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العربية

prove cos(2B)=(1-tan^2(B))/(1+tan^2(B))

الحلّ

prove

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

صحيح

خطوات الحلّ

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

قم بالعمل على الطرف الأيمن

\frac{1 - tan ^{2} ( B )}{1 + tan ^{2} ( B )}

sin, cos :

عبّر بواسطة

\frac{1 - tan ^{2} ( B )}{1 + tan ^{2} ( B )}

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

:Use the basic trigonometric identity

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

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

بسّط:

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

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

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

:فعّل قانون القوى

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

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

:فعّل قانون القوى

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

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

وحّد:

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

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

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

:حوّل الأعداد لكسور

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

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

:بما أنّ المقامات متساوية، اجمع البسوط

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

1 \cdot cos^{2} (B) = cos^{2} (B) :

اضرب

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

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

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

وحّد:

\frac{cos ^{2} ( B ) - sin ^{2} ( B )}{cos ^{2} ( B )}

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

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

:حوّل الأعداد لكسور

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

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

:بما أنّ المقامات متساوية، اجمع البسوط

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

1 \cdot cos^{2} (B) = cos^{2} (B) :

اضرب

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

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

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

:اقسم الكسور

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

cos^{2} (B) :

إلغ العوامل المشتركة

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

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

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

Rewrite using trig identities

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

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

:فعّل نطريّة فيتاغوروس

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

\frac{a}{1} = a

فعّل القانون

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

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

:فعّل متطابقة الزاوية المضاعفة

= cos (2 B)

= cos (2 B)

أظهرنا بأنّ الطرفين من الممكن أن تكون لهما نفس الصورة

\Rightarrow صحيح

p ro v e \frac{1 + sec ( x )}{1 + cos ( x )} = sec (x)

p ro v e \frac{1 - sin ^{2} ( x )}{sin ( x )} = cot (x) cos (x)

p ro v e 7 cos^{2} (x) + 5 sin^{2} (x) = cos (2 x) + 6

p ro v e 1 + sec^{2} (x) = 2 + tan^{2} (x)

p ro v e 9 (1 + sin (y)) (1 + sin (- y)) = 9 cos^{2} (y)