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شائع علم المثلثات >

prove cot(x+pi)=cot(x)

الحلّ

prove

cot (x + π) = cot (x)

الحلّ

صحيح

خطوات الحلّ

cot (x + π) = cot (x)

قم بالعمل على الطرف الأيسر

cot (x + π)

Rewrite using trig identities

cot (x + π)

cot (x) = \frac{cos ( x )}{sin ( x )}

:Use the basic trigonometric identity

= \frac{cos ( x + π )}{sin ( x + π )}

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

:فعّل متطابقة الجمع لزوايا

= \frac{cos ( x + π )}{sin ( x ) cos ( π ) + cos ( x ) sin ( π )}

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

:فعّل متطابقة الجمع لزوايا

= \frac{cos ( x ) cos ( π ) - sin ( x ) sin ( π )}{sin ( x ) cos ( π ) + cos ( x ) sin ( π )}

\frac{cos ( x ) cos ( π ) - sin ( x ) sin ( π )}{sin ( x ) cos ( π ) + cos ( x ) sin ( π )}

بسّط:

\frac{cos ( x )}{sin ( x )}

\frac{cos ( x ) cos ( π ) - sin ( x ) sin ( π )}{sin ( x ) cos ( π ) + cos ( x ) sin ( π )}

cos (x) cos (π) - sin (x) sin (π) = - cos (x)

cos (x) cos (π) - sin (x) sin (π)

cos (x) cos (π) = - cos (x)

cos (x) cos (π)

cos (π)

بسّط:

- 1

cos (π)

استخدم المتطابقة الأساسيّة التالية:

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) cos (x)

بسّط

= - cos (x)

= - cos (x) - sin (π) sin (x)

sin (x) sin (π) = 0

sin (x) sin (π)

sin (π)

بسّط:

0

sin (π)

استخدم المتطابقة الأساسيّة التالية:

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 sin (x)

0 \cdot a = 0

فعّل القانون

= 0

= - cos (x) - 0

- cos (x) - 0 = - cos (x)

= - cos (x)

= \frac{- cos ( x )}{cos ( π ) sin ( x ) + sin ( π ) cos ( x )}

sin (x) cos (π) + cos (x) sin (π) = - sin (x)

sin (x) cos (π) + cos (x) sin (π)

sin (x) cos (π) = - sin (x)

sin (x) cos (π)

cos (π)

بسّط:

- 1

cos (π)

استخدم المتطابقة الأساسيّة التالية:

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) sin (x)

بسّط

= - sin (x)

= - sin (x) + sin (π) cos (x)

cos (x) sin (π) = 0

cos (x) sin (π)

sin (π)

بسّط:

0

sin (π)

استخدم المتطابقة الأساسيّة التالية:

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 (x)

0 \cdot a = 0

فعّل القانون

= 0

= - sin (x) + 0

- sin (x) + 0 = - sin (x)

= - sin (x)

= \frac{- cos ( x )}{- sin ( x )}

\frac{- a}{- b} = \frac{a}{b}

: استخدم ميزات الكسور التالية

= \frac{cos ( x )}{sin ( x )}

= \frac{cos ( x )}{sin ( x )}

= \frac{cos ( x )}{sin ( x )}

\frac{cos ( x )}{sin ( x )} = cot (x)

:Use the basic trigonometric identity

= cot (x)

أظهرنا بأنّ الطرفين من الممكن أن تكون لهما نفس الصورة

\Rightarrow صحيح

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