wiz-icon
MyQuestionIcon
MyQuestionIcon
4
You visited us 4 times! Enjoying our articles? Unlock Full Access!
Question

Prove:
sin2(π4x)+sin2(π4+x)=1

Open in App
Solution

LHS=sin2(π4x)+sin2(π4+x)=[sin(π4x)]2+[sin(π4+x)]2=[sinπ4cosxcosπ4sinx]2+[sinπ4cosx+cosπ4sinx]2=(sinπ4cosx)2+(cosπ4sinx)2+(sinπ4cosx)2+(cosπ4sinx)2=2(sinπ4cosx)2+2(cosπ4sinx)2=2(12cosx)2+2(12sinx)2=1=RHS

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
General Solutions
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon