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Math Help - Complicated Trigo Identity

  1. #1
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    Complicated Trigo Identity

    Hi Guys,

    Am Stuck with this few trigo identtes for days !

    1) Prove Identity 1-[ (sinx)^6 + (cosx)^6 ] = 3/4(sin2x)^2

    2) Prove Identity (sin2x)^2 [ (cotx)^2 - (tanx)^2 ] = 4cos2x

    3) If cosec2x + cot2x = cotx , prove that cosec2x + cosec4x + cosec8x = cotx -cot8x

    4) Prove that cosx + cos(x+120) + cos (x+240) = 0

    5) Prove that sinx[sin(x+120)] + sinx[sin(x-120)] + [sin(x+120)sin(x-120)] = -3/4

    6) Prove that [sin(x-120)]^2 + (sinx)^2 + [sin(x+120)]^2 = 3/2

    Please help guys thanks alot !
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  2. #2
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    Quote Originally Posted by liukawa View Post
    1) Prove Identity 1-[ (sinx)^6 + (cosx)^6 ] = 3/4(sin2x)^2
    Suppose a = \sin{x} and b = \cos{x}, then we have:

    a^6+b^6 = (a^2+b^2) (a^4+b^4-a^2 b^2) = (a^2+b^2) [(a^2+b^2)^2-2a^2b^2-a^2 b^2]

    .............  = (a^2+b^2) [(a^2+b^2)^2-3a^2 b^2] = 1-3a^2 b^2 = 1-3\left(ab\right)^2.

    So 1-\left(\sin^6{x}+\cos^6{x}\right) = 1-\left[1-3\left(\sin{x}\cos{x}\right)^2\right]= 3\left(\frac{1}{2}\sin{2x}\right)^2 = \frac{3}{4}\sin^2{2x}.
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  3. #3
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    Hello, liukawa!

    Here's the first one . . .


    \text{1) Prove: }\;1-\left(\sin^6\!x + \cos^6\!x\right) \;=\; \frac{3}{4}\sin^2\!2x

    \text{Factor: }\;1 - \underbrace{(\sin^2\!x + \cos^2\!x)}_{\text{This is 1}}(\sin^4\!x - \sin^2\!x\cos^2\!x + \cos^4\!x)

    . . . . . . . . =\;1 - \bigg(\sin^4\!x - \sin^2\!x\cos^2\!x + \cos^4\!x\bigg)


    Add and subtract 3\sin^2\!x\cos^2\!x\!:

    . . 1 - \bigg(\sin^4\!x + 2\sin^2\!x\cos^2\!x + \cos^4\!x - 3\sin^2\!x\cos^2\!x\bigg)

    . . =\; 1 - \bigg(\left[\sin^2\!x + \cos^2\!x\right]^2 - 3\sin^2\!x\cos^2\!x\bigg)

    . . =\;1 - \left(1 - 3\sin^2\!x\cos^2\!x\right)

    . . =\;1 - 1 + 3\sin^2\!x\cos^2\!x

    . . =\;3\sin^2\!x\cos^2\!x

    . . =\;\frac{3}{4}\cdot 4\sin^2\!x\cos^2\!x

    . . =\;\frac{3}{4}(2\sin x\cos x)^2

    . . =\;\frac{3}{4}\sin^2\!2x



    Edit: Ah, TheCoffeeMachine beat me to it . . .
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  4. #4
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    Thanks guys !

    You guys are genius !

    Do help out with the rest of the identities also if possible !

    thanks alot in advance !
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  5. #5
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    Quote Originally Posted by liukawa View Post
    Hi Guys,

    Am Stuck with this few trigo identtes for days !

    1) Prove Identity 1-[ (sinx)^6 + (cosx)^6 ] = 3/4(sin2x)^2

    2) Prove Identity (sin2x)^2 [ (cotx)^2 - (tanx)^2 ] = 4cos2x

    3) If cosec2x + cot2x = cotx , prove that cosec2x + cosec4x + cosec8x = cotx -cot8x

    4) Prove that cosx + cos(x+120) + cos (x+240) = 0

    5) Prove that sinx[sin(x+120)] + sinx[sin(x-120)] + [sin(x+120)sin(x-120)] = -3/4

    6) Prove that [sin(x-120)]^2 + (sinx)^2 + [sin(x+120)]^2 = 3/2

    Please help guys thanks alot !

