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شائع علم المثلثات >

prove (sin(θ))/(1+cos(θ))+cot(θ)=csc(θ)

الحلّ

prove

\frac{sin ( θ )}{1 + cos ( θ )} + cot (θ) = csc (θ)

الحلّ

صحيح

خطوات الحلّ

\frac{sin ( θ )}{1 + cos ( θ )} + cot (θ) = csc (θ)

قم بالعمل على الطرف الأيسر

\frac{sin ( θ )}{1 + cos ( θ )} + cot (θ)

sin, cos :

عبّر بواسطة

cot (θ) + \frac{sin ( θ )}{1 + cos ( θ )}

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

:Use the basic trigonometric identity

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

\frac{cos ( θ )}{sin ( θ )} + \frac{sin ( θ )}{1 + cos ( θ )}

بسّط:

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

\frac{cos ( θ )}{sin ( θ )} + \frac{sin ( θ )}{1 + cos ( θ )}

sin (θ), 1 + cos (θ)

المضاعف المشترك الأصغر لـ:

sin (θ) (cos (θ) + 1)

sin (θ), 1 + cos (θ)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

sin (θ)

1 + cos (θ)

= sin (θ) (cos (θ) + 1)

اكتب مجددًا الكسور بحيث يكون المقام مشترك

sin (θ) (cos (θ) + 1)

اضرب كل بسط ومقام بتعبير الذي يؤدّي إلى مقام مشترك

For

\frac{cos ( θ )}{sin ( θ )} :

multiply the denominator and numerator by

cos (θ) + 1

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

For

\frac{sin ( θ )}{1 + cos ( θ )} :

multiply the denominator and numerator by

sin (θ)

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

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

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

:بما أنّ المقامات متساوية، اجمع البسوط

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

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

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

Rewrite using trig identities

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

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

:فعّل نطريّة فيتاغوروس

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

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

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

بسّط:

\frac{1}{sin ( θ )}

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

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

وسٌع:

cos (θ) + 1

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

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

cos (θ) (1 + cos (θ))

وسٌع:

cos (θ) + cos^{2} (θ)

cos (θ) (1 + cos (θ))

a (b + c) = ab + a c

: افتح أقواس بالاستعانة بـ

a = cos (θ), b = 1, c = cos (θ)

= cos (θ) \cdot 1 + cos (θ) cos (θ)

= 1 \cdot cos (θ) + cos (θ) cos (θ)

1 \cdot cos (θ) + cos (θ) cos (θ)

بسّط:

cos (θ) + cos^{2} (θ)

1 \cdot cos (θ) + cos (θ) cos (θ)

1 \cdot cos (θ) = cos (θ)

1 \cdot cos (θ)

1 \cdot cos (θ) = cos (θ) :

اضرب

= cos (θ)

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

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

:فعّل قانون القوى

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

1 + 1 = 2 :

اجمع الأعداد

= cos^{2} (θ)

= cos (θ) + cos^{2} (θ)

= cos (θ) + cos^{2} (θ)

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

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

بسّط:

cos (θ) + 1

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

جمّع التعابير المتشابهة

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

- cos^{2} (θ) + cos^{2} (θ) = 0 :

اجمع العناصر المتشابهة

= cos (θ) + 1

= cos (θ) + 1

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

cos (θ) + 1 :

إلغ العوامل المشتركة

= \frac{1}{sin ( θ )}

= \frac{1}{sin ( θ )}

= \frac{1}{sin ( θ )}

Rewrite using trig identities

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

:Use the basic trigonometric identity

\frac{1}{\frac{1}{c s c ( θ )}}

بسّط

\frac{1}{\frac{1}{c s c ( θ )}}

\frac{1}{\frac{b}{c}} = \frac{c}{b}

: استخدم ميزات الكسور التالية

= \frac{csc ( θ )}{1}

\frac{a}{1} = a

فعّل القانون

= csc (θ)

csc (θ)

csc (θ)

أظهرنا بأنّ الطرفين من الممكن أن تكون لهما نفس الصورة

\Rightarrow صحيح

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