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u0860867

  • 2 years ago

pls help urgent help needed

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  1. u0860867
    • 2 years ago
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  2. RadEn
    • 2 years ago
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    for Q.a use the identity : cosAcosB + sinAsinB = cos(A-B)

  3. RadEn
    • 2 years ago
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    what u get ?

  4. u0860867
    • 2 years ago
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    i got 24 which is right

  5. u0860867
    • 2 years ago
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    @RadEn

  6. RadEn
    • 2 years ago
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    yes, it is cos24

  7. RadEn
    • 2 years ago
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    for Q.b use the identity : sinAcosB - cosAsinB = sin(A-B), multiply by -1 to both sides, giving us cosAsinB - sinAcosB = - sin(A-B)

  8. RadEn
    • 2 years ago
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    in this case, given A = 17 and B = 7

  9. u0860867
    • 2 years ago
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    for b will the answer be -10

  10. RadEn
    • 2 years ago
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    hmmm.... cosAsinB - sinAcosB = - sin(A-B) put A= 17 and B=7 cos17sin7 - sin17cos7 = - sin(17-7) cos17sin7 - sin17cos7 = - sin10 then use the identity : -sinx = sin(-x) so, we get -sin10 = sin(-10) you were right :)

  11. u0860867
    • 2 years ago
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    and how do we do part 3 i got the wrong answer for it just have once chance left on it

  12. RadEn
    • 2 years ago
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    do like Q.b for Q.c i think u can solve this, now :)

  13. u0860867
    • 2 years ago
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    @RadEn i got the wrong answer again for part 3

  14. RadEn
    • 2 years ago
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    hmmm.. cos3sin(-2) - cos2sin3 = -cos3sin2 - cos2sin3 = - {cos3sin2 + cos2sin3} we have to use the identity cosAsinB + cosBsinA = sin(A+B) so, - {cos3sin2 + cos2sin3} = - sin(3+2) = -sin5 again, use the identity -sinx =sinx now we have -sin5 = sin(-5)

  15. u0860867
    • 2 years ago
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    thanks @RadEn

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