Is 1+sin(2θ)=(sin(θ)+cos(θ))2 ?

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

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prove 1+sin(2θ)=(sin(θ)+cos(θ))2

prove

1 + sin (2 θ) = (sin (θ) + cos (θ)) 2

False

Solution steps

1 + sin (2 θ) = (sin (θ) + cos (θ)) \cdot 2

Show that the two sides are not equal

For

1 + sin (2 θ) = (sin (θ) + cos (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

(sin (0) + cos (0)) \cdot 2 = 2

(sin (0) + cos (0)) \cdot 2

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

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

= (0 + 1) \cdot 2

Simplify

= 2

The two sides are not equal

\Rightarrow False

tan^{2} (x) - tan (x) + 1 = sec^{2} (x) - tan (x)

\frac{1 - sin ( a )}{1 + sin ( a )} = \frac{cos ( a )}{1 + sin ( a )}

2 \cdot cos (x) \cdot sin (x) = sin (2 x)

2 tan (x^{2}) + 1 = 1

\frac{tan ( x )}{1} = \frac{sin ( x )}{cos ( x )}

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