sin 12° sin 48° sin 54° =
Answers
Answer:
Now, let's try the manual method. sin12°∗sin48°∗sin54°=sin54°∗(sin12°∗sin48°)=sin54°∗12∗2∗(sin12°∗sin48°)=12sin54°∗(2sin12°∗sin48°)sin12°∗sin48°∗sin54°=sin54°∗(sin12°∗sin48°)=sin54°∗12∗2∗(sin12°∗sin48°)=12sin54°∗(2sin12°∗sin48°) Now, 2sinAsinB=cos(A−B)−cos(A+B)2sinAsinB=cos(A−B)−cos(A+B) And, sin54°=sinx=cos(90°−x)=cos(90°−54°)=cos36°sin54°=sinx=cos(90°−x)=cos(90°−54°)=cos36° ⇒sin12°∗sin48°∗sin54°=12cos36°∗[cos(12°−48°)−cos(12°+48°)]=12cos36°∗[cos(36°)−cos(60°)]⇒sin12°∗sin48°∗sin54°=12cos36°∗[cos(12°−48°)−cos(12°+48°)]=12cos36°∗[cos(36°)−cos(60°)] We know that cos60°=12cos60°=12 What's the value of cos36°cos36°? LetA=18°A=18° 5A=90°5A=90° 2A=90−3ASin2A=Sin(90−3A)=Cos3A2A=90−3ASin2A=Sin(90−3A)=Cos3A 2SinACosA=4Cos³A−3CosA2SinACosA=4Cos³A−3CosA 2SinA=4Cos²A−3=4(1−Sin2A)−3=1−4Sin2A2SinA=4Cos²A−3=4(1−Sin2A)−3=1−4Sin2A 4Sin²A+2SinA−1=04Sin²A+2SinA−1=0 This is a quadratic equation in Sin A. Then, SinA=−2±22−4∗4∗(−1)√2∗4SinA=−2±22−4∗4∗(−1)2∗4 ⇒SinA=−2±22+4∗4∗1√2∗4=−2±20√8=−2±25√8=−1±5√4⇒SinA=−2±22+4∗4∗12∗4=−2±208=−2±258=−1±54 Solving quadrati equation: SinA=−1±5√4SinA=−1±54 Sin 18° lies in quad 1 thus we take the positive value. SinA=Sin18°=5√−14SinA=Sin18°=5−14 Then cos36°=1−2Sin2(18°)=1−2∗(5√−14)2=1−∗(5√−1)28=8−5+25√−18=5√+14cos36°=1−2Sin2(18°)=1−2∗(5−14)2=1−∗(5−1)28=8−5+25−18=5+14 ⇒sin12°∗sin48°∗sin54°=12cos36°∗[cos(36°)−cos(60°)]=12∗5√+14∗[5√+14−12]=12∗(5√+14)2∗−12∗(5√+14)∗12=3+5√16−5√+116=3+5√−5√−116=216=18=0.125⇒sin12°∗sin48°∗sin54°=12cos36°∗[cos(36°)−cos(60°)]=12∗5+14∗[5+14−12]=12∗(5+14)2∗−12∗(5+14)∗12=3+516−5+116=3+5−5−116=216=18=0.125 Thus, sin12°∗sin48°∗sin54°=0.125sin12°∗sin48°∗sin54°=0.125Now,let
′
strythemanualmethod.sin12°∗sin48°∗sin54°=sin54°∗(sin12°∗sin48°)=sin54°∗12∗2∗(sin12°∗sin48°)=12sin54°∗(2sin12°∗sin48°)sin12°∗sin48°∗sin54°=sin54°∗(sin12°∗sin48°)=sin54°∗12∗2∗(sin12°∗sin48°)=12sin54°∗(2sin12°∗sin48°)Now, 2sinAsinB=cos(A−B)−cos(A+B)2sinAsinB=cos(A−B)−cos(A+B)And, sin54°=sinx=cos(90°−x)=cos(90°−54°)=cos36°sin54°=sinx=cos(90°−x)=cos(90°−54°)=cos36°⇒sin12°∗sin48°∗sin54°=12cos36°∗[cos(12°−48°)−cos(12°+48°)]=12cos36°∗[cos(36°)−cos(60°)]⇒sin12°∗sin48°∗sin54°=12cos36°∗[cos(12°−48°)−cos(12°+48°)]=12cos36°∗[cos(36°)−cos(60°)]Weknowthat cos60°=12cos60°=12What
′
sthevalueof cos36°cos36°?LetA=18°A=18°5A=90°5A=90°2A=90−3ASin2A=Sin(90−3A)=Cos3A2A=90−3ASin2A=Sin(90−3A)=Cos3A2SinACosA=4Cos³A−3CosA2SinACosA=4Cos³A−3CosA2SinA=4Cos²A−3=4(1−Sin2A)−3=1−4Sin2A2SinA=4Cos²A−3=4(1−Sin2A)−3=1−4Sin2A4Sin²A+2SinA−1=04Sin²A+2SinA−1=0ThisisaquadraticequationinSinA.Then,SinA=−2±22−4∗4∗(−1)√2∗4SinA=−2±22−4∗4∗(−1)2∗4⇒SinA=−2±22+4∗4∗1√2∗4=−2±20√8=−2±25√8=−1±5√4⇒SinA=−2±22+4∗4∗12∗4=−2±208=−2±258=−1±54Solvingquadratiequation: SinA=−1±5√4SinA=−1±54Sin18°liesinquad1thuswetakethepositivevalue.SinA=Sin18°=5√−14SinA=Sin18°=5−14Then cos36°=1−2Sin2(18°)=1−2∗(5√−14)2=1−∗(5√−1)28=8−5+25√−18=5√+14cos36°=1−2Sin2(18°)=1−2∗(5−14)2=1−∗(5−1)28=8−5+25−18=5+14⇒sin12°∗sin48°∗sin54°=12cos36°∗[cos(36°)−cos(60°)]=12∗5√+14∗[5√+14−12]=12∗(5√+14)2∗−12∗(5√+14)∗12=3+5√16−5√+116=3+5√−5√−116=216=18=0.125⇒sin12°∗sin48°∗sin54°=12cos36°∗[cos(36°)−cos(60°)]=12∗5+14∗[5+14−12]=12∗(5+14)2∗−12∗(5+14)∗12=3+516−5+116=3+5−5−116=216=18=0.125Thus, sin12°∗sin48°∗sin54°=0.125sin12°∗sin48°∗sin54°=0.125