anonymous
  • anonymous
pls help urgent help needed
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
1 Attachment
RadEn
  • RadEn
for Q.a use the identity : cosAcosB + sinAsinB = cos(A-B)
RadEn
  • RadEn
what u get ?

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anonymous
  • anonymous
i got 24 which is right
anonymous
  • anonymous
RadEn
  • RadEn
yes, it is cos24
RadEn
  • RadEn
for Q.b use the identity : sinAcosB - cosAsinB = sin(A-B), multiply by -1 to both sides, giving us cosAsinB - sinAcosB = - sin(A-B)
RadEn
  • RadEn
in this case, given A = 17 and B = 7
anonymous
  • anonymous
for b will the answer be -10
RadEn
  • RadEn
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 :)
anonymous
  • anonymous
and how do we do part 3 i got the wrong answer for it just have once chance left on it
RadEn
  • RadEn
do like Q.b for Q.c i think u can solve this, now :)
anonymous
  • anonymous
@RadEn i got the wrong answer again for part 3
RadEn
  • RadEn
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)
anonymous
  • anonymous
thanks @RadEn

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