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((1-tan^2(a)))/((tan(a)))=2cot^2(a)

الحلّ

\frac{( 1 - tan ^{2} ( a ) )}{( tan ( a ) )} = 2 cot^{2} (a)

الحلّ

a = 2.15228 \dots + πn

درجات

a = 123.31684 \dots^{\circ} + 18 0^{\circ} n

خطوات الحلّ

\frac{( 1 - tan ^{2} ( a ) )}{( tan ( a ) )} = 2 cot^{2} (a)

من الطرفين

2 cot^{2} (a)

اطرح

\frac{1 - tan ^{2} ( a )}{tan ( a )} - 2 cot^{2} (a) = 0

\frac{1 - tan ^{2} ( a )}{tan ( a )} - 2 cot^{2} (a)

بسّط:

\frac{1 - tan ^{2} ( a ) - 2 cot ^{2} ( a ) tan ( a )}{tan ( a )}

\frac{1 - tan ^{2} ( a )}{tan ( a )} - 2 cot^{2} (a)

2 cot^{2} (a) = \frac{2 cot ^{2} ( a ) tan ( a )}{tan ( a )}

:حوّل الأعداد لكسور

= \frac{1 - tan ^{2} ( a )}{tan ( a )} - \frac{2 cot ^{2} ( a ) tan ( a )}{tan ( a )}

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

:بما أنّ المقامات متساوية، اجمع البسوط

= \frac{1 - tan ^{2} ( a ) - 2 cot ^{2} ( a ) tan ( a )}{tan ( a )}

\frac{1 - tan ^{2} ( a ) - 2 cot ^{2} ( a ) tan ( a )}{tan ( a )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

1 - tan^{2} (a) - 2 cot^{2} (a) tan (a) = 0

Rewrite using trig identities

1 - tan^{2} (a) - 2 cot^{2} (a) tan (a)

tan (x) = \frac{1}{cot ( x )}

:Use the basic trigonometric identity

= 1 - (\frac{1}{cot ( a )})^{2} - 2 cot^{2} (a) \frac{1}{cot ( a )}

1 - (\frac{1}{cot ( a )})^{2} - 2 cot^{2} (a) \frac{1}{cot ( a )}

بسّط:

1 - \frac{1}{cot ^{2} ( a )} - 2 cot (a)

1 - (\frac{1}{cot ( a )})^{2} - 2 cot^{2} (a) \frac{1}{cot ( a )}

(\frac{1}{cot ( a )})^{2} = \frac{1}{cot ^{2} ( a )}

(\frac{1}{cot ( a )})^{2}

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

:فعّل قانون القوى

= \frac{1 ^{2}}{cot ^{2} ( a )}

1^{a} = 1

فعّل القانون

1^{2} = 1

= \frac{1}{cot ^{2} ( a )}

2 cot^{2} (a) \frac{1}{cot ( a )} = 2 cot (a)

2 cot^{2} (a) \frac{1}{cot ( a )}

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

:اضرب كسور

= \frac{1 \cdot 2 cot ^{2} ( a )}{cot ( a )}

1 \cdot 2 = 2 :

اضرب الأعداد

= \frac{2 cot ^{2} ( a )}{cot ( a )}

cot (a) :

إلغ العوامل المشتركة

= 2 cot (a)

= 1 - \frac{1}{cot ^{2} ( a )} - 2 cot (a)

= 1 - \frac{1}{cot ^{2} ( a )} - 2 cot (a)

1 - \frac{1}{cot ^{2} ( a )} - 2 cot (a) = 0

بالاستعانة بطريقة التعويض

1 - \frac{1}{cot ^{2} ( a )} - 2 cot (a) = 0

cot (a) = u :

على افتراض أنّ

1 - \frac{1}{u ^{2}} - 2 u = 0

1 - \frac{1}{u ^{2}} - 2 u = 0 : u \approx - 0.65729 \dots

1 - \frac{1}{u ^{2}} - 2 u = 0

u^{2}

اضرب الطرفين بـ

1 - \frac{1}{u ^{2}} - 2 u = 0

u^{2}

اضرب الطرفين بـ

1 \cdot u^{2} - \frac{1}{u ^{2}} u^{2} - 2 u u^{2} = 0 \cdot u^{2}

بسّط

1 \cdot u^{2} - \frac{1}{u ^{2}} u^{2} - 2 u u^{2} = 0 \cdot u^{2}

1 \cdot u^{2}

بسّط:

u^{2}

1 \cdot u^{2}

1 \cdot u^{2} = u^{2} :

