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Math Help - Verifying trigonometric identities?

  1. #1
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    Verifying trigonometric identities?

    Can anyone help me verify these two trigonometric identities

    (sinA+cosA)^2= 2+ secAcscA
    ___________
    secAcscA
    2/(secAcscA)+ (secAcscA/secAcscA)
    2/(secAcscA)+1
    2(1/sec+1/csc)+1
    2cosAsinA+1
    Unfortunately I am stuck and do not know how to proceed

    For my second problem
    sinx+cosx= (cosx/1-tanx)+(sinx/1-cot)
    (cosx/(1-sin/cos)+(sinx/(1-cosx/sinx)
    I am unsure how to proceed

    Can anyone please assist me.
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  2. #2
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    2.

    \displaystyle \frac{\cos{x}}{1 - \tan{x}} + \frac{\sin{x}}{1 - \cot{x}} = \frac{\cos{x}}{1 - \frac{\sin{x}}{\cos{x}}} + \frac{\sin{x}}{1 - \frac{\cos{x}}{\sin{x}}}

    \displaystyle = \frac{\cos{x}}{\frac{\cos{x} - \sin{x}}{\cos{x}}} + \frac{\sin{x}}{\frac{\sin{x} - \cos{x}}{\sin{x}}}

    \displaystyle = \frac{\cos^2{x}}{\cos{x} - \sin{x}} + \frac{\sin^2{x}}{\sin{x} - \cos{x}}

    \displaystyle = \frac{\cos^2{x}}{\cos{x} - \sin{x}} - \frac{\sin^2{x}}{\cos{x} - \sin{x}}

    \displaystyle = \frac{\cos^2{x} - \sin^2{x}}{\cos{x} - \sin{x}}

    \displaystyle = \frac{(\cos{x} - \sin{x})(\cos{x} + \sin{x})}{\cos{x} - \sin{x}}

    \displaystyle = \cos{x} + \sin{x}.
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  3. #3
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    Hello, homeylova223!

    (\sin A+\cos A)^2 \:=\:\dfrac{2+ \sec A\csc A}{\sec A\csc A}

    (\sin A + \cos A)^2 \;=\;\sin^2\!A + 2\sin A\cos A + \cos^2\!A

    . . . . . . . . . . . . =\; 2\sin A\cos A + \underbrace{\sin^2\!A + \cos^2\!A}}_{\text{This is 1}}
    . . . . . . . . . . . . =\;2\sin A\cos A + 1

    . . . . . . . . . . . . =\;\dfrac{2}{\sec A\csc A} + 1

    . . . . . . . . . . . . =\;\dfrac{2 + \sec A\csc A}{\sec A\csc A}

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  4. #4
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    Quick question on number 2 on the bottom part where you have (1-sinx/cosx) times cos x
    How do you get cosx-sinx/cosx I understand the rest of the problem I am just slightly confused on that part.
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  5. #5
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    Quote Originally Posted by homeylova223 View Post
    Quick question on number 2 on the bottom part where you have (1-sinx/cosx) times cos x
    How do you get cosx-sinx/cosx I understand the rest of the problem I am just slightly confused on that part.
    Sorry but without brackets in the correct spots I can't understand what you're writing...
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  6. #6
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    Like when you have
    (cosx/ 1-sinx/cosx) how do you get I

    (cosx/(cosx-sinx/cosx)

    I am slightly confused on that part
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  7. #7
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    Do you mean \frac{\cos(x)}{1-\frac{\sin(x)}{\cos(x)}} = \frac{\cos(x)}{\frac{\cos(x)-\sin(x)}{\cos(x)}}???

    Note that:

    1-\dfrac{\sin(x)}{\cos(x)}= \dfrac{\cos(x)}{\cos(x)}-\dfrac{\sin(x)}{\cos(x)}=\dfrac{\cos(x)-\sin(x)}{\cos(x)}

    Therefore, \dfrac{\cos(x)}{1-\dfrac{\sin(x)}{\cos(x)}} = \dfrac{\cos(x)}{\dfrac{\cos(x)-\sin(x)}{\cos(x)}}
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    Right now I have two more problems that are similar to the ones I just did can anyone help me with these?

    It is like this

    (1+tanx)/(sinx+cosx)= secx
    (secx)/(1)=secx
    (1/cosx/1)= secx
    secx=secx
    Did I do this right?

    My second problem is
    sinx+cosx= (2 sin^2x-1)/(sinx-cosx) I am not sure how to this one
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  9. #9
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    Quote Originally Posted by homeylova223 View Post
    Right now I have two more problems that are similar to the ones I just did can anyone help me with these?

    It is like this

    (1+tanx)/(sinx+cosx)= secx
    (secx)/(1)=secx ... no, 1+tanx is not secx, 1 + tan^2x = sec^2x. also sinx+cosx is not 1, it's sin^2x+cos^2x = 1
    (1/cosx/1)= secx
    secx=secx
    Did I do this right?

    My second problem is
    sinx+cosx= (2 sin^2x-1)/(sinx-cosx) I am not sure how to this one

    on the RHS, numerator ... 2sin^2x - 1 = 2sin^2x - (sin^2x + cos^2x) = sin^2x - cos^2x ... which will factor.
    ...
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  10. #10
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    So on
    (1+tanx)/(sinx+cosx)= secx

    Would I do it like this

    (1/sinx+cosx)+(sinx/cosx)/(sinx+cosx)= secx
    Then I am not sure what to do?
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  11. #11
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    Quote Originally Posted by homeylova223 View Post
    So on
    (1+tanx)/(sinx+cosx)= secx

    Would I do it like this

    (1/sinx+cosx)+(sinx/cosx)/(sinx+cosx)= secx
    Then I am not sure what to do?
    on the left side ...

    \dfrac{1+\tan{x}}{\sin{x}+\cos{x}} \cdot \dfrac{\cos{x}}{\cos{x}} = \dfrac{\cos{x}+\sin{x}}{(\sin{x}+\cos{x})\cos{x}}

    finish it
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