    1)

    1-\left[sin^6x+\left(1-sin^2x\right)^3\right]=1-\left[\left(1-2sin^2x+sin^4x\right)\left(1-sin^2x\right)+sin^6x\right]

    =1-\left[1-sin^2x-2sin^2x+2sin^4x+sin^4x-sin^6x+sin^6x\right]=-3sin^4x+3sin^2x

    =3sin^2x\left(1-sin^2x\right)=3\left(sinxcosx\right)^2=3\left(\fra  c{sin2x}{2}\right)^2


    2)

    cos2x=cos^2x-sin^2x

    sin2x=2sinxcosx

    \left(sin2x\right)^2\left[\frac{cos^2x}{sin^2x}-\frac{sin^2x}{cos^2x}\right]=4sin^2xcos^2x\left[\frac{cos^2x}{sin^2x}-\frac{sin^2x}{cos^2x}\right]=4\left[cos^4x-sin^4x\right]

    =4\left[cos^2x-sin^2x\right]\left[cos^2x+sin^2x\right]
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  6. #6
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    hi guys.

    please help with the other identities if possible thanks !
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  7. #7
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    3) Oh well...

    cosec 2x + \cot 2x = \cot x

    Then, we also have:

    cosec 4x + \cot 4x = \cot 2x

    cosec 8x + \cot 8x = \cot 4x

    Combining those three, we get:

    cosec 2x + cosec 4x + cosec 8x + \cot 2x  + \cot 4x + \cot 8x = \cot x + \cot 2x + \cot 4x

    Can you simplify this?
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  8. #8
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    Hello, liukawa!

    There's always some fool who will do all these problems for you.

    Okay, it's me . . .


    \text{4) Prove: }\:\cos x + \cos(x+120) + \cos (x+240) \:=\: 0

    \cos x + \bigg[\cos x\cos120 - \sin x\sin120\bigg] + \bigg[\cos x\cos240 - \sin x\sin240\bigg]

    . . =\;\cos x + \bigg[\cos x\left(\text{-}\frac{1}{2}\right) - \sin x\left(\frac{\sqrt{3}}{2}\right)\bigg] + \bigg[\cos x\left(\text{-}\frac{1}{2}\right) - \sin x\left(\text{-}\frac{\sqrt{3}}{2}\right)\bigg]

    . . =\;\cos x - \frac{1}{2}\cos x - \frac{\sqrt{3}}{2}\sin x - \frac{1}{2}\cos x + \frac{\sqrt{3}}{2}\sin x

    . . =\quad 0




    \text{5) Prove that:}
    . . \sin x[\sin(x\!+\!120)] + \sin x[\sin(x\!-\!120)] + [\sin(x\!+\!120)\sin(x\!-\!120)] \:=\: -\frac{3}{4}

    \sin x\bigg[\sin x\cos120 + \cos x\sin120\bigg] + \sin x\bigg[\sin x\cos120 - \cos x\sin120 \bigg]

    . . . . . . + \bigg[\sin x\cos120 - \cos x\sin120\bigg]\,\bigg[\sin x\cos120 + \cos x\sin120\bigg]


    . . =\;\sin x\bigg[\sin x\left(\text{-}\frac{1}{2}\right) + \cos x\left(\frac{\sqrt{3}}{2}\right)\bigg] + \sin x\bigg[\sin x\left(\text{-}\frac{1}{2}\right) - \cos x\left(\frac{\sqrt{3}}{2}\right)\bigg]

    . . . . . . + \bigg[\sin x\left(\text{-}\frac{1}{2}\right) + \cos x\left(\frac{\sqrt{3}}{2}\right)\bigg]\,\bigg[\sin x\left(\text{-}\frac{1}{2}\right) - \cos x\left(\frac{\sqrt{3}}{2}\right)\bigg]


    . . =\;\text{-}\frac{1}{2}\sin^2\!x + \frac{\sqrt{3}}{2}\sin x\cos x - \frac{1}{2}\sin^2\!x - \frac{\sqrt{3}}{2}\sin x\cos x + \frac{1}{4}\sin^2x - \frac{3}{4}\cos^2\!x


    . . =\;\text{-}\frac{3}{4}\sin^2\!x - \frac{3}{4}\cos^2\!x \;\;=\;\;-\frac{3}{4}\underbrace{(\sin^2\!x + \cos^2\!x)}_{\text{This is 1}} \;\;=\;\;-\frac{3}{4}




    \text{6) Prove: }\:\sin^2(x-120) + \sin^2\!x+ \sin^2(x+120) \:=\: \frac{3}{2}

    \bigg[\sin x\cos120 - \cos x\sin120\bigg]^2 + \sin^2\!x + \bigg[\sin x\cos120 + \cos x\sin120\bigg]^2


    . . =\;\sin^2\1x\cos^2120 - 2\sin x\cos x\sin120\cos120 + \cos^2\!x\sin^2\!120 + \sin^2\!x

    . . . . . . + \sin^2\!x\cos^2\!120 + 2\sin x\cos x\sin120\cos120 + \cos^2\!x\sin^2\!120


    . . =\;2\sin^2\!x\cos^2\!120 + 2\cos^2\!x\sin^2\!120 + \sin^2\!x


    . . =\;2\sin^2\!x\left(\text{-}\frac{1}{2}\right)^2 + 2\cos^2\!x\left(\frac{\sqrt{3}}{2}\right)^2 + \sin^2\1x


    . . =\;\frac{1}{2}\sin^2\!x + \frac{3}{2}\cos^2\!x + \sin^2\!x \;=\;\frac{3}{2}\sin^2\!x + \frac{3}{2}\cos^2\!x


    . . =\;\frac{3}{2}\left(\sin^2\!x + \cos^2\1x\right) \;=\;\frac{3}{2}

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  9. #9
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    Quote Originally Posted by Soroban View Post
    Hello, liukawa!

