Question

Will fan and medal Hard Trig Question

Answer 1

\[\sin^3A \cos(B-C)+\sin^3B \cos (C-A) + \sin^3C \cos(A-B) = 3sinAsinBsinC\]
Provided that A+B+C=180

Answer 2

by the way it ends with =sinAsinBsinC

Answer 3

@ganeshie8 @ikram002p @mathmate

Answer 4

If we start by breaking it into :\[\sin^2A \times sinA \times \cos(B-C)\] etc. how do we simplify it

Answer 5

sinA= sin (180 - B-C) = sin(B+C)

Answer 6

Yes but apparantly, it is meant to be done as above

Answer 7

sin^2 A×sin(B+C)×cos(B−C)

Answer 8

2sin(B+C)×cos(B−C)= sin2B+sin2C

Answer 9

Yea so it is (1-cos2A) x (sin2B+sin2C) right?

Answer 10

sin^2A= (1-cos2A)/2

Answer 11

wait all that on four

Answer 12

yea but isnt there meant to be another 1/2 in front of the sin2B+sin2C

Answer 13

sin^2 A×sin(B+C)×cos(B−C)=1/4 * (1-cos2A)*(sin2B+sin2C)

Answer 14

yes ok

Answer 15

similarly hence for other terms too
and then add up

Answer 16

sorry can you simplify it more because it gets very very messy

Answer 17

@matricked

Answer 18

1/4 *{ (1-cos2A)*(sin2B+sin2C) + (1-cos2B)*(sin2C+sin2A) + (1-cos2C)*(sin2B+sin2A) )

Answer 19

Yes I'm there but can we simplify 1/4(1-cos2A)(Sin2B+sin2C) more?

Answer 20

@matricked

Answer 21

1/4(1-cos2A)(Sin2B+sin2C) =1/4 *(Sin2B+sin2C -cos2ASin2B -cos2Asin2C)

Answer 22

that cant really be simplified further without messing up the equation. is there a faster way?

Answer 23

@matricked

Answer 24

you need to simplify further...

Answer 25

right now it is too messy to simplify but i could try, the result wont be great

Answer 26

yup..
need to find some special trick used here

Answer 27

do you noe? i wont waste anymore of your time if you dont

Answer 28

further simplification may help here ...

Answer 29

but what do you simplify

Answer 30

@matricked

Answer 31

=1/4 * ( 2*(sin2A +Sin2B+sin2C ) - (cos2ASin2B + cos2Asin2C +cos2BSin2A + cos2Bsin2C +cos2CSin2B + cos2Csin2A))

Answer 32

ok nevermind thanks anyway @matricked

Answer 33

=1/4 * ( 2*(sin2A +Sin2B+sin2C ) - (cos2ASin2B +cos2BSin2A + cos2Bsin2C +cos2CSin2B + cos2Asin2C + cos2Csin2A))
=1/4 * ( 2*(sin2A +Sin2B+sin2C ) - sin(2A+2B) - sin(2C+2B) -sin(2A+2C) )
now
sin(2A+2B) = sin 2(A+B)= sin 2( 180- C)) = -sin2C
=1/4 * (( 2*(sin2A +Sin2B+sin2C ) - sin(2A+2B) - sin(2C+2B) -sin(2A+2C) )
=1/4 *( ( 2*(sin2A +Sin2B+sin2C ) + sin2C+sin2A +Sin2B )
=3/4 * (sin2A +Sin2B+sin2C )

Answer 34

=3/4 * (sin2A +Sin2B+sin2C )
=3/4 * (2sin(A+B)cos(A-B)+2sinC cosC)
=3/4*2sinC (cos(A-B) +cos C)
cosC=cos (180-A-B)=-cos(A+B)
=3/4*2sinC (cos(A-B)-cos(A+B) )
=3/4 * 2sinC *2sinAsinB
=3sinAsinBsinC

Answer 35

hope its clear to you...

Answer 36

oh thanks so much!!!