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Expert: - 10/2/2009


Question
The problem starts with a circle, like the unit circle, that has a point in the first qudrant on the circle labeled (5,8) and is called A. The problem is to translate point A by:

a: -theta + pi
b: pi + theta
c: (3/2)pi + theta

I have figured out the hypotenuse of the triangle that the point forms with the circle and axis in the circle is the square root of 89 but I don't know if this helps. I really just don't understand what the problem is asking me to do.

Answer

5-8
Questioner: Margaret
Country: United States
Category: Advanced Math
Private: No
Subject: trig
Question: The problem starts with a circle, like the unit circle, that has a point in the first qudrant on the circle labeled (5,8) and is called A. The problem is to translate point A by:

a: -theta + pi
b: pi + theta
c: (3/2)pi + theta

I have figured out the hypotenuse of the triangle that the point forms with the circle and axis in the circle is the square root of 89 but I don't know if this helps. I really just don't understand what the problem is asking me to do.
.................................................
Hi, Margie,

(Now don't get all upset because I called you that.)

I think the problem is asking where this point 'goes' when you change its 'standard angle'.

b. pi + theta.   
(Yes, I am doing b before a.)

This adds pi (exactly 180 deg), which is two quadrants, to theta.  That should put it into Quad 3.  Moving it 180 degrees 'reflects' it through the origin, changing the sign of both coordinates.  That gets you from (5,8) to (-5,-8).  That is your new point.

a. -theta + pi.

First you do a '-theta', then add pi.  Changing the sign of theta reflects the point across the x-axis.  It keeps its 'x', but flips the sign of 'y'.  You get from (5,8) to (5,-8).

Then you add pi.  We already know what that does -- flips both signs.  So you go from (5,-8) to (-5,8).  That is your new point.

c. (3/2)pi + theta.

This is pi/2 + pi + theta.

What does adding pi/2 (90deg) do?  It sends it into the next quadrant (Quad II), exchanges the x and y, and then flips one of the signs.

Exchange:  (5,8) becomes (8,5).
Flip a sign:  (8,5) becomes, er....  (-8,5) or (8,-5)?   Which one?  Oh, yes -- we are in quadrant II, where x is negative.  So it is (-8,5).  That is your new point.

............................................

Now you say you don't like this? You wanted to do some trig?  OK, we can do it without any pictures like this:

If A is (5,8), then OA = sqrt(89), as you found, which I shall call r.

So sin theta = 8/r,  and  cos theta = 5/r.  (From now on, t = theta)
And, if we need them,  r sin t = 8,  r cos t = 5   << VERY IMPORTANT

Now for any point  P, on the same circle, as you observed, if it is at an angle phi, then the coordinates are  x = r cos phi,   y = r sin phi.

So let's do b first, like I did before.

Sorry, I mean AS I did before.  


b. phi = pi + theta.

x = r cos phi
x = r cos (pi + t)
x = r (cos pi cos t - sin pi sin t)
x = r ( (-1) cos t  -    (0) sin t)
x = - r cos t
x = - 5

y = r sin phi
y = r sin (pi + t)
y = r (sin pi cos t + cos pi sin t)
y = r ( (0)   cos t +   (-1) sin t)
y = - r sin t
y = - 8

Our coordinates are P(-5,-8).

Now that you see the scheme, you should be able to do the others.  Keep in mind these facts, too:

cos (-t) = cos t
sin (-t) = - sin t

Trig is fun.

Paul Klarreich

Expertise

I can answer questions in basic to advanced algebra (theory of equations, complex numbers), precalculus (functions, graphs, exponential, logarithmic, and trigonometric functions and identities), basic probability, and finite mathematics, including mathematical induction. I can also try (but not guarantee) to answer questions on Abstract Algebra -- groups, rings, etc. and Analysis -- sequences, limits, continuity. I won't understand specialized engineering or business jargon.

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I taught at a two-year college for 25 years, including all subjects from algebra to third-semester calculus.

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