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

= 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) + cos (2 θ)

We showed that the two sides could take the same form

\Rightarrow True

\frac{tan ^{4} ( x ) - 1}{tan ^{2} ( x ) - 1} = sec^{2} (x)

cos (27 0^{\circ} - θ) = - sin (θ)

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

cot (2 a) = \frac{1}{2} (cot (a) - tan (a))

sec (t) = \frac{cot ( t ) + tan ( t )}{csc ( t )}

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