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Homework Help: Sin cos problem finding exact value

  1. Jan 9, 2005 #1
    sin cos problem finding exact value....

    I have to find the exact value of : [tex] \frac {\sin^2 45 degrees + \cos ^2 45 degrees} {\sin 60 degrees \cos 30 degrees}[/tex]

    The squares are throwing me off , without them I got
    [tex] \frac {\frac {1} {sqrt2} + \frac {1} {sqrt2}} {3/2} [/tex]

    what do I do with the sqares on sin and cos in the numerator?
     
    Last edited: Jan 9, 2005
  2. jcsd
  3. Jan 9, 2005 #2
    Try this:

    [tex] sin^2 45 degrees [tex], Sin(45) = radical 2 / 2, so sin^2(45) = 2/4
     
  4. Jan 9, 2005 #3

    dextercioby

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    If u use these formulas
    [tex] \sin^{2}x+\cos^{2}x=1[/tex]
    ,for all "x"
    and
    [tex] \sin(90-x)=\cos x [/tex]
    ,u'd have the problem solved.

    Besides,your calculations are wrong,since at the numerator they didn't include squaring thevalues for sine and cosine,which were supposed to be done.

    And the denominator is 3/4.

    Daniel.
     
  5. Jan 9, 2005 #4
    ok without the square roots on the sine and cosine I get 2/3 but the solution says the answer is 4/3?
     
  6. Jan 9, 2005 #5

    dextercioby

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    If the denominator is 3/4 and the numerator is 1,what is the whole fraction equal to??

    Daniel.
     
  7. Jan 9, 2005 #6
    [tex] \sin 60 = \frac {sqrt3} {2} [/tex]

    and doesnt [tex] \cos 30 = \frac {sqrt3} {2} [/tex] also?

    these multiplied gives 3/2 ??

    How do u get the denominator to equal 3/4?
     
  8. Jan 9, 2005 #7

    dextercioby

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    The multiplication is wrong.U multiplied only the numerators.U need to multiply the denominators as well.I'm sure you'll get
    [tex] 2\cdot 2 =4[/tex]

    Daniel.
     
  9. Jan 9, 2005 #8
    (sqrt3/2) * (sqrt3/2) = 3/4

    you forgot to multiply the two TWO's.

    I've done that before, =\.
     
  10. Jan 9, 2005 #9
    LOL omg silly me :rofl: Thanks Holy cant believe I missed that thanks
     
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