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Math Help - Help in Calculating a Value - Trigonometry

  1. #1
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    Help in Calculating a Value - Trigonometry

    Hi, I had this sum for an exam and I couldn't get it, so I just wanted to know how to solve it:

    Given: sinA + sin^2A = 1

    Calculate: cos^12A + 3cos^10A + 3cos^8A cos^6A 2cos^4A 2cos^2A - 2

    I didn't know how to use the code, so I used ^ for power. Sorry for that.

    This is what I did:
    sinA + sin^2A = 1
    sinA = 1 - sin^2A
    But, 1 - sin^2A = cos^2A
    Therefore, sinA = cos^2A

    ...And then I substituted cos^2A as sinA throughout this:
    cos^12A + 3cos^10A + 3cos^8A cos^6A 2cos^4A 2cos^2A - 2

    Yet, I could not arrive at an answer. All help is greatly appreciated.

    Thanks,
    RoH.
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by Ruler of Hell View Post
    Given: sinA + sin^2A = 1

    Calculate: cos^12A + 3cos^10A + 3cos^8A cos^6A 2cos^4A 2cos^2A - 2
    Are you sure it is not :
    \cos^{12}A + 3\cos^{10}A + 3\cos^8A + \cos^6A + 2\cos^4A + 2\cos^2A  - 2

    Ok, when substituting, you should have :
    \sin^6A+3\sin^5A+3\sin^4A+\sin^3A+2\sin^2A+2\sin A-2

    Develop (\sin A+\sin^2A)^3=1^3, use \sin A+\sin^2A=1 and see what can be simplified =)
    Last edited by Moo; September 4th 2008 at 10:16 AM. Reason: added some info, in regard to the following :s
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  3. #3
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    Hello, Ruler of Hell!

    Your game plan is a good one.
    And I agree with Moo . . . you left out some plus-signs.


    Given: . \sin A + \sin^2\!A \:=\:1 \quad\Rightarrow\quad {\color{blue}\sin A(\sin A + 1) \:=\:1}

    Calculate: . \cos^{12}\!A + 3\cos^{10}\!A + 3\cos^8\!A \:{\color{red}+}\: \cos^6\!A \:{\color{red}+}\:  2\cos^4\!A \:{\color{red}+}\: 2\cos^2\!A  - 2

    This is what I did: . \sin A + \sin^2\!A \:=\: 1 \quad\Rightarrow\quad \sin A \:= \:1 - \sin^2\!A

    But . 1 - \sin^2\!A \:= \:\cos^2\!A
    Hence, \sin A \:= \:\cos^2\!A

    Substituting, we get: . \sin^6\!A + 3\sin^5\!A + 3\sin^4\!A + \sin^3\!A + 2\sin^2\!A + 2\sin A - 2

    Factor: . \sin^3\!A\left(\sin^3\!A + 3\sin^2\!A + 3\sin A + 1\right) + 2\sin x(\sin x + 1) - 2

    Factor: . \sin^3\!A(\sin A + 1)^3 + 2\sin x(\sin x + 1) - 2

    We have: . \underbrace{\bigg[\sin A(\sin A + 1)\bigg]^3}_{\text{This is 1}} + 2\underbrace{\bigg[\sin A(\sin A + 1)\bigg]}_{\text{This is 1}} - 2


    Therefore: . 1^3 + 2\!\cdot\!1 - 2 \;\;=\;\;\boxed{1}

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  4. #4
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    <br />
  \sin A + \sin ^2 A = 1 \hfill \\
     \Rightarrow \sin A = 1 - \sin ^2 A \hfill \\
       \Rightarrow \sin A = \cos ^2 A \hfill \\  <br />
\

    we will use this value in the given expression:
    Substitute:
    <br />
= \sin ^6 A + 3\sin ^5 A + 3\sin ^4 A + \sin ^3 A + 2\sin ^2 A + 2\sin A - 2<br />
\

    Factoring:
    <br />
=\sin^3\!A\left(\sin^3\!A + 3\sin^2\!A + 3\sin A + 1\right) + 2\sin A(\sin A + 1) - 2<br />

    <br />
=\sin^3\!A(\sin A + 1)^3 + 2\sin A(\sin A + 1) - 2<br />

    <br />
=[\sin A(\sin A + 1)]^3 + 2[\sin A(\sin A + 1)] - 2<br />

    <br />
=[\sin^2 A+\sin A)]^3 + 2[\sin^2 A+\sin A)] - 2<br />

    = 1^3+2(1)-2\

    =1
    Last edited by Shyam; September 4th 2008 at 11:59 AM.
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