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Math Help - Need Help:Inverse Trigonmetry Ratios

  1. #1
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    Need Help:Inverse Trigonmetry Ratios

    How do I do these trig ratio's, and what formulas is the easiest, thanks..

    Find all the angles x that satisfy the following equations;0˚ ≤ X < 360˚ Show your work:

    1)sin(x)=0.4478.........................x=________ __and___________
    2)cot(x)=-2.112........................x=__________and______ _____
    3)csc(x)=5.2145........................x=_________ _and___________
    4)cos(x)=0.7884........................x=_________ _and___________
    5)sec(x)=-1.4557........................x=__________and_____ ______
    6)cos(x)=-0.2298........................x=__________and_____ ______
    7)sin(x)=-1.00 ........................x=__________and___________
    8)tan(x)=0.6678........................x=_________ _and___________
    9)sin(x)=0.0000........................x=_________ _and___________
    10)cot(x)=0.8897........................x=________ __and___________
    11)csc(x)=-2.0021........................x=__________and_____ ______
    12)sin(x)=-0.1124........................x=__________and_____ ______
    13)cos(x)=0.0000........................x=________ __and___________
    14)tan(x)=-3.000........................x=__________and______ _____
    15)cos(x)=1.000........................x=_________ _and___________
    16)sec(x)=2.4012........................x=________ __and___________
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  2. #2
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    How dare you!
    I am not a textbook. I do not supply answers, I explain them!
    ---
    I am not going to answer all questions.
    Rather tell you how to do them.
    You need to use the inverse-trigonometric functions.

    In the first example,
    \sin x = .4478
    Use the inverse trigonometric function on the calculator,
    \boxed{\boxed{ \sin^{-1}}}
    Therefore you get,
    x=26.60^o
    The other value is in the second quadrant,
    x=153.39^o.

    Note: when you reach a problem like,
    \sec x = 2.4012
    Is the same as,
    \cos x = .416
    Cuz, secant is reciprocal of cosine.
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  3. #3
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    thank you for your help but how do you get the second quadrant?
    and for the 3 question, how do you find csc(x)=5.2145. Ive tried <br />
\boxed{\boxed{ \sin^{-1}}}<br />
and i got error on my calculator
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  4. #4
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    Quote Originally Posted by bigstarz
    thank you for your help but how do you get the second quadrant?
    Use the following formula.
    Let x be in the first quadrant.
    Then, 180-x---> Second quadrant.
    Then, 180+x---> Third quadrant.
    Then, 360-x---> Fourth quadrant.


    It gets tricker when after you use the inverse function and get a negative number. If you cannot do those, first do the ones that do not give negative number and if you still do not understand how to work with negatives post your complaint here and someone will answer.
    ---
    Because you first need to find the reciprocal, that is,
    \frac{1}{5.2145}\approx .192
    Then you use the \boxed{\boxed{ \sin^{-1}}}math]
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  5. #5
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    Hello, bigstarz!

    I'll do one of each function . . . and hope it helps.


    Find all the angles x that satisfy the equations: . 0^o \leq x < 360^o

    1)\;\sin x \:=\:0.4478

    Enter: \boxed{\sin^{-1}}\;0.4478\;\boxed{=} . . . and get: 26.60262161\:\approx\:26.6^o

    \text{sine} is positive in Quadrants 1 and 2.
    . . Therefore: .  x\:=\:26.6,\;153.4^o



    4)\;\cos x\:=\:0.7884

    Enter: \boxed{\cos^{-1}}\;0.7884\;\boxed{=} . . . and get: 37.96376059 \:\approx\:38.0^o

    \text{cosine} is positive in Quadrants 1 and 4.
    . . Therefore: . x \:=\:38.0^o,\;322.0^o


    8)\;\tan(x)\:=\:0.6678

    Enter: \boxed{\tan^{-1}}\;0.6678\;\boxed{=} . . . and get: 33.73499916 \:\approx\:33.7^o

    \text{tangent} is positive in Quadrants 1 and 3.
    . . Therefore: . x \:=\:33.7^o,\;213.7^o



    10)\;\cot x \:=\:0.8897

    Since \cot x \:=\:\frac{1}{\tan x}, enter: 0.8897\;\boxed{x^{-1}} . . . and get: 1.123974373

    Then enter: \boxed{\tan^{-1}}\;\boxed{\text{ans}}\;\boxed{=} . . . and get: 48.3405102\:\approx\:48.3^o

    \text{cotangent} is positive in Quadrants 1 and 3.
    . . Therefore: . x \:=\:48.3^o,\;228.3^o



    (5)\;\sec x \:= \:-1.4557

    Since \sec x \:= \:\frac{1}{\cos x}, enter:  -1.4557\;\boxed{x^{-1}} . . . and get: -0.68695473

    Then enter: \boxed{\tan^{-1}}\;\boxed{\text{ans}}\;\boxed{=} . . . and get: 133.3895302\:\approx\:133.4^o

    \text{secant} is negative in Quadrants 2 and 3.
    . . Therefore: . x \:=\:133.4^o,\;226.6^o



    3)\;\csc x \:= \:5.2145

    Since \csc x \:=\:\frac{1}{\sin x}, enter: 5.2145\;\boxed{x^{-1}} . . . and get: 0.191772941

    Then enter: \boxed{\sin^{-1}}\;\boxed{\text{ans}}\;\boxed{=} . . . and get: 11.05626911\:\approx\:11.1^o

    \text{cosecant} is positive in Quadrants 1 and 2.
    . . Therefore: . x \:= \:11.1^o,\;168.9^o

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