اضرب

= u^{2}

- \frac{1}{u ^{2}} u^{2}

بسّط:

- 1

- \frac{1}{u ^{2}} u^{2}

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

:اضرب كسور

= - \frac{1 \cdot u ^{2}}{u ^{2}}

u^{2} :

إلغ العوامل المشتركة

= - 1

- 2 u u^{2}

بسّط:

- 2 u^{3}

- 2 u u^{2}

a^{b} \cdot a^{c} = a^{b + c}

:فعّل قانون القوى

u u^{2} = u^{1 + 2}

= - 2 u^{1 + 2}

1 + 2 = 3 :

اجمع الأعداد

= - 2 u^{3}

0 \cdot u^{2}

بسّط:

0

0 \cdot u^{2}

0 \cdot a = 0

فعّل القانون

= 0

u^{2} - 1 - 2 u^{3} = 0

u^{2} - 1 - 2 u^{3} = 0

u^{2} - 1 - 2 u^{3} = 0

u^{2} - 1 - 2 u^{3} = 0

حلّ:

u \approx - 0.65729 \dots

u^{2} - 1 - 2 u^{3} = 0

a_{n} x^{n} + \dots + a_{1} x + a = 0

اكتب بالصورة الاعتياديّة

- 2 u^{3} + u^{2} - 1 = 0

بطريقة نيوتون ريبسون

- 2 u^{3} + u^{2} - 1 = 0

جدّ حلًا لـ:

u \approx - 0.65729 \dots

- 2 u^{3} + u^{2} - 1 = 0

تعريف تقريب نيوتن-ريبسون

f (u) = - 2 u^{3} + u^{2} - 1

f^{^{'}} (u)

جد:

- 6 u^{2} + 2 u

\frac{d}{d u} (- 2 u^{3} + u^{2} - 1)

(f \pm g)^{'} = f^{'} \pm g^{'}

:استعمل قانون الجمع

= - \frac{d}{d u} (2 u^{3}) + \frac{d}{d u} (u^{2}) - \frac{d}{d u} (1)

\frac{d}{d u} (2 u^{3}) = 6 u^{2}

\frac{d}{d u} (2 u^{3})

(a \cdot f)^{'} = a \cdot f^{'}

:استخرج الثابت

= 2 \frac{d}{d u} (u^{3})

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

:استعمل قانون الأسس

= 2 \cdot 3 u^{3 - 1}

بسّط

= 6 u^{2}

\frac{d}{d u} (u^{2}) = 2 u

\frac{d}{d u} (u^{2})

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

:استعمل قانون الأسس

= 2 u^{2 - 1}

بسّط

= 2 u

\frac{d}{d u} (1) = 0

\frac{d}{d u} (1)

\frac{d}{d x} (a) = 0

:مشتقة الثابت

= 0

= - 6 u^{2} + 2 u - 0

بسّط

= - 6 u^{2} + 2 u

u_{0} = - 1

استبدل

Δ u_{n + 1} < 0.000001

حتّى

u_{n + 1}

احسب

u_{1} = - 0.75 : Δ u_{1} = 0.25

f (u_{0}) = - 2 (- 1)^{3} + (- 1)^{2} - 1 = 2

f^{^{'}} (u_{0}) = - 6 (- 1)^{2} + 2 (- 1) = - 8

u_{1} = - 0.75

Δ u_{1} = ∣ - 0.75 - (- 1) ∣ = 0.25

Δ u_{1} = 0.25

u_{2} = - 0.66666 \dots : Δ u_{2} = 0.08333 \dots

f (u_{1}) = - 2 (- 0.75)^{3} + (- 0.75)^{2} - 1 = 0.40625

f^{^{'}} (u_{1}) = - 6 (- 0.75)^{2} + 2 (- 0.75) = - 4.875

u_{2} = - 0.66666 \dots

Δ u_{2} = ∣ - 0.66666 \dots - (- 0.75) ∣ = 0.08333 \dots

Δ u_{2} = 0.08333 \dots

u_{3} = - 0.65740 \dots : Δ u_{3} = 0.00925 \dots

f (u_{2}) = - 2 (- 0.66666 \dots)^{3} + (- 0.66666 \dots)^{2} - 1 = 0.03703 \dots

