The answer to whether 0=sin(x)+cos(x) is False

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prove 0=sin(x)+cos(x)

Solution

prove

0 = sin (x) + cos (x)

False

Solution steps

0 = sin (x) + cos (x)

Show that the two sides are not equal

For

0 = sin (x) + cos (x)

plug in

x = 0

sin (0) + cos (0) = 1

sin (0) + cos (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

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

Simplify

= 1

The two sides are not equal

\Rightarrow False

\frac{2 cot ( x )}{csc ^{2} ( x )} = 2 cos (x) sin (x)

cos (4 x) = 8 cos^{4} (x)

((1 + cot (a) - csc (a)) (1 + tan (a) + sec (a))) = 2

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

\frac{sin ^{2} ( x )}{sin ( x )} = \frac{2}{sec ( x )}