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RadEn Group TitleBest ResponseYou've already chosen the best response.1
for Q.a use the identity : cosAcosB + sinAsinB = cos(AB)
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
what u get ?
 one year ago

u0860867 Group TitleBest ResponseYou've already chosen the best response.0
i got 24 which is right
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
yes, it is cos24
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
for Q.b use the identity : sinAcosB  cosAsinB = sin(AB), multiply by 1 to both sides, giving us cosAsinB  sinAcosB =  sin(AB)
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
in this case, given A = 17 and B = 7
 one year ago

u0860867 Group TitleBest ResponseYou've already chosen the best response.0
for b will the answer be 10
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
hmmm.... cosAsinB  sinAcosB =  sin(AB) put A= 17 and B=7 cos17sin7  sin17cos7 =  sin(177) cos17sin7  sin17cos7 =  sin10 then use the identity : sinx = sin(x) so, we get sin10 = sin(10) you were right :)
 one year ago

u0860867 Group TitleBest ResponseYou've already chosen the best response.0
and how do we do part 3 i got the wrong answer for it just have once chance left on it
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
do like Q.b for Q.c i think u can solve this, now :)
 one year ago

u0860867 Group TitleBest ResponseYou've already chosen the best response.0
@RadEn i got the wrong answer again for part 3
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
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)
 one year ago

u0860867 Group TitleBest ResponseYou've already chosen the best response.0
thanks @RadEn
 one year ago
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