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trig. problem
trig. problem  
I've been having a problem with the trig. problem enclosed.  I'm not really sure even where to start.  Would you take a look at it and give see some advice or maybe explain it to me as to how to find the answers?.....thanks

Start by noting that for each point on the graph of y = cos x, you always have
P(x, cos x) as its coordinates.

So you have P(a, cos a) and Q(b, cos b)

Therefore, the line PQ has slope equal to
   y2 - y1
m = --------
   x2 - x1

   cos b - cos a
m = -------------
      b  -  a

Now your next step is apply it to  a = pi/4,  b = 4pi/3

I leave it to you to review your 'special triangle' work and find that:

cos a = sqrt(2)/2

cos b = - 1/2

Now you can get the slope:

   - 1/2 - (sqrt(2)/2)
m = -------------------
        4pi/3 - pi/2

Simplify a little (not so easy working with fractions, I know)

   - 1/2 - (sqrt(2)/2)   6
m = ------------------- ----
        4pi/3 - pi/2     6

   - 3 -  3sqrt(2)
m = -----------------
        8pi - 3pi

   - 3(1 -  sqrt(2))
m = -----------------

I'll leave it to you to do the decimal stuff, but:

The equation will use the 'point-slope' form:

y - y0 = m(x - x0)

You can take x0 = a; that's simpler

y - a = m (x - a)

y - sqrt(2)/2 = m (x - pi/4), where m is as above.

You can finish up now.

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Paul Klarreich


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.


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


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