in the lower half of the unit circle, which means that the "opposite" line goes down into the negative side of the y axis. This means that sin(4pi/3) will have a negative value. If our point on the unit circle were on the positive side of the y axis, our value for sine woudl be positive.
Combining the two pieces of information (that sine(60) = sqrt(3)/2 AND that sin(4pi/3) must be negative), we can finally respond that the answer is -sqrt(3)/2.
OT: You will probably have to memorize some double angle/half angle formulae soon (I am assuming this is a precalc class). Euler's identity will be very useful to you: e^ix=cos(x) + isin(x). Thus e^7ix=cos(7x)+isin(7x) which is also [e^(ix)]^7 = [cos(x)+isin(x)]^7. Then the real part is cox(7x) and the imaginary part sin(7x). If you get quick with the binomial theorem and use this, you have much less to memorize.
Thanks oog, I visualize what you're saying now. I just have a problem with actually applying what I see in my head into real numbers, which is because I didn't know what sin 30, 45, 60, etc was equal to. I'm going to have to memorize that whole chart of sin, cos, tan angles and what they are equal to.
well, 16pi/3 is the same as 5pi + 1/3pi...so that's 4pi + 4pi/3.
Sin(4pi) we can ignore...b/c that's basically travelling the unit circle twice. So, sin(4pi/3) = -sqrt(3)/2
And...7pi/6 is waaaay larger than one. Arcsin is limited to x values between -1 and 1...so arcsin(7pi/6) is undefined.
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