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شائع علم المثلثات >

sin^2(a)=1-cos(2a)

الحلّ

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

الحلّ

a = 2 πn, a = π + 2 πn

+1

درجات

a = 0^{\circ} + 36 0^{\circ} n, a = 18 0^{\circ} + 36 0^{\circ} n

خطوات الحلّ

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

من الطرفين

1 - cos (2 a)

اطرح

sin^{2} (a) - 1 + cos (2 a) = 0

Rewrite using trig identities

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

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

:فعّل متطابقة الزاوية المضاعفة

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

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

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

بسّط:

- 1 + cos^{2} (a)

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

- (cos (2 a) - cos^{2} (a)) : - cos (2 a) + cos^{2} (a)

- (cos (2 a) - cos^{2} (a))

افتح أقواس

= - (cos (2 a)) - (- cos^{2} (a))

فعّل قوانين سالب-موجب

- (- a) = a, - (a) = - a

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

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

cos (2 a) - cos (2 a) = 0 :

اجمع العناصر المتشابهة

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

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

- 1 + cos^{2} (a) = 0

بالاستعانة بطريقة التعويض

- 1 + cos^{2} (a) = 0

cos (a) = u :

على افتراض أنّ

- 1 + u^{2} = 0

- 1 + u^{2} = 0 : u = 1, u = - 1

- 1 + u^{2} = 0

انقل

1

إلى الجانب الأيمن

- 1 + u^{2} = 0

للطرفين

1

أضف

- 1 + u^{2} + 1 = 0 + 1

بسّط

u^{2} = 1

u^{2} = 1

x = f (a), - f (a)

الحلول هي

x^{2} = f (a)

لـ

u = 1, u = - 1

1 = 1

1

1 = 1

فعّل القانون

= 1

- 1 = - 1

- 1

1 = 1

فعّل القانون

= - 1

u = 1, u = - 1

u = cos (a)

استبدل مجددًا

cos (a) = 1, cos (a) = - 1

cos (a) = 1, cos (a) = - 1

cos (a) = 1 : a = 2 πn

cos (a) = 1

cos (a) = 1 :

حلول عامّة لـ

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}

a = 0 + 2 πn

a = 0 + 2 πn

a = 0 + 2 πn

حلّ:

a = 2 πn

a = 0 + 2 πn

0 + 2 πn = 2 πn

a = 2 πn

a = 2 πn

cos (a) = - 1 : a = π + 2 πn

cos (a) = - 1

cos (a) = - 1 :

حلول عامّة لـ

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}

a = π + 2 πn

a = π + 2 πn

وحّد الحلول

a = 2 πn, a = π + 2 πn

رسم

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