Is (cos(θ)-sin(θ))^2=1+sin(2θ) ?

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

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prove (cos(θ)-sin(θ))^2=1+sin(2θ)

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

θ = 1

(cos (1) - sin (1))^{2} = 0.09070 \dots

(cos (1) - sin (1))^{2}

Refine to a decimal form

= 0.09070 \dots

1 + sin (2 \cdot 1) = 1.90929 \dots

1 + sin (2 \cdot 1)

Refine to a decimal form

= 1.90929 \dots

The two sides are not equal

\Rightarrow False

(4 sin (θ) + 4 cos (θ)) = 16 + 16 sin (2 θ)

5 sec^{2} (x) = 5 sec^{2} (- x)

tan (θ) + 1 = \frac{cos ( θ ) + sin ( θ )}{cos ( θ )}

- \frac{1}{2} = 2 sin (- \frac{π}{12}) cos (- \frac{π}{12})

sec^{4} (x) = tan^{4} (x)

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