If sinx+sin2x=1, then cos12x+3cos10x+3cos8x+cos6x+2cos4x+cos2x−2=
A
0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
sin2x
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
cos2x
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is Csin2x sinx+sin2x=1⇒sinx=cos2x∴cos12x+3cos10x+3cos8x+cos6x+2cos4x+cos2x−2 =sin6x+3sin5x+3sin4x+sin3x+sin2x+sinx−2 =[sin6x+sin3x+3sin5x+3sin4x]+[sin2x+sinx]−2+sin2x =[(sin2x)3+(sinx)3+3(sin2x)(sinx)(sin2x+sinx)]+1−2+sin2x=(sinx+sin2x)3−1+sin2x=1+1−2+sin2x=sin2x