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Math Help - help correcting my trig equations and proving one

  1. #1
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    help correcting my trig equations and proving one

    hi, i have this HW for my class assessment mark so i really want to get it all perfect, its some tedious work, but help is greatly appreciated.Thanks in advance. pls help me check if i made any mistakes in my calculations

    1) solve 5sinX - 2cos^2X - 1 = 0 range(for 0 <= x <= 2pie) <= is great equal sign

    i worked out and got sinX = 1/2 and sinX = -3

    for sin = 1/2 converting to degress(using calculator inverse sin) is 30,
    using quadrant diagram x1 = 30 degrees
    x2 = 180-30 = 150
    sinx = -3 is undefined? though i don't know if i did any mistake in my previous calculations

    btw if the range is given in terms of pie, does the teacher expects the answer to be given in terms of pie?, i can't figure out what 150 is in terms of pie


    2) sec^2X - 4tanX = 0 for (0<= x <= 180)

    using identities sec^2X = tan^2X +1 thus
    tan^2 X - 4tan X +1 = 0
    since it can't be factored mentally , i used the quadratic equation to factor

    got tan X = 3.732

    using calculator inverse tan and quadrant diagram got
    X1 = 75
    x2 = 180 + 75 = 255
    (all in terms of degees)

    next tanX = 0.268
    using above methods got X1= 15.5
    X2 = 180 - 15.5 = 164.5

    3) 3sinX = 5sinXcosX (for -90 <x<90)
    completely lost, tried diveding everything by cosX, didn't work
    and where do i start of finish in the quadrant diagram if its -90

    4) 3 sinX = 5sinXcosX (for -90<x<90)

    got 3/5 = cosx
    inverse of cos 3/5 = 53.1(degrees)

    stuck on the quadrant diagram, becasue of the range.

    help is greatly appreciated, i tried, just got stuck on some, and its a very important piece of HW
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  2. #2
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    last one,how do you prove this ?

    (cotX + cosecX)^2 = 1+cosX/1-cosX

    some people are just born to fail maths
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  3. #3
    A riddle wrapped in an enigma
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    Quote Originally Posted by llkkjj24 View Post
    last one,how do you prove this ?

    (cotX + cosecX)^2 = 1+cosX/1-cosX

    some people are just born to fail maths
    Hi llkkjj24,

    I'll jump on this one. Working with left side, only:

    (\cot x+\csc x)^2=\frac{1+\cos x}{1-\cos x}

    \left(\frac{\cos x}{\sin x}+\frac{1}{\sin x}\right)^2=

    \frac{\cos^2 x}{\sin^2 x}+\frac{2\cos x}{\sin^2 x}+\frac{1}{\sin^2 x}=

    \frac{\cos^2 x+2\cos x+1}{\sin^2 x}=

    \frac{(\cos x+1)(\cos x+1)}{1-\cos^2 x}=

    \frac{(\cos x+1)(\cos x+1)}{(1-\cos x)(1+\cos x)}=

    {\color{red}\frac{1+\cos x}{1-\cos x}}
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  4. #4
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    Quote Originally Posted by llkkjj24 View Post
    hi, i have this HW for my class assessment mark so i really want to get it all perfect, its some tedious work, but help is greatly appreciated.Thanks in advance. pls help me check if i made any mistakes in my calculations

    1) solve 5sinX - 2cos^2X - 1 = 0 range(for 0 <= x <= 2pie) <= is great equal sign

    i worked out and got sinX = 1/2 and sinX = -3

    for sin = 1/2 converting to degress(using calculator inverse sin) is 30,
    using quadrant diagram x1 = 30 degrees
    x2 = 180-30 = 150
    sinx = -3 is undefined? though i don't know if i did any mistake in my previous calculations

    btw if the range is given in terms of pie, does the teacher expects the answer to be given in terms of pie?, i can't figure out what 150 is in terms of pie
    30^{\circ}=\frac{30}{1}\cdot \frac{\pi}{180}=\frac{\pi}{6} radians.

    150^{\circ}=\frac{150}{1}\cdot \frac{\pi}{180}=\frac{5\pi}{6} radians.
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  5. #5
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    Quote Originally Posted by llkkjj24 View Post
    3) 3sinX = 5sinXcosX (for -90 <x<90)
    completely lost, tried diveding everything by cosX, didn't work
    and where do i start of finish in the quadrant diagram if its -90

    4) 3 sinX = 5sinXcosX (for -90<x<90)

    got 3/5 = cosx
    inverse of cos 3/5 = 53.1(degrees)

    stuck on the quadrant diagram, becasue of the range.

    help is greatly appreciated, i tried, just got stuck on some, and its a very important piece of HW
    3) and 4) appear to be the same problem.

    -53.1 works and is within the range.
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  6. #6
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    thank you very much for the help
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  7. #7
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    Hello, llkkjj24!

    Another approach . . .


    Prove: . (\cot x + \csc x)^2 \:=\:\frac{1+\cos x}{1-\cos x}

    Multiply the right side by \frac{1+\cos x}{1 +\cos x}


    . . \frac{1+\cos x}{1-\cos x}\cdot{\color{blue}\frac{1+\cos x}{1 + \cos x}} \;=\;\frac{(1+\cos x)^2}{1-\cos^2\!x} \;=\;\frac{(1+\cos x)^2}{\sin^2\!x}


    . . . = \;\left(\frac{1+\cos x}{\sin x}\right)^2  \;= \;\left(\frac{1}{\sin x} + \frac{\cos x}{\sin x}\right)^2 \;=\;(\csc x + \cot x)^2

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