Given
sin(4x+π4)+cos(4x5π4)=√2
Let us now consider L.H.S
L.H.S =sin[4x+π4]+cos[4x+5π4]→eq(1)
as we know that sin(A+B)=sinAcosB+cosAsinB
cos(A+B)=cosAcosB−sinAsinB
Let us now apply the above froms formation to eq(1)
⇔sin4xcosπ4+cos4xsinπ4+cos4xcos5π4−sin4xsin5π4
⇔sin4xcosπ4+cos4xsinπ4+cosx4cos[ππ4]−sin4xsin[π+π4]
[∵180+θ is in QIII cos+vesin−ve]
⇔sin4xcosπ4+cos4xsinπ4+cos4xcosπ4+sin4xsinπ4
[∵sin(90−θ)=cosθ&cos(90−θ)=sinθ]
sin4xcosπ4+cos4xsinπ4+cos4xcos(π2−π4)+sin4xsin(π2−π4)
⇔sin4xcosπ4+cos4xsinπ4+cos4xsinπ4+sin4xcosπ4
(∵sinπ4=cosπ4=1/√2)
⇔2sin4x[1√2]+2cos4x[1√2]
[∵2=√2.√2]
⇒2√2[sin4x+cos4x]⇔√2[sin4x+cos4x]→eq(2)
[∵sin2A=2sincosA,cos2A=cos2A−sin2]
Let as apply the above transformation to eq(2)
⇔√2[sin2(2x)+cos2(2x)]
⇔√2[2sin2xcos2x+cos22x−sin22x]
[∵(a+b)2=a2+b2+2ab]
⇔√2[2sin2xcos2x+cos22x−sin22x]
⇔√2[(sin2x+cos2x)2]
⇔√2[(0−1)2]⇒√2(−1)2
[ x is considered as 180∘ & as both the fuction ile in Q]
[∵sin180∘=0cos180∘=−1]
⇒√2⇒ R.H.S
∴ L.H.S = R.H.S