Is (sin(x)+cos(x))^2=1+cos(2x) ?

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

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prove (sin(x)+cos(x))^2=1+cos(2x)

prove

(sin (x) + cos (x))^{2} = 1 + cos (2 x)

False

Solution steps

(sin (x) + cos (x))^{2} = 1 + cos (2 x)

Show that the two sides are not equal

For

(sin (x) + cos (x))^{2} = 1 + cos (2 x)

plug in

x = 0

(sin (0) + cos (0))^{2} = 1

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

Simplify

= 1

1 + cos (2 \cdot 0) = 2

1 + cos (2 \cdot 0)

Simplify

= 1 + cos (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

= 1 + 1

Simplify

= 2

The two sides are not equal

\Rightarrow False

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

sin (θ) \cdot tan (θ) \cdot cos (θ) = sin^{2} (θ)

cos^{2} (x) + sin (x) tan (x) = sec (x)

\frac{1}{1 - sin ( x )} = - csc (x)

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

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