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شائع >

sec^2(θ)+csc^2(θ)=4

الحلّ

sec^{2} (θ) + csc^{2} (θ) = 4

الحلّ

θ = \frac{3 π}{4} + πn, θ = \frac{π}{4} + πn

درجات

θ = 13 5^{\circ} + 18 0^{\circ} n, θ = 4 5^{\circ} + 18 0^{\circ} n

خطوات الحلّ

sec^{2} (θ) + csc^{2} (θ) = 4

من الطرفين

4

اطرح

sec^{2} (θ) + csc^{2} (θ) - 4 = 0

sin, cos :

عبّر بواسطة

- 4 + csc^{2} (θ) + sec^{2} (θ)

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

:Use the basic trigonometric identity

= - 4 + (\frac{1}{sin ( θ )})^{2} + sec^{2} (θ)

sec (x) = \frac{1}{cos ( x )}

:Use the basic trigonometric identity

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

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

بسّط:

\frac{- 4 sin ^{2} ( θ ) cos ^{2} ( θ ) + cos ^{2} ( θ ) + sin ^{2} ( θ )}{sin ^{2} ( θ ) cos ^{2} ( θ )}

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

(\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 )}

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

(\frac{1}{cos ( θ )})^{2}

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

:فعّل قانون القوى

= \frac{1 ^{2}}{cos ^{2} ( θ )}

1^{a} = 1

فعّل القانون

1^{2} = 1

= \frac{1}{cos ^{2} ( θ )}

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

4 = \frac{4}{1}

:حوّل الأعداد لكسور

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

1, sin^{2} (θ), cos^{2} (θ)

المضاعف المشترك الأصغر لـ:

sin^{2} (θ) cos^{2} (θ)

1, sin^{2} (θ), cos^{2} (θ)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear in at least one of the factored expressions

= sin^{2} (θ) cos^{2} (θ)

اكتب مجددًا الكسور بحيث يكون المقام مشترك

sin^{2} (θ) cos^{2} (θ)

اضرب كل بسط ومقام بتعبير الذي يؤدّي إلى مقام مشترك

For

\frac{4}{1} :

multiply the denominator and numerator by

sin^{2} (θ) cos^{2} (θ)

\frac{4}{1} = \frac{4 sin ^{2} ( θ ) cos ^{2} ( θ )}{1 \cdot sin ^{2} ( θ ) cos ^{2} ( θ )} = \frac{4 sin ^{2} ( θ ) cos ^{2} ( θ )}{sin ^{2} ( θ ) cos ^{2} ( θ )}

For

\frac{1}{sin ^{2} ( θ )} :

multiply the denominator and numerator by

cos^{2} (θ)

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

For

\frac{1}{cos ^{2} ( θ )} :

multiply the denominator and numerator by

sin^{2} (θ)

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

= - \frac{4 sin ^{2} ( θ ) cos ^{2} ( θ )}{sin ^{2} ( θ ) cos ^{2} ( θ )} + \frac{cos ^{2} ( θ )}{sin ^{2} ( θ ) cos ^{2} ( θ )} + \frac{sin ^{2} ( θ )}{sin ^{2} ( θ ) cos ^{2} ( θ )}

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

:بما أنّ المقامات متساوية، اجمع البسوط

= \frac{- 4 sin ^{2} ( θ ) cos ^{2} ( θ ) + cos ^{2} ( θ ) + sin ^{2} ( θ )}{sin ^{2} ( θ ) cos ^{2} ( θ )}

= \frac{- 4 sin ^{2} ( θ ) cos ^{2} ( θ ) + cos ^{2} ( θ ) + sin ^{2} ( θ )}{sin ^{2} ( θ ) cos ^{2} ( θ )}

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

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

cos^{2} (θ) + sin^{2} (θ) - 4 cos^{2} (θ) sin^{2} (θ) = 0

Rewrite using trig identities

cos^{2} (θ) + sin^{2} (θ) - 4 cos^{2} (θ) sin^{2} (θ)

cos^{2} (x) + sin^{2} (x) = 1

:فعّل نطريّة فيتاغوروس

= - 4 cos^{2} (θ) sin^{2} (θ) + 1

1 - 4 cos^{2} (θ) sin^{2} (θ) = 0

1 - 4 cos^{2} (θ) sin^{2} (θ)

حلل إلى عوامل:

(1 + 2 sin (θ) cos (θ)) (1 - 2 sin (θ) cos (θ))

1 - 4 cos^{2} (θ) sin^{2} (θ)

1 - (2 sin (θ) cos (θ))^{2}

كـ

1 - 4 sin^{2} (θ) cos^{2} (θ)

اكتب مجددًا

1 - 4 sin^{2} (θ) cos^{2} (θ)

2^{2}

كـ

4

اكتب مجددًا

= 1 - 2^{2} sin^{2} (θ) cos^{2} (θ)

a^{m} b^{m} = (ab)^{m}

:فعّل قانون القوى

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

= 1 - (2 sin (θ) cos (θ))^{2}

= 1 - (2 sin (θ) cos (θ))^{2}

x^{2} - y^{2} = (x + y) (x - y)

