Question

Buktikan bahwa
1+cos2C - cos2A - cos 2B = 4sinAsinBsinC

Mohon bantuannya yah, kalau gak bisa jawab gak ush jawab, saya laporkan ke penyalahgunaan

Answer 1

1 + cos2C - cos2A - cos2B = 4sinA.sinB.sinC

1 - cos2A - cos2B + cos2C = 4sinA.sinB.sinC

cos2A + cos2B - cos2C = 1 - 4sinA.sinB.sinC

akan kita buktikan ruas kiri dan ruas kanan pada persamaan di atas sama!

cos2A + cos2B - cos2C

= cos2A - 2sin[(2B + 2C)/2].sin[(2B - 2C)/2]

= cos2A - 2sin(B + C).sin(B - C)

kita tahu bahwa :
A + B + C = 180°
=> B + C = 180° - A

maka,

= cos2A - 2sin(180° - A ).sin(B - C)

= cos2A - 2(sin180°.cosA - sinA.cos180°).sin(B - C)

= cos2A + 2sinA.sin(B - C)

= 1 - 2sin²A + 2sinA.sin(B - C)

= 1 - 2sinA [sinA - sin(B - C)]

= 1 - 2sinA ( 2cos[(A + B - C)/2].sin[(A - B + C)/2] )

= 1 - 4sinA.cos( A + B - C )/2.sin(A - B + C ) /2

padahal:
A + B + C = 180°
=> A + B = 180° - C
dan A + C = 180° - B

maka,

= 1 - 4sinA.cos[(180° - C - C)/2].sin[(180° - B - B)/2]

= 1 - 4sinA.cos(-2C)/2.sin(-2B)/2

= 1 - 4sinA.cos(-C).sin(-B)

= 1 - 4sinA.sinB.cosC

terbukti

semoga membantu ya :)