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العربية

prove cos(2α)=-1+2/(1+tan^2(α))

الحلّ

prove

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

صحيح

خطوات الحلّ

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

قم بالعمل على الطرف الأيمن

- 1 + \frac{2}{1 + tan ^{2} ( α )}

sin, cos :

عبّر بواسطة

- 1 + \frac{2}{1 + tan ^{2} ( α )}

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

:Use the basic trigonometric identity

= - 1 + \frac{2}{1 + ( \frac{s i n ( α )}{c o s ( α )} ) ^{2}}

- 1 + \frac{2}{1 + ( \frac{s i n ( α )}{c o s ( α )} ) ^{2}}

بسّط:

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

- 1 + \frac{2}{1 + ( \frac{s i n ( α )}{c o s ( α )} ) ^{2}}

\frac{2}{1 + ( \frac{s i n ( α )}{c o s ( α )} ) ^{2}} = \frac{2 cos ^{2} ( α )}{cos ^{2} ( α ) + sin ^{2} ( α )}

\frac{2}{1 + ( \frac{s i n ( α )}{c o s ( α )} ) ^{2}}

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

:فعّل قانون القوى

= \frac{2}{1 + \frac{s i n ^{2} ( α )}{c o s ^{2} ( α )}}

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

وحّد:

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

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

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

:حوّل الأعداد لكسور

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

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

:بما أنّ المقامات متساوية، اجمع البسوط

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

1 \cdot cos^{2} (α) = cos^{2} (α) :

اضرب

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

= \frac{2}{\frac{c o s ^{2} ( α ) + s i n ^{2} ( α )}{c o s ^{2} ( α )}}

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

: استخدم ميزات الكسور التالية

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

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

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

:حوّل الأعداد لكسور

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

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

:بما أنّ المقامات متساوية، اجمع البسوط

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

1 \cdot (cos^{2} (α) + sin^{2} (α)) = (cos^{2} (α) + sin^{2} (α)) :

اضرب

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

- (cos^{2} (α) + sin^{2} (α)) + 2 cos^{2} (α)

وسٌع:

cos^{2} (α) - sin^{2} (α)

- (cos^{2} (α) + sin^{2} (α)) + 2 cos^{2} (α)

- (cos^{2} (α) + sin^{2} (α)) : - cos^{2} (α) - sin^{2} (α)

- (cos^{2} (α) + sin^{2} (α))

افتح أقواس

= - (cos^{2} (α)) - (sin^{2} (α))

فعّل قوانين سالب-موجب

+ (- a) = - a

= - cos^{2} (α) - sin^{2} (α)

= - cos^{2} (α) - sin^{2} (α) + 2 cos^{2} (α)

- cos^{2} (α) + 2 cos^{2} (α) = cos^{2} (α) :

اجمع العناصر المتشابهة

= cos^{2} (α) - sin^{2} (α)

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

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

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

Rewrite using trig identities

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

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

:فعّل نطريّة فيتاغوروس

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

\frac{a}{1} = a

فعّل القانون

= cos^{2} (α) - sin^{2} (α)

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

:فعّل متطابقة الزاوية المضاعفة

= cos (2 α)

= cos (2 α)

أظهرنا بأنّ الطرفين من الممكن أن تكون لهما نفس الصورة

\Rightarrow صحيح

p ro v e (1 + sec (x)) (csc (x) - cot (x)) = tan (x)

p ro v e 2 - csc^{2} (z) = 1 - cot^{2} (z)

p ro v e csc (\frac{π}{2} - u) = sec (u)

p ro v e - 2 sin (k - l) sin (k + l) = cos (2 k) - cos (2 l)

p ro v e cos (θ - 6 0^{\circ}) + cos (θ + 6 0^{\circ}) = cos (θ)