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Math Help - Trig identities, exact values

  1. #1
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    Trig identities, exact values

    Hey, I was absent the day my class went over how to solve these problems. So I was wondering if I could get some help.

    -Use the fundamental identities to find the exact values of the remaining trig functions of x...

    1)cos x= square root 7/4 and cot x= - square root 7/3

    Would the sec x= 1/square root 7/4...? Or is that not the way to think of it?

    -Simplify each expression using the fundamental identities

    2) Sec theta X cos theta

    3) 1/csc^2x + 1/sec^2x

    4) 1-sin^2u/cosu

    5) cos^2x+ (sin x +1)^2/sin x + 1


    Thank you so much if you can help me. Also, is there a chat room or anything of the sort on this site, I'm new.


    -and yes, I'm horrible at math haha
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by zerostumbleine33 View Post
    1)cos x= square root 7/4 and cot x= - square root 7/3
    I presume these are
    cos(x) = \frac{\sqrt{7}}{4}
    and
    cot(x) = - \frac{\sqrt{7}}{3}

    We know that cosine is positive and cotangent is negative. Where does this happen? In Quadrant IV. Thus we know where the angle is.

    We want sin(x). Well, we know that sin^2(x) + cos^2(x) = 1, so we can get that easily enough. (Note that sine is negative in QIV.)

    Getting tan(x) is pretty easy:
    tan(x) = \frac{1}{cot(x)} = -\frac{3}{\sqrt{7}}

    Now the last two are also easy:
    csc(x) = \frac{1}{sin(x)}
    and
    sec(x) = \frac{1}{cos(x)}

    -Dan

    Edit: I just thought of another way to get sin(x):
    sin(x) = \frac{cos(x)}{cot(x)}
    This is a faster method, if you know how to deal with the complex fraction.
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  3. #3
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    In the first equation, the square root goes over the 7/4, same with the - being outside the square root over 7/3...so that would still make it in QIV, but what else would that affect?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by zerostumbleine33 View Post
    In the first equation, the square root goes over the 7/4, same with the - being outside the square root over 7/3...so that would still make it in QIV, but what else would that affect?
    Nothing really. You will have to do the problem over again using the correct values, but the method is still the same.

    -Dan
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