English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

prove sin^2(θ)(1+cot^2(θ))=1

الحلّ

prove

sin^{2} (θ) (1 + cot^{2} (θ)) = 1

صحيح

خطوات الحلّ

sin^{2} (θ) (1 + cot^{2} (θ)) = 1

قم بالعمل على الطرف الأيسر

sin^{2} (θ) (1 + cot^{2} (θ))

sin, cos :

عبّر بواسطة

(1 + cot^{2} (θ)) sin^{2} (θ)

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

:Use the basic trigonometric identity

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

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

بسّط:

sin^{2} (θ) + cos^{2} (θ)

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

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

:فعّل قانون القوى

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

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

وحّد:

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

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

1 \cdot sin^{2} (θ) = sin^{2} (θ) :

اضرب

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

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

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

:اضرب كسور

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

sin^{2} (θ) :

إلغ العوامل المشتركة

= sin^{2} (θ) + cos^{2} (θ)

= sin^{2} (θ) + cos^{2} (θ)

= cos^{2} (θ) + sin^{2} (θ)

Rewrite using trig identities

cos^{2} (θ) + sin^{2} (θ)

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

:فعّل نطريّة فيتاغوروس

= 1

= 1

أظهرنا بأنّ الطرفين من الممكن أن تكون لهما نفس الصورة

\Rightarrow صحيح

5 cos (6 x) + 6 = 9

sin (- \frac{5 π}{12})

4 sin (2 x) = 4 cos (x)

2 sin (x) - 3 cos (x) = 2

sec (0)