Trig - Exact Values, Max & Mins

K.ourt

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Joined
Jul 20, 2005
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17
The questions says algebraically solve for 1 + 2cosx = 5cosx, where x is between 0º and 360º, rounded to the nearest tenth. Then in brackets it states "first find exact value answers, then round".

So! I let w=cosx
1 + 2w = 5w
1=3w
1/3=w, sub w=cosx back in, and we get:
1/3=cosx

But I can't use one of my reference triangles to find exact value for x, so did I do my math wrong? Or am I just missing something?



And another says: Graph f(x)=(1-cos^2X)/(tanx), for x[-2pi,2pi,pi/2], label all minimums and maximums.

But when I graph it on my calculator, there are no minimums or maximums because the graph has asymptotes so it just keeps going, never touching the asymptote (it resembles the tan graph). Uhm..so did I punch it in wrong? I know that tanx=sinx/cosx, does that have something to do with it?
 
On the first:
Now that you have seen the w-substitution, you never need to do that again.

cos(x) = 1/3, No you won't find that 'x' on your reference chart. Just plug it in. Are you SURE you wrote it down correctly?

On the second:
Have you noticed that 1 - (cos(x))^2 = (sin(x))^2
 
Yes, Im sure I wrote it down right. But the thing is, there was an error on our assignment sheet. We had a sub. yesterday that doesnt know Trig. But out teacher fixed it today. Thank you though.

The bottom one is supposed to be (1-sin^2X)/tanx and we're supposed to solve it graphically.

But there are still no mins and maxes when I type it into my calculator, and the question asks for mins and maxes. There are asymptotes at: 0,pi,-pi,

What am I doing wrong?!
 
K.ourt said:
The bottom one is supposed to be (1-sin^2X)/tanx and we're supposed to solve it graphically.

But there are still no mins and maxes when I type it into my calculator, and the question asks for mins and maxes. There are asymptotes at: 0,pi,-pi,

What am I doing wrong?!
You're fine. When there are none of what it asks for, say, "There are none." If it is the correct response, write it. Why question?

(1-sin^2(x))/tan(x) = cos^2(x)/tan(x) = cos^3(x)/sin(x)

No min or max. You should be able to find a Point of Inflection, but that's a different story.
 
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