English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

شائع علم المثلثات >

prove tan(-sqrt(7))=tan(pi-sqrt(7))

الحلّ

prove

tan (- 7) = tan (π - 7)

الحلّ

صحيح

خطوات الحلّ

tan (- 7) = tan (π - 7)

قم بالعمل على الطرف الأيسر

tan (- 7)

tan (- x) = - tan (x)

:Use the negative angle identity

= - tan (7)

sin, cos :

عبّر بواسطة

- tan (7)

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

:Use the basic trigonometric identity

= - \frac{sin ( 7 )}{cos ( 7 )}

= - \frac{sin ( 7 )}{cos ( 7 )}

قم بالعمل على الطرف الأيمن

tan (π - 7)

Rewrite using trig identities

tan (π - 7)

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

:Use the basic trigonometric identity

= \frac{sin ( π - 7 )}{cos ( π - 7 )}

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

:فعّل متطابقة الطرح لزوايا

= \frac{sin ( π ) cos ( 7 ) - cos ( π ) sin ( 7 )}{cos ( π - 7 )}

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

:فعّل متطابقة الطرح لزوايا

= \frac{sin ( π ) cos ( 7 ) - cos ( π ) sin ( 7 )}{cos ( π ) cos ( 7 ) + sin ( π ) sin ( 7 )}

\frac{sin ( π ) cos ( 7 ) - cos ( π ) sin ( 7 )}{cos ( π ) cos ( 7 ) + sin ( π ) sin ( 7 )} = - \frac{sin ( 7 )}{cos ( 7 )}

\frac{sin ( π ) cos ( 7 ) - cos ( π ) sin ( 7 )}{cos ( π ) cos ( 7 ) + sin ( π ) sin ( 7 )}

sin (π) cos (7) - cos (π) sin (7) = sin (7)

sin (π) cos (7) - cos (π) sin (7)

sin (π) cos (7) = 0

sin (π) cos (7)

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 (7)

0 \cdot a = 0

فعّل القانون

= 0

cos (π) sin (7) = - sin (7)

cos (π) sin (7)

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 \cdot sin (7)

1 \cdot sin (7) = sin (7) :

اضرب

= - sin (7)

= 0 - (- sin (7))

- (- a) = a

فعّل القانون

= 0 + sin (7)

0 + sin (7) = sin (7)

= sin (7)

= \frac{sin ( 7 )}{cos ( π ) cos ( 7 ) + sin ( π ) sin ( 7 )}

cos (π) cos (7) + sin (π) sin (7) = - cos (7)

cos (π) cos (7) + sin (π) sin (7)

cos (π) cos (7) = - cos (7)

cos (π) cos (7)

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 \cdot cos (7)

1 \cdot cos (7) = cos (7) :

اضرب

= - cos (7)

= - cos (7) + sin (π) sin (7)

sin (π) sin (7) = 0

sin (π) sin (7)

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 (7)

0 \cdot a = 0

فعّل القانون

= 0

= - cos (7) + 0

- cos (7) + 0 = - cos (7)

= - cos (7)

= \frac{sin ( 7 )}{- cos ( 7 )}

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

: استخدم ميزات الكسور التالية

= - \frac{sin ( 7 )}{cos ( 7 )}

= - \frac{sin ( 7 )}{cos ( 7 )}

= - \frac{sin ( 7 )}{cos ( 7 )}

أظهرنا بأنّ الطرفين من الممكن أن تكون لهما نفس الصورة

\Rightarrow صحيح

أمثلة شائعة

prove 25sec^2(5x)=25+25tan^2(5x)

p ro v e 25 sec^{2} (5 x) = 25 + 25 tan^{2} (5 x)

prove (sin(x)cos(x))/(1-cos^2(x))=cot(x)

p ro v e \frac{sin ( x ) cos ( x )}{1 - cos ^{2} ( x )} = cot (x)

prove tan(x-pi)=tan(x)

p ro v e tan (x - π) = tan (x)

prove 1/(csc(y)-cot(y))=csc(y)+cot(y)

p ro v e \frac{1}{csc ( y ) - cot ( y )} = csc (y) + cot (y)

prove tan^4(t)+tan^2(t)=sec^4(t)-sec^2(t)

p ro v e tan^{4} (t) + tan^{2} (t) = sec^{4} (t) - sec^{2} (t)