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anonymous

  • one year ago

anyone pls help me .....I will become ur fan and give u a medal?????

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  1. anonymous
    • one year ago
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    cos (a+45)cos (a-45)=1/2 -sin^2 a

  2. anonymous
    • one year ago
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    that is the real q

  3. zepdrix
    • one year ago
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    Hey Panda :) Is this an identity that we want to prove? Or are we simply trying to solve for a?

  4. anonymous
    • one year ago
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    ithis is an identity

  5. zepdrix
    • one year ago
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    \[\large\rm \cos(a+45^o)\cos(a-45^o)=\frac{1}{2}-\sin^2a\]Hmm there is probably some shortcut for doing this.. I can't seem to remember though. I guess what I would do is apply your Angle Sum Formula to both cosines.\[\large\rm \cos(\alpha+\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta\]\[\large\rm \cos(\alpha-\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta\]

  6. anonymous
    • one year ago
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    then????

  7. zepdrix
    • one year ago
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    So on the left side, applying both identities would give you\[\large\rm (\cos a \cos45-\sin a \sin45)(\cos a \cos45+\sin a \sin45)\]Recognize that here we have conjugates being multiplied together. So we'll get the difference of squares,\[\large\rm \cos^2a \cos^245-\sin^2a \sin^245=\frac{1}{2}-\sin^2a\]

  8. zepdrix
    • one year ago
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    Recall that at 45 degrees, you're getting the same value for sine and cosine, sqrt2/2.\[\large\rm \left(\frac{\sqrt2}{2}\right)^2\cos^2a -\left(\frac{\sqrt2}{2}\right)^2\sin^2a=\frac{1}{2}-\sin^2a\]Which simplifies a bit,\[\large\rm \frac{1}{2}\cos^2a -\frac{1}{2}\sin^2a=\frac{1}{2}-\sin^2a\]Which is quite a bit closer to where we want to be :)

  9. zepdrix
    • one year ago
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    What do you think? :) Too confusing doing it this way?

  10. anonymous
    • one year ago
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    yup

  11. anonymous
    • one year ago
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    another q pls...how to solve these q

  12. LynFran
    • one year ago
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    ok we need to express in terms of \[R \cos(x+\alpha)\]to find R\[\sqrt{2+1}\]so R=\[\sqrt{3}\]and to get alpha we find tan of it so \[\tan^{-1}\frac{ 1 }{ 2 }=26.565 \]now we can express it in the given form... \[R \cos(x+\alpha)\]\[\sqrt{3}\cos(x+26.565)\]and then to solve the second part we let \[\sqrt{3}\cos(x+26.565)=1\]can u continue from there?

  13. anonymous
    • one year ago
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    HOW DID U GET THE R

  14. LynFran
    • one year ago
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    i took the amplitude of 2cosx..which is 2...and the amplitude of sinx..which is 1...and then \[\sqrt{2+1}\]

  15. anonymous
    • one year ago
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    OOOH

  16. anonymous
    • one year ago
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    I CANNOT SOLVE IT........

  17. LynFran
    • one year ago
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    \[\sqrt{3}\cos (x+26.565)=1\]\[\cos(x+26.565)=\frac{ 1 }{ \sqrt{3} }\]we rationalise the denominator \[\cos(x+26.565)=\frac{ \sqrt{3} }{ 3 }\]we take cos inverse..\[(x+26.565)=\cos^{-1} \frac{ \sqrt{3} }{ 3 }+2n \pi\]ok since ur using degree and not radians we have\[(x+26.565)=\pm54.736+360\]\[x=-26.565\pm54.736+360\]..so can u find the solution from here? now u must look at the given \[0\le x \le 180\]

  18. LynFran
    • one year ago
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    can u solve for x from there?

  19. LynFran
    • one year ago
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    ok so \[x=-26.565+54.736\]now the +360 is the period here thats y we continue to add the period for given domain cause ur answer would vary of that domain so.. \[x=28.171\]now if we add the period which is 360 here we get\[x=28.171+360\]\[x=388.171\]now this 388.171 is out of the given domain so thats not a solution so we have 1 solution to this problem for the given domain \[x=28.171\]...now if we check the other \[x=-26.565-54.736\]\[x=-81.301\]...this is out of domain if we add 360 we get \[x=-81.301+360\]\[x=278.699\]so it would not be a solution... the only solution is \[x=28.171\]

  20. anonymous
    • one year ago
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    SORRY FOR THE LATE REPLY....I JUST ATE MY BREAKFAST...

  21. anonymous
    • one year ago
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    OOOH NOW I GET IT ..... TQ VERY MUCH....

  22. LynFran
    • one year ago
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    ok

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is replying to Can someone tell me what button the professor is hitting...

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