    There's always some fool who will do all these problems for you.

    Okay, it's me . . .



    \cos x + \bigg[\cos x\cos120 - \sin x\sin120\bigg] + \bigg[\cos x\cos240 - \sin x\sin240\bigg]

    . . =\;\cos x + \bigg[\cos x\left(\text{-}\frac{1}{2}\right) - \sin x\left(\frac{\sqrt{3}}{2}\right)\bigg] + \bigg[\cos x\left(\text{-}\frac{1}{2}\right) - \sin x\left(\text{-}\frac{\sqrt{3}}{2}\right)\bigg]

    . . =\;\cos x - \frac{1}{2}\cos x - \frac{\sqrt{3}}{2}\sin x - \frac{1}{2}\cos x + \frac{\sqrt{3}}{2}\sin x

    . . =\quad 0





    \sin x\bigg[\sin x\cos120 + \cos x\sin120\bigg] + \sin x\bigg[\sin x\cos120 - \cos x\sin120 \bigg]

    . . . . . . + \bigg[\sin x\cos120 - \cos x\sin120\bigg]\,\bigg[\sin x\cos120 + \cos x\sin120\bigg]


    . . =\;\sin x\bigg[\sin x\left(\text{-}\frac{1}{2}\right) + \cos x\left(\frac{\sqrt{3}}{2}\right)\bigg] + \sin x\bigg[\sin x\left(\text{-}\frac{1}{2}\right) - \cos x\left(\frac{\sqrt{3}}{2}\right)\bigg]

    . . . . . . + \bigg[\sin x\left(\text{-}\frac{1}{2}\right) + \cos x\left(\frac{\sqrt{3}}{2}\right)\bigg]\,\bigg[\sin x\left(\text{-}\frac{1}{2}\right) - \cos x\left(\frac{\sqrt{3}}{2}\right)\bigg]


    . . =\;\text{-}\frac{1}{2}\sin^2\!x + \frac{\sqrt{3}}{2}\sin x\cos x - \frac{1}{2}\sin^2\!x - \frac{\sqrt{3}}{2}\sin x\cos x + \frac{1}{4}\sin^2x - \frac{3}{4}\cos^2\!x


    . . =\;\text{-}\frac{3}{4}\sin^2\!x - \frac{3}{4}\cos^2\!x \;\;=\;\;-\frac{3}{4}\underbrace{(\sin^2\!x + \cos^2\!x)}_{\text{This is 1}} \;\;=\;\;-\frac{3}{4}





    \bigg[\sin x\cos120 - \cos x\sin120\bigg]^2 + \sin^2\!x + \bigg[\sin x\cos120 + \cos x\sin120\bigg]^2


    . . =\;\sin^2\1x\cos^2120 - 2\sin x\cos x\sin120\cos120 + \cos^2\!x\sin^2\!120 + \sin^2\!x

    . . . . . . + \sin^2\!x\cos^2\!120 + 2\sin x\cos x\sin120\cos120 + \cos^2\!x\sin^2\!120


    . . =\;2\sin^2\!x\cos^2\!120 + 2\cos^2\!x\sin^2\!120 + \sin^2\!x


    . . =\;2\sin^2\!x\left(\text{-}\frac{1}{2}\right)^2 + 2\cos^2\!x\left(\frac{\sqrt{3}}{2}\right)^2 + \sin^2\1x


    . . =\;\frac{1}{2}\sin^2\!x + \frac{3}{2}\cos^2\!x + \sin^2\!x \;=\;\frac{3}{2}\sin^2\!x + \frac{3}{2}\cos^2\!x


    . . =\;\frac{3}{2}\left(\sin^2\!x + \cos^2\1x\right) \;=\;\frac{3}{2}

    Thanks dude !

    nah you are not an idiot.

    I am the idiot =P

    You are a genius dude !

    Thanks alot !

    Appreciate your effort and time spent thanks !
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  10. #10
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    Quote Originally Posted by Unknown008 View Post
    3) Oh well...

    cosec 2x + \cot 2x = \cot x

    Then, we also have:

    cosec 4x + \cot 4x = \cot 2x

    cosec 8x + \cot 8x = \cot 4x

    Combining those three, we get:

    cosec 2x + cosec 4x + cosec 8x + \cot 2x  + \cot 4x + \cot 8x = \cot x + \cot 2x + \cot 4x

    Can you simplify this?
    Yup .

    I can carry on from here.

    Thanks for your enlightenment !
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