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شائع علم المثلثات >

prove tan(pi/4-θ)=(cos(2θ))/(1+sin(2θ))

الحلّ

prove

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

الحلّ

صحيح

خطوات الحلّ

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

قم بالعمل على الطرف الأيسر

tan (\frac{π}{4} - θ)

Rewrite using trig identities

tan (\frac{π}{4} - θ)

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

:Use the basic trigonometric identity

= \frac{sin ( \frac{π}{4} - θ )}{cos ( \frac{π}{4} - θ )}

sin (s - t) = sin (s) cos (t) - cos (s) sin (t)

:فعّل متطابقة الطرح لزوايا

= \frac{sin ( \frac{π}{4} ) cos ( θ ) - cos ( \frac{π}{4} ) sin ( θ )}{cos ( \frac{π}{4} - θ )}

cos (s - t) = cos (s) cos (t) + sin (s) sin (t)

:فعّل متطابقة الطرح لزوايا

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

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

بسّط:

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

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

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

sin (\frac{π}{4}) cos (θ) - cos (\frac{π}{4}) sin (θ)

sin (\frac{π}{4})

بسّط:

\frac{2}{2}

sin (\frac{π}{4})

استخدم المتطابقة الأساسيّة التالية:

sin (\frac{π}{4}) = \frac{2}{2}

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}

= \frac{2}{2}

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

cos (\frac{π}{4})

بسّط:

\frac{2}{2}

cos (\frac{π}{4})

استخدم المتطابقة الأساسيّة التالية:

cos (\frac{π}{4}) = \frac{2}{2}

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{2}{2}

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

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

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

cos (\frac{π}{4}) cos (θ) + sin (\frac{π}{4}) sin (θ)

cos (\frac{π}{4})

بسّط:

\frac{2}{2}

cos (\frac{π}{4})

استخدم المتطابقة الأساسيّة التالية:

cos (\frac{π}{4}) = \frac{2}{2}

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{2}{2}

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

sin (\frac{π}{4})

بسّط:

\frac{2}{2}

sin (\frac{π}{4})

استخدم المتطابقة الأساسيّة التالية:

sin (\frac{π}{4}) = \frac{2}{2}

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}

= \frac{2}{2}

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

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

\frac{2}{2} cos (θ)

اضرب بـ:

\frac{2 cos ( θ )}{2}

\frac{2}{2} cos (θ)

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

:اضرب كسور

= \frac{2 cos ( θ )}{2}

= \frac{\frac{2}{2} cos ( θ ) - \frac{2}{2} sin ( θ )}{\frac{2 c o s ( θ )}{2} + \frac{2}{2} sin ( θ )}

\frac{2}{2} sin (θ)

اضرب بـ:

\frac{2 sin ( θ )}{2}

\frac{2}{2} sin (θ)

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

:اضرب كسور

= \frac{2 sin ( θ )}{2}

= \frac{\frac{2}{2} cos ( θ ) - \frac{2}{2} sin ( θ )}{\frac{2 c o s ( θ )}{2} + \frac{2 s i n ( θ )}{2}}

\frac{2}{2} cos (θ)

اضرب بـ:

\frac{2 cos ( θ )}{2}

\frac{2}{2} cos (θ)

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

:اضرب كسور

= \frac{2 cos ( θ )}{2}

= \frac{\frac{2 c o s ( θ )}{2} - \frac{2}{2} sin ( θ )}{\frac{2 c o s ( θ )}{2} + \frac{2 s i n ( θ )}{2}}

\frac{2}{2} sin (θ)

اضرب بـ:

\frac{2 sin ( θ )}{2}

\frac{2}{2} sin (θ)

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

:اضرب كسور

= \frac{2 sin ( θ )}{2}

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

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

وحّد الكسور:

\frac{2 cos ( θ ) + 2 sin ( θ )}{2}

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

فعّل القانون

= \frac{2 cos ( θ ) + 2 sin ( θ )}{2}

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

\frac{2 cos ( θ )}{2} - \frac{2 sin ( θ )}{2}

وحّد الكسور:

\frac{2 cos ( θ ) - 2 sin ( θ )}{2}

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

فعّل القانون

= \frac{2 cos ( θ ) - 2 sin ( θ )}{2}

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

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

:اقسم الكسور

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

2 :

إلغ العوامل المشتركة

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

2

قم باخراج العامل المشترك

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

2

قم باخراج العامل المشترك

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

2 :

إلغ العوامل المشتركة

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

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

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

\frac{cos ( 2 θ ) ( cos ( θ ) - sin ( θ ) )}{cos ( 2 θ ) ( cos ( θ ) - sin ( θ ) )}

اضرب بـ

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

(cos (θ) - sin (θ)) (cos (θ) - sin (θ)) cos (2 θ)

وسٌع:

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

(cos (θ) - sin (θ)) (cos (θ) - sin (θ)) cos (2 θ)

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

:فعّل قانون القوى

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

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

1 + 1 = 2 :

اجمع الأعداد

= (cos (θ) - sin (θ))^{2} cos (2 θ)

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

(cos (θ) - sin (θ))^{2}

(a - b)^{2} = a^{2} - 2 ab + b^{2}

:فعّل صيغة الضرب المختصر

a = cos (θ), b = sin (θ)

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

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

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

فعّل قانون ضرب الأقواس

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

فعّل قوانين سالب-موجب

+ (- a) = - a

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

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

Rewrite using trig identities

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

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

:فعّل نطريّة فيتاغوروس

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

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

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

بسّط:

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

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

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

حلل إلى عوامل:

cos (2 θ) (- 2 cos (θ) sin (θ) + 1)

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

cos (2 θ)

قم باخراج العامل المشترك

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

بسّط

= cos (2 θ) (- 2 cos (θ) sin (θ) + 1)

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

cos (2 θ) :

إلغ العوامل المشتركة

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

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

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

:فعّل متطابقة الزاوية المضاعفة

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

(cos (θ) + sin (θ)) (cos (θ) - sin (θ))

وسٌع:

cos^{2} (θ) - sin^{2} (θ)

(cos (θ) + sin (θ)) (cos (θ) - sin (θ))

(a + b) (a - b) = a^{2} - b^{2}

فعّل قانون فرق المربّعات

a = cos (θ), b = sin (θ)

= cos^{2} (θ) - sin^{2} (θ)

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

cos^{2} (θ) - sin^{2} (θ) = cos (2 θ)

:فعّل متطابقة الزاوية المضاعفة

= \frac{1 - sin ( 2 θ )}{cos ( 2 θ )}

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

اضرب بـ

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

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

فعّل قانون فرق المربّعات

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

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

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

:فعّل نطريّة فيتاغوروس

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

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

cos (2 θ) :

إلغ العوامل المشتركة

= \frac{cos ( 2 θ )}{1 + sin ( 2 θ )}

= \frac{cos ( 2 θ )}{1 + sin ( 2 θ )}

أظهرنا بأنّ الطرفين من الممكن أن تكون لهما نفس الصورة

\Rightarrow صحيح

أمثلة شائعة

prove (1-sin(2x))/(cos(2x))=(cos(2x))/(1+sin(2x))

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

prove (csc(α)+cot(α))(sec(α)-1)=tan(α)

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

prove 1-2sin^2(θ)=2cos^2(θ)-1

p ro v e 1 - 2 sin^{2} (θ) = 2 cos^{2} (θ) - 1

prove sin(pi/4)=cos(pi/4)

p ro v e sin (\frac{π}{4}) = cos (\frac{π}{4})

prove sin(3a)+sin(a)=4sin(a)-4sin(3a)

p ro v e sin (3 a) + sin (a) = 4 sin (a) - 4 sin (3 a)