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

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

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

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

(sin (x) + cos (x))^{2} = 1 + 2 cos (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 + 2 cos (0) = 3

1 + 2 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 + 2 \cdot 1

Simplify

= 3

The two sides are not equal

\Rightarrow False

\frac{csc ( x ) - 1}{csc ^{2} ( x )} = cos^{2} (x)

csc (s) - sin (s) = cos (s) cot (s)

\frac{( 4 sin ^{2} ( x ) )}{tan ^{2} ( x )} = 4 cos^{2} (x)

\frac{tan ( 2 x ) + tan ( x )}{1 - tan ( 2 x ) tan ( x )} = tan (3 x)

csc (x) \cdot (sin (x) + tan (x)) = 1 + sec (x)

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