f^{^{'}} (u_{2}) = - 6 (- 0.66666 \dots)^{2} + 2 (- 0.66666 \dots) = - 4

u_{3} = - 0.65740 \dots

Δ u_{3} = ∣ - 0.65740 \dots - (- 0.66666 \dots) ∣ = 0.00925 \dots

Δ u_{3} = 0.00925 \dots

u_{4} = - 0.65729 \dots : Δ u_{4} = 0.00010 \dots

f (u_{3}) = - 2 (- 0.65740 \dots)^{3} + (- 0.65740 \dots)^{2} - 1 = 0.00042 \dots

f^{^{'}} (u_{3}) = - 6 (- 0.65740 \dots)^{2} + 2 (- 0.65740 \dots) = - 3.90792 \dots

u_{4} = - 0.65729 \dots

Δ u_{4} = ∣ - 0.65729 \dots - (- 0.65740 \dots) ∣ = 0.00010 \dots

Δ u_{4} = 0.00010 \dots

u_{5} = - 0.65729 \dots : Δ u_{5} = 1.51148 E - 8

f (u_{4}) = - 2 (- 0.65729 \dots)^{3} + (- 0.65729 \dots)^{2} - 1 = 5.90512 E - 8

f^{^{'}} (u_{4}) = - 6 (- 0.65729 \dots)^{2} + 2 (- 0.65729 \dots) = - 3.90684 \dots

u_{5} = - 0.65729 \dots

Δ u_{5} = ∣ - 0.65729 \dots - (- 0.65729 \dots) ∣ = 1.51148 E - 8

Δ u_{5} = 1.51148 E - 8

u \approx - 0.65729 \dots

فعّل القسمة الطويلة:

\frac{- 2 u ^{3} + u ^{2} - 1}{u + 0.65729 \dots} = - 2 u^{2} + 2.31459 \dots u - 1.52137 \dots

- 2 u^{2} + 2.31459 \dots u - 1.52137 \dots \approx 0

بطريقة نيوتون ريبسون

- 2 u^{2} + 2.31459 \dots u - 1.52137 \dots = 0

جدّ حلًا لـ:

u \in R

لا يوجد حلّ لـ

- 2 u^{2} + 2.31459 \dots u - 1.52137 \dots = 0

تعريف تقريب نيوتن-ريبسون

f (u) = - 2 u^{2} + 2.31459 \dots u - 1.52137 \dots

f^{^{'}} (u)

جد:

- 4 u + 2.31459 \dots

\frac{d}{d u} (- 2 u^{2} + 2.31459 \dots u - 1.52137 \dots)

(f \pm g)^{'} = f^{'} \pm g^{'}

:استعمل قانون الجمع

= - \frac{d}{d u} (2 u^{2}) + \frac{d}{d u} (2.31459 \dots u) - \frac{d}{d u} (1.52137 \dots)

\frac{d}{d u} (2 u^{2}) = 4 u

\frac{d}{d u} (2 u^{2})

(a \cdot f)^{'} = a \cdot f^{'}

:استخرج الثابت

= 2 \frac{d}{d u} (u^{2})

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

:استعمل قانون الأسس

= 2 \cdot 2 u^{2 - 1}

بسّط

= 4 u

\frac{d}{d u} (2.31459 \dots u) = 2.31459 \dots

\frac{d}{d u} (2.31459 \dots u)

(a \cdot f)^{'} = a \cdot f^{'}

:استخرج الثابت

= 2.31459 \dots \frac{d u}{d u}

\frac{d u}{d u} = 1

:استعمل المشتقة الأساسية

= 2.31459 \dots \cdot 1

بسّط

= 2.31459 \dots

\frac{d}{d u} (1.52137 \dots) = 0

\frac{d}{d u} (1.52137 \dots)