فعّل قانون فرق المربّعات

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

= (2 sin (θ) cos (θ) + 1) (- 2 sin (θ) cos (θ) + 1)

(1 + 2 sin (θ) cos (θ)) (1 - 2 sin (θ) cos (θ)) = 0

حلّ كل جزء على حدة

1 + 2 sin (θ) cos (θ) = 0 or 1 - 2 sin (θ) cos (θ) = 0

1 + 2 sin (θ) cos (θ) = 0 : θ = \frac{3 π}{4} + πn

1 + 2 sin (θ) cos (θ) = 0

Rewrite using trig identities

1 + 2 sin (θ) cos (θ)

2 sin (x) cos (x) = sin (2 x)

:فعّل متطابقة الزاوية المضاعفة

= 1 + sin (2 θ)

1 + sin (2 θ) = 0

انقل

1

إلى الجانب الأيمن

1 + sin (2 θ) = 0

من الطرفين

1

اطرح

1 + sin (2 θ) - 1 = 0 - 1

بسّط

sin (2 θ) = - 1

sin (2 θ) = - 1

sin (2 θ) = - 1 :

حلول عامّة لـ

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}

2 θ = \frac{3 π}{2} + 2 πn

2 θ = \frac{3 π}{2} + 2 πn

2 θ = \frac{3 π}{2} + 2 πn

حلّ:

θ = \frac{3 π}{4} + πn

2 θ = \frac{3 π}{2} + 2 πn

2

اقسم الطرفين على

2 θ = \frac{3 π}{2} + 2 πn

2

اقسم الطرفين على

\frac{2 θ}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

بسّط

\frac{2 θ}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

\frac{2 θ}{2}

بسّط:

θ

\frac{2 θ}{2}

\frac{2}{2} = 1 :

اقسم الأعداد

= θ

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

بسّط:

\frac{3 π}{4} + πn

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{3 π}{2}}{2} = \frac{3 π}{4}

\frac{\frac{3 π}{2}}{2}

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

: استخدم ميزات الكسور التالية

= \frac{3 π}{2 \cdot 2}

2 \cdot 2 = 4 :

اضرب الأعداد

= \frac{3 π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

\frac{2}{2} = 1 :

اقسم الأعداد

= πn

= \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

1 - 2 sin (θ) cos (θ) = 0 : θ = \frac{π}{4} + πn

1 - 2 sin (θ) cos (θ) = 0

Rewrite using trig identities

1 - 2 sin (θ) cos (θ)

2 sin (x) cos (x) = sin (2 x)

:فعّل متطابقة الزاوية المضاعفة

= 1 - sin (2 θ)

1 - sin (2 θ) = 0

انقل

1

إلى الجانب الأيمن

1 - sin (2 θ) = 0

من الطرفين

1

اطرح

1 - sin (2 θ) - 1 = 0 - 1

بسّط

- sin (2 θ) = - 1

- sin (2 θ) = - 1

- 1

اقسم الطرفين على

- sin (2 θ) = - 1

- 1

اقسم الطرفين على

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

بسّط

sin (2 θ) = 1

sin (2 θ) = 1

sin (2 θ) = 1 :

حلول عامّة لـ

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}

2 θ = \frac{π}{2} + 2 πn

2 θ = \frac{π}{2} + 2 πn

2 θ = \frac{π}{2} + 2 πn

حلّ:

θ = \frac{π}{4} + πn

2 θ = \frac{π}{2} + 2 πn

2

اقسم الطرفين على

2 θ = \frac{π}{2} + 2 πn

2

اقسم الطرفين على

\frac{2 θ}{2} = \frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

بسّط

\frac{2 θ}{2} = \frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

\frac{2 θ}{2}

بسّط:

θ

\frac{2 θ}{2}

\frac{2}{2} = 1 :

اقسم الأعداد

= θ

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

بسّط:

\frac{π}{4} + πn

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

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

: استخدم ميزات الكسور التالية

= \frac{π}{2 \cdot 2}

2 \cdot 2 = 4 :

اضرب الأعداد

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

\frac{2}{2} = 1 :

اقسم الأعداد

= πn

= \frac{π}{4} + πn

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

وحّد الحلول

θ = \frac{3 π}{4} + πn, θ = \frac{π}{4} + πn

رسم

أمثلة شائعة

cos(θ/3)=sin(θ)

cos (\frac{θ}{3}) = sin (θ)

sin(x)-cos(x)= 1/5

sin (x) - cos (x) = \frac{1}{5}

tan(θ)=(-2)/3

tan (θ) = \frac{- 2}{3}

cos(105)cos(x)=-1/4

cos (10 5^{\circ}) cos (x^{\circ}) = - \frac{1}{4}

cos(θ)tan^2(θ)=3cos(θ)

cos (θ) tan^{2} (θ) = 3 cos (θ)