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Y=sec^2x at pi/4

  1. Nov 3, 2015 #1
    • Post was misplaced in a technical forum section, so is missing the homework template
    I need to solve y = sec^2 x - 2sin x, i'm fine with the 2sin x, but for the sec^2 x...

    I can't seem to understand how to do it... I can get it to...

    sin^2 x + cos^2 x / cos^2 x or 1 / cos^2x

    but I don't understand how to solve it when the square comes right after the cos/sin etc.
     
  2. jcsd
  3. Nov 4, 2015 #2

    Mark44

    Staff: Mentor

    What do you mean by "solving" this equation?
    To solve an equation that involves a single variable, you isolate the variable on one side of the equation, and everything else on the other side.
    When you solve an equation that involves two variables, you isolate the variable you're solving for on one side, with everything else on the other.


    ##\sec x = \frac{1}{\cos x}##, so ##\sec^2 x = \frac{1}{\cos^2 x}##. What is ##\cos(\pi/4)##?
    Your equation is already solved for y, and I don't think it's possible to solve for x in that equation.

    Edit: Ah, now I understand. You want to evaluate ##\sec^2 x - 2\sin x## at ##x = \pi/4##.
    This information should be in the body of the post, not just in the thread title.
     
    Last edited: Nov 4, 2015
  4. Nov 4, 2015 #3

    HallsofIvy

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    Now I am confused! abm7, do you want to solve [itex]y= sec^2(x)- 2sin(x)[/itex] for x, in terms of y, or do you want to evaluate it at [itex]x= \pi/4[/itex]?

    "Solving" the equation will be hard- it reduces to a cubic equation for sin(x).
     
  5. Nov 4, 2015 #4

    SammyS

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    You need to enclose the entire numerator in parentheses as follows/
    (sin^2 x + cos^2 x) / cos^2 x​

    I get the sense that you don';t understand what is meant by using the exponent with the trig functions such as
    sin2(x)​

    All it means is to take sin(x) and square it:
    sin2(x) = (sin(x))2
     
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