## u0860867 2 years ago pls help urgent help needed

1. u0860867

for Q.a use the identity : cosAcosB + sinAsinB = cos(A-B)

what u get ?

4. u0860867

i got 24 which is right

5. u0860867

yes, it is cos24

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

in this case, given A = 17 and B = 7

9. u0860867

for b will the answer be -10

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

and how do we do part 3 i got the wrong answer for it just have once chance left on it

do like Q.b for Q.c i think u can solve this, now :)

13. u0860867

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