Is sin(x+pi)-sin(x)+1=0 ?

The answer to whether sin(x+pi)-sin(x)+1=0 is False

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

prove sin(x+pi)-sin(x)+1=0

prove

sin (x + π) - sin (x) + 1 = 0

False

Solution steps

sin (x + π) - sin (x) + 1 = 0

Show that the two sides are not equal

For

sin (x + π) - sin (x) + 1 = 0

plug in

x = 0

sin (0 + π) - sin (0) + 1 = 1

sin (0 + π) - sin (0) + 1

Rewrite using trig identities:

sin (0 + π) = 0

sin (0 + π)

Simplify

= sin (π)

Use the following trivial identity:

sin (π) = 0

sin (π)

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

= 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

= 0 - 0 + 1

Simplify

= 1

The two sides are not equal

\Rightarrow False

p ro v e csc (x) = csc (x) cos^{2} (x) + sin (x)

p ro v e (sin (x) + cos (x))^{2} = 1 + cos (2 x)

p ro v e sin (θ) \cdot tan (θ) \cdot cos (θ) = sin^{2} (θ)

p ro v e - sin (θ) = - sin (- θ)

p ro v e cos (2 B) = \frac{1 - tan ^{2} ( B )}{1 + tan ^{2} ( B )}