Is sin(2θ)+cos(2θ)=2sin(θ)cos(θ)+2cos^2(θ)-1 ?

The answer to whether sin(2θ)+cos(2θ)=2sin(θ)cos(θ)+2cos^2(θ)-1 is True

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prove sin(2θ)+cos(2θ)=2sin(θ)cos(θ)+2cos^2(θ)-1

prove

sin (2 θ) + cos (2 θ) = 2 sin (θ) cos (θ) + 2 cos^{2} (θ) - 1

True

Solution steps

sin (2 θ) + cos (2 θ) = 2 sin (θ) cos (θ) + 2 cos^{2} (θ) - 1

Manipulating left side

sin (2 θ) + cos (2 θ)

Rewrite using trig identities

sin (2 θ) + cos (2 θ)

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

= 2 sin (θ) cos (θ) + cos (2 θ)

= 2 sin (θ) cos (θ) + cos (2 θ)

Manipulating right side

2 sin (θ) cos (θ) + 2 cos^{2} (θ) - 1

Rewrite using trig identities

2 sin (θ) cos (θ) + 2 cos^{2} (θ) - 1

Use the Double Angle identity:

2 cos^{2} (x) - 1 = cos (2 x)

= 2 sin (θ) cos (θ) + cos (2 θ)

= 2 sin (θ) cos (θ) + cos (2 θ)

We showed that the two sides could take the same form

\Rightarrow True

Is sin(2θ)+cos(2θ)=2sin(θ)cos(θ)+2cos^2(θ)-1 ?
The answer to whether sin(2θ)+cos(2θ)=2sin(θ)cos(θ)+2cos^2(θ)-1 is True