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

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

Solution

prove

sin (x) + cos (x) = 0

False

Solution steps

sin (x) + cos (x) = 0

Show that the two sides are not equal

For

sin (x) + cos (x) = 0

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{sec ^{2} ( x ) - 1}{sin ^{2} ( x )} = sec^{2} (x)

\frac{1 + sec ( β )}{tan ( β ) + sin ( β )} = csc (β)

sin^{2} (x) \cdot (1 + cot^{2} (x)) = 1

3 sec (x) - 3 cos (x) = 3 sin (x) tan (x)

1 + tan^{2} (a) = \frac{1}{cos ^{2} ( a )}