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anonymous

  • 5 years ago

Without using a calculator determine the exact value of sin(x - y) given that sin x = -1/3) , tan y = (1/3) , x is a quadrant II angle and y is a quadrant III angle.

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  1. anonymous
    • 5 years ago
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    sin(x) shouldn't be negative if x is in the second quadrant. Is there a typo?

  2. anonymous
    • 5 years ago
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    Im sorry the problem is Without using a calculator determine the exact value of cos(x - y) given that sin x = (1/3) , tan y = (1/3) , x is a quadrant II angle and y is a quadrant III angle.

  3. anonymous
    • 5 years ago
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    its easy

  4. anonymous
    • 5 years ago
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    use sum/diff of angle expansion

  5. anonymous
    • 5 years ago
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    = cos(x)cos(y)+sin(x)sin(y)

  6. anonymous
    • 5 years ago
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    if sin(x) = (1/3) , draw up a general right angle triangle , mark an angle x ( obviously not the right angle ) , and label the opposite 1 and hypotenuse 3

  7. anonymous
    • 5 years ago
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    then you can find the adjacent side by pythagorus , which works out as sqrt(8) therefore cos(x) = - sqrt(8) / 3 ( because x is negative in the 2nd quadrant

  8. anonymous
    • 5 years ago
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    now you just need to find siny and cosy

  9. anonymous
    • 5 years ago
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    Possible answers are \[A. 0 B. (12\sqrt{5} - \sqrt{10}) /30 C. (-12\sqrt{5} - \sqrt{10}) /30 D. (-12\sqrt{5} + \sqrt{10}) /30\] \[E. (12\sqrt{5} + \sqrt{10}) /30\]

  10. anonymous
    • 5 years ago
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    with tan(y)=(1/3) , draw up a general right angle triangle , mark opposite as 1 and adjacent as 3

  11. anonymous
    • 5 years ago
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    therefore the hypotenuse is sqrt(10) so sin(y) = - 1 / sqrt(10) and cosy =- 3/sqrt(10)

  12. anonymous
    • 5 years ago
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    then sub in

  13. anonymous
    • 5 years ago
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    So my equation should look like (-sqrt(8)/3 * -3/sqrt(10) ) + (1/3 * -1/sqrt(10)) ???

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