Question

if theta=45 degrees,verify the following

1.sin2theta=2sintheta.costheta
2.cos2theta=cos^2theta-sin^2theta

Answer 1

sin2θ = 2sinθ.cosθ
sin 2×45 ° = 2×sin 45° .cos 45°
sin90° = 2×sin45°.cos 45°
(put the values )
1 =2×1/√2×1/√2
1= 2×1/2
1=1


ans2. cos 2θ =cos^2θ-sin^2θ
cos 2×45° = cos ^2(45°) -sin ^2(45°)
cos 90° = (1/√2)^2 -(1/√2)^2
0 = 1/2 -12
0=0
I HOPE THIS WILL HELP YOU

Answer 2

GIVEN
=> θ=45°
_____________________________

(1) sin2θ= 2sinθcosθ

➡LHS
=sin2(45)
=sin90 =1

➡RHS
=2sin(45) cos(45)
=2×(1/√2)×(1/√2)
=2×1/2
=1=LHS(proved)
_______________________________

(2) cos2θ=cos^2θ -sin^2θ

➡LHS
=cos2θ
=cos2(45)
=cos 90
=0

➡RHS
=cos^2θ-sin^2θ
=cos^2(45) -sin^2(45)
=(1/√2)^2 -(1/√2)^2
=(1/2)-(1/2)
=0= RHS (proved)
_____________________________

hope it helps you....
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