\frac{d}{d x} (a) = 0

:مشتقة الثابت

= 0

= - 4 u + 2.31459 \dots - 0

بسّط

= - 4 u + 2.31459 \dots

u_{0} = 1

استبدل

Δ u_{n + 1} < 0.000001

حتّى

u_{n + 1}

احسب

u_{1} = 0.28397 \dots : Δ u_{1} = 0.71602 \dots

f (u_{0}) = - 2 \cdot 1^{2} + 2.31459 \dots \cdot 1 - 1.52137 \dots = - 1.20678 \dots

f^{^{'}} (u_{0}) = - 4 \cdot 1 + 2.31459 \dots = - 1.68540 \dots

u_{1} = 0.28397 \dots

Δ u_{1} = ∣ 0.28397 \dots - 1 ∣ = 0.71602 \dots

Δ u_{1} = 0.71602 \dots

u_{2} = 1.15391 \dots : Δ u_{2} = 0.86993 \dots

f (u_{1}) = - 2 \cdot 0.28397 \dots^{2} + 2.31459 \dots \cdot 0.28397 \dots - 1.52137 \dots = - 1.02537 \dots

f^{^{'}} (u_{1}) = - 4 \cdot 0.28397 \dots + 2.31459 \dots = 1.17867 \dots

u_{2} = 1.15391 \dots

Δ u_{2} = ∣ 1.15391 \dots - 0.28397 \dots ∣ = 0.86993 \dots

Δ u_{2} = 0.86993 \dots

u_{3} = 0.49614 \dots : Δ u_{3} = 0.65777 \dots

f (u_{2}) = - 2 \cdot 1.15391 \dots^{2} + 2.31459 \dots \cdot 1.15391 \dots - 1.52137 \dots = - 1.51356 \dots

f^{^{'}} (u_{2}) = - 4 \cdot 1.15391 \dots + 2.31459 \dots = - 2.30105 \dots

u_{3} = 0.49614 \dots

Δ u_{3} = ∣ 0.49614 \dots - 1.15391 \dots ∣ = 0.65777 \dots

Δ u_{3} = 0.65777 \dots

u_{4} = 3.11809 \dots : Δ u_{4} = 2.62195 \dots

f (u_{3}) = - 2 \cdot 0.49614 \dots^{2} + 2.31459 \dots \cdot 0.49614 \dots - 1.52137 \dots = - 0.86532 \dots

f^{^{'}} (u_{3}) = - 4 \cdot 0.49614 \dots + 2.31459 \dots = 0.33003 \dots

u_{4} = 3.11809 \dots

Δ u_{4} = ∣ 3.11809 \dots - 0.49614 \dots ∣ = 2.62195 \dots

Δ u_{4} = 2.62195 \dots

u_{5} = 1.76452 \dots : Δ u_{5} = 1.35357 \dots

f (u_{4}) = - 2 \cdot 3.11809 \dots^{2} + 2.31459 \dots \cdot 3.11809 \dots - 1.52137 \dots = - 13.74928 \dots

f^{^{'}} (u_{4}) = - 4 \cdot 3.11809 \dots + 2.31459 \dots = - 10.15778 \dots

u_{5} = 1.76452 \dots

Δ u_{5} = ∣ 1.76452 \dots - 3.11809 \dots ∣ = 1.35357 \dots

Δ u_{5} = 1.35357 \dots

u_{6} = 0.99203 \dots : Δ u_{6} = 0.77249 \dots

f (u_{5}) = - 2 \cdot 1.76452 \dots^{2} + 2.31459 \dots \cdot 1.76452 \dots - 1.52137 \dots = - 3.66430 \dots

f^{^{'}} (u_{5}) = - 4 \cdot 1.76452 \dots + 2.31459 \dots = - 4.74350 \dots

u_{6} = 0.99203 \dots

Δ u_{6} = ∣ 0.99203 \dots - 1.76452 \dots ∣ = 0.77249 \dots

Δ u_{6} = 0.77249 \dots

u_{7} = 0.27025 \dots : Δ u_{7} = 0.72177 \dots

f (u_{6}) = - 2 \cdot 0.99203 \dots^{2} + 2.31459 \dots \cdot 0.99203 \dots - 1.52137 \dots = - 1.19348 \dots

