Is sin(pi+θ)=-sin(θ) ?

The answer to whether sin(pi+θ)=-sin(θ) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove sin(pi+θ)=-sin(θ)

prove

sin (π + θ) = - sin (θ)

True

Solution steps

sin (π + θ) = - sin (θ)

Manipulating left side

sin (π + θ)

Rewrite using trig identities

sin (π + θ)

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= sin (π) cos (θ) + cos (π) sin (θ)

Simplify

sin (π) cos (θ) + cos (π) sin (θ) : - sin (θ)

sin (π) cos (θ) + cos (π) sin (θ)

sin (π) cos (θ) = 0

sin (π) cos (θ)

Simplify

sin (π) : 0

sin (π)

Use the following trivial identity:

sin (π) = 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 \cdot cos (θ)

Apply rule

0 \cdot a = 0

= 0

cos (π) sin (θ) = - sin (θ)

cos (π) sin (θ)

Simplify

cos (π) : - 1

cos (π)

Use the following trivial identity:

cos (π) = (- 1)

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 \cdot sin (θ)

Multiply:

1 \cdot sin (θ) = sin (θ)

= - sin (θ)

= 0 - sin (θ)

0 - sin (θ) = - sin (θ)

= - sin (θ)

= - sin (θ)

= - sin (θ)

We showed that the two sides could take the same form

\Rightarrow True

\frac{tan ( x ) cot ( x )}{sin ( x )} = csc (x)

csc^{2} (\frac{θ}{2}) = \frac{2 sec ( θ )}{sec ( θ ) - 1}

csc^{2} (x) \cdot tan^{2} (x) - 1 = tan^{2} (x)

\frac{1 + csc ( θ )}{sec ( θ )} = cos (θ) + cot (θ)

\frac{cos ( x )}{sec ( 2 x )} - \frac{sin ( x )}{csc ( 2 x )} = cos (3 x)