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شائع علم المثلثات >

cot^2(θ)-1=csc^2(θ)

الحلّ

cot^{2} (θ) - 1 = csc^{2} (θ)

الحلّ

θ \in R لا يوجد حلّ لـ

خطوات الحلّ

cot^{2} (θ) - 1 = csc^{2} (θ)

من الطرفين

csc^{2} (θ)

اطرح

cot^{2} (θ) - 1 - csc^{2} (θ) = 0

sin, cos :

عبّر بواسطة

- 1 + cot^{2} (θ) - csc^{2} (θ)

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

:Use the basic trigonometric identity

= - 1 + (\frac{cos ( θ )}{sin ( θ )})^{2} - csc^{2} (θ)

csc (x) = \frac{1}{sin ( x )}

:Use the basic trigonometric identity

= - 1 + (\frac{cos ( θ )}{sin ( θ )})^{2} - (\frac{1}{sin ( θ )})^{2}

- 1 + (\frac{cos ( θ )}{sin ( θ )})^{2} - (\frac{1}{sin ( θ )})^{2}

بسّط:

\frac{- sin ^{2} ( θ ) + cos ^{2} ( θ ) - 1}{sin ^{2} ( θ )}

- 1 + (\frac{cos ( θ )}{sin ( θ )})^{2} - (\frac{1}{sin ( θ )})^{2}

(\frac{cos ( θ )}{sin ( θ )})^{2} = \frac{cos ^{2} ( θ )}{sin ^{2} ( θ )}

(\frac{cos ( θ )}{sin ( θ )})^{2}

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

:فعّل قانون القوى

= \frac{cos ^{2} ( θ )}{sin ^{2} ( θ )}

(\frac{1}{sin ( θ )})^{2} = \frac{1}{sin ^{2} ( θ )}

(\frac{1}{sin ( θ )})^{2}

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

:فعّل قانون القوى

= \frac{1 ^{2}}{sin ^{2} ( θ )}

1^{a} = 1

فعّل القانون

1^{2} = 1

= \frac{1}{sin ^{2} ( θ )}

= - 1 + \frac{cos ^{2} ( θ )}{sin ^{2} ( θ )} - \frac{1}{sin ^{2} ( θ )}

\frac{cos ^{2} ( θ )}{sin ^{2} ( θ )} - \frac{1}{sin ^{2} ( θ )}

وحّد الكسور:

\frac{cos ^{2} ( θ ) - 1}{sin ^{2} ( θ )}

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

فعّل القانون

= \frac{cos ^{2} ( θ ) - 1}{sin ^{2} ( θ )}

= - 1 + \frac{cos ^{2} ( θ ) - 1}{sin ^{2} ( θ )}

1 = \frac{1 sin ^{2} ( θ )}{sin ^{2} ( θ )}

:حوّل الأعداد لكسور

= - \frac{1 \cdot sin ^{2} ( θ )}{sin ^{2} ( θ )} + \frac{cos ^{2} ( θ ) - 1}{sin ^{2} ( θ )}

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

:بما أنّ المقامات متساوية، اجمع البسوط

= \frac{- 1 \cdot sin ^{2} ( θ ) + cos ^{2} ( θ ) - 1}{sin ^{2} ( θ )}

1 \cdot sin^{2} (θ) = sin^{2} (θ) :

اضرب

= \frac{- sin ^{2} ( θ ) + cos ^{2} ( θ ) - 1}{sin ^{2} ( θ )}

= \frac{- sin ^{2} ( θ ) + cos ^{2} ( θ ) - 1}{sin ^{2} ( θ )}

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

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

- 1 + cos^{2} (θ) - sin^{2} (θ) = 0

Rewrite using trig identities

- 1 + cos^{2} (θ) - sin^{2} (θ)

cos^{2} (x) - sin^{2} (x) = cos (2 x)

:فعّل متطابقة الزاوية المضاعفة

= - 1 + cos (2 θ)

- 1 + cos (2 θ) = 0

انقل

1

إلى الجانب الأيمن

- 1 + cos (2 θ) = 0

للطرفين

1

أضف

- 1 + cos (2 θ) + 1 = 0 + 1

بسّط

cos (2 θ) = 1

cos (2 θ) = 1

cos (2 θ) = 1 :

حلول عامّة لـ

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}

2 θ = 0 + 2 πn

2 θ = 0 + 2 πn

2 θ = 0 + 2 πn

حلّ:

θ = πn

2 θ = 0 + 2 πn

0 + 2 πn = 2 πn

2 θ = 2 πn

2

اقسم الطرفين على

2 θ = 2 πn

2

اقسم الطرفين على

\frac{2 θ}{2} = \frac{2 πn}{2}

بسّط

θ = πn

θ = πn

θ = πn

πn :

بما أنّ المعادلة غير معرّفة لـ

θ \in R لا يوجد حلّ لـ

رسم

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