Is 1+sin(2θ)=(cos(θ)sin(θ))^2 ?

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

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prove 1+sin(2θ)=(cos(θ)sin(θ))^2

prove

1 + sin (2 θ) = (cos (θ) sin (θ))^{2}

False

Solution steps

1 + sin (2 θ) = (cos (θ) sin (θ))^{2}

Show that the two sides are not equal

For

1 + sin (2 θ) = (cos (θ) sin (θ))^{2}

plug in

θ = 0

1 + sin (2 \cdot 0) = 1

1 + sin (2 \cdot 0)

Simplify

= 1 + sin (0)

Use the following trivial identity:

sin (0) = 0

sin (0)

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

= 1 + 0

Simplify

= 1

(cos (0) sin (0))^{2} = 0

(cos (0) sin (0))^{2}

Use the following trivial identity:

cos (0) = 1

cos (0)

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 1

Use the following trivial identity:

sin (0) = 0

sin (0)

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

= (1 \cdot 0)^{2}

Simplify

= 0

The two sides are not equal

\Rightarrow False

p ro v e cos^{2} (x) = \frac{1}{2} (cos (2 x) + 1)

p ro v e (csc (x) - 1) (csc (x) + 1) = cot^{2} (x)

p ro v e tan^{2} (y) (csc^{2} (y) - 1) = 1

p ro v e cos (2 π + x) = cos (x)

p ro v e cos (a - b) - cos (a + b) = 2 sin (a) sin (b)

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