f^{^{'}} (u_{6}) = - 4 \cdot 0.99203 \dots + 2.31459 \dots = - 1.65353 \dots

u_{7} = 0.27025 \dots

Δ u_{7} = ∣ 0.27025 \dots - 0.99203 \dots ∣ = 0.72177 \dots

Δ u_{7} = 0.72177 \dots

u_{8} = 1.11489 \dots : Δ u_{8} = 0.84464 \dots

f (u_{7}) = - 2 \cdot 0.27025 \dots^{2} + 2.31459 \dots \cdot 0.27025 \dots - 1.52137 \dots = - 1.04192 \dots

f^{^{'}} (u_{7}) = - 4 \cdot 0.27025 \dots + 2.31459 \dots = 1.23356 \dots

u_{8} = 1.11489 \dots

Δ u_{8} = ∣ 1.11489 \dots - 0.27025 \dots ∣ = 0.84464 \dots

Δ u_{8} = 0.84464 \dots

u_{9} = 0.44970 \dots : Δ u_{9} = 0.66519 \dots

f (u_{8}) = - 2 \cdot 1.11489 \dots^{2} + 2.31459 \dots \cdot 1.11489 \dots - 1.52137 \dots = - 1.42683 \dots

f^{^{'}} (u_{8}) = - 4 \cdot 1.11489 \dots + 2.31459 \dots = - 2.14500 \dots

u_{9} = 0.44970 \dots

Δ u_{9} = ∣ 0.44970 \dots - 1.11489 \dots ∣ = 0.66519 \dots

Δ u_{9} = 0.66519 \dots

u_{10} = 2.16551 \dots : Δ u_{10} = 1.71580 \dots

f (u_{9}) = - 2 \cdot 0.44970 \dots^{2} + 2.31459 \dots \cdot 0.44970 \dots - 1.52137 \dots = - 0.88496 \dots

f^{^{'}} (u_{9}) = - 4 \cdot 0.44970 \dots + 2.31459 \dots = 0.51576 \dots

u_{10} = 2.16551 \dots

Δ u_{10} = ∣ 2.16551 \dots - 0.44970 \dots ∣ = 1.71580 \dots

Δ u_{10} = 1.71580 \dots

لا يمكن إيجاد حلّ

الحل للمعادلة هو

u \approx - 0.65729 \dots

u \approx - 0.65729 \dots

افحص الإجبات

جد نقاط غير معرّفة:

u = 0

وقم بمساواتها لصفر

1 - \frac{1}{u ^{2}} - 2 u

خذ المقامات في

u^{2} = 0

حلّ:

u = 0

u^{2} = 0

x^{n} = 0 \Rightarrow x = 0

فعّل القانون

u = 0

النقاط التالية غير معرّفة

u = 0

ضمّ النقاط غير المعرّفة مع الحلول

u \approx - 0.65729 \dots

u = cot (a)

استبدل مجددًا

cot (a) \approx - 0.65729 \dots

cot (a) \approx - 0.65729 \dots

cot (a) = - 0.65729 \dots : a = arccot (- 0.65729 \dots) + πn

cot (a) = - 0.65729 \dots

Apply trig inverse properties

cot (a) = - 0.65729 \dots

cot (a) = - 0.65729 \dots :

حلول عامّة لـ

cot (x) = - a \Rightarrow x = arccot (- a) + πn

a = arccot (- 0.65729 \dots) + πn

a = arccot (- 0.65729 \dots) + πn

وحّد الحلول

a = arccot (- 0.65729 \dots) + πn

أظهر الحلّ بالتمثيل العشريّ

a = 2.15228 \dots + πn

رسم

أمثلة شائعة

cos(6x)=cos(2x)

cos (6 x) = cos (2 x)

cot(x)cos(x)-sin(x)=1

cot (x) cos (x) - sin (x) = 1

cos^4(x)+cos^3(x)-2=0

cos^{4} (x) + cos^{3} (x) - 2 = 0

tan^2(x)= 1/(cos(x)+1)

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

(sin^2(x)-2cos(x)+1)/4 =0

\frac{sin ^{2} ( x ) - 2 cos ( x ) + 1}{4} = 0