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شائع علم المثلثات >

2cos^2(a)=1+cos(120)

الحلّ

2 cos^{2} (a) = 1 + cos (12 0^{\circ})

الحلّ

a = 6 0^{\circ} + 36 0^{\circ} n, a = 30 0^{\circ} + 36 0^{\circ} n, a = 12 0^{\circ} + 36 0^{\circ} n, a = 24 0^{\circ} + 36 0^{\circ} n

راديان

a = \frac{π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn, a = \frac{2 π}{3} + 2 πn, a = \frac{4 π}{3} + 2 πn

خطوات الحلّ

2 cos^{2} (a) = 1 + cos (12 0^{\circ})

cos (12 0^{\circ}) = - \frac{1}{2}

cos (12 0^{\circ})

استخدم المتطابقة الأساسيّة التالية:

cos (12 0^{\circ}) = - \frac{1}{2}

cos (12 0^{\circ})

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= - \frac{1}{2}

= - \frac{1}{2}

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

بالاستعانة بطريقة التعويض

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

cos (a) = u :

على افتراض أنّ

2 u^{2} = 1 + - \frac{1}{2}

2 u^{2} = 1 + - \frac{1}{2} : u = \frac{1}{2}, u = - \frac{1}{2}

2 u^{2} = 1 + - \frac{1}{2}

2

اقسم الطرفين على

2 u^{2} = 1 + - \frac{1}{2}

2

اقسم الطرفين على

\frac{2 u ^{2}}{2} = \frac{1}{2} - \frac{\frac{1}{2}}{2}

بسّط

\frac{2 u ^{2}}{2} = \frac{1}{2} - \frac{\frac{1}{2}}{2}

\frac{2 u ^{2}}{2}

بسّط:

u^{2}

\frac{2 u ^{2}}{2}

\frac{2}{2} = 1 :

اقسم الأعداد

= u^{2}

\frac{1}{2} - \frac{\frac{1}{2}}{2}

بسّط:

\frac{1}{4}

\frac{1}{2} - \frac{\frac{1}{2}}{2}

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

فعّل القانون

= \frac{1 - \frac{1}{2}}{2}

1 - \frac{1}{2}

وحّد:

\frac{1}{2}

1 - \frac{1}{2}

1 = \frac{1 \cdot 2}{2}

:حوّل الأعداد لكسور

= \frac{1 \cdot 2}{2} - \frac{1}{2}

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

:بما أنّ المقامات متساوية، اجمع البسوط

= \frac{1 \cdot 2 - 1}{2}

1 \cdot 2 - 1 = 1

1 \cdot 2 - 1

1 \cdot 2 = 2 :

اضرب الأعداد

= 2 - 1

2 - 1 = 1 :

اطرح الأعداد

= 1

= \frac{1}{2}

= \frac{\frac{1}{2}}{2}

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

: استخدم ميزات الكسور التالية

= \frac{1}{2 \cdot 2}

2 \cdot 2 = 4 :

اضرب الأعداد

= \frac{1}{4}

u^{2} = \frac{1}{4}

u^{2} = \frac{1}{4}

u^{2} = \frac{1}{4}

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

الحلول هي

x^{2} = f (a)

لـ

u = \frac{1}{4}, u = - \frac{1}{4}

\frac{1}{4} = \frac{1}{2}

\frac{1}{4}

a \geq 0, b \geq 0

بافتراض أنّ

n \frac{a}{b} = \frac{n a}{n b}

:فعّل قانون الجذور

= \frac{1}{4}

4 = 2

4

4 = 2^{2} :

حلّل العدد لعوامله أوّليّة

= 2^{2}

n a^{n} = a

:فعْل قانون الجذور

2^{2} = 2

= 2

= \frac{1}{2}

1 = 1

فعّل القانون

= \frac{1}{2}

- \frac{1}{4} = - \frac{1}{2}

- \frac{1}{4}

\frac{1}{4}

بسّط:

\frac{1}{2}

\frac{1}{4}

a \geq 0, b \geq 0

بافتراض أنّ

n \frac{a}{b} = \frac{n a}{n b}

:فعّل قانون الجذور

= \frac{1}{4}

4 = 2

4

4 = 2^{2} :

حلّل العدد لعوامله أوّليّة

= 2^{2}

n a^{n} = a

:فعْل قانون الجذور

2^{2} = 2

= 2

= \frac{1}{2}

= - \frac{1}{2}

1 = 1

فعّل القانون

= - \frac{1}{2}

u = \frac{1}{2}, u = - \frac{1}{2}

u = cos (a)

استبدل مجددًا

cos (a) = \frac{1}{2}, cos (a) = - \frac{1}{2}

cos (a) = \frac{1}{2}, cos (a) = - \frac{1}{2}

cos (a) = \frac{1}{2} : a = 6 0^{\circ} + 36 0^{\circ} n, a = 30 0^{\circ} + 36 0^{\circ} n

cos (a) = \frac{1}{2}

cos (a) = \frac{1}{2} :

حلول عامّة لـ

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

a = 6 0^{\circ} + 36 0^{\circ} n, a = 30 0^{\circ} + 36 0^{\circ} n

a = 6 0^{\circ} + 36 0^{\circ} n, a = 30 0^{\circ} + 36 0^{\circ} n

cos (a) = - \frac{1}{2} : a = 12 0^{\circ} + 36 0^{\circ} n, a = 24 0^{\circ} + 36 0^{\circ} n

cos (a) = - \frac{1}{2}

cos (a) = - \frac{1}{2} :

حلول عامّة لـ

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

a = 12 0^{\circ} + 36 0^{\circ} n, a = 24 0^{\circ} + 36 0^{\circ} n

a = 12 0^{\circ} + 36 0^{\circ} n, a = 24 0^{\circ} + 36 0^{\circ} n

وحّد الحلول

a = 6 0^{\circ} + 36 0^{\circ} n, a = 30 0^{\circ} + 36 0^{\circ} n, a = 12 0^{\circ} + 36 0^{\circ} n, a = 24 0^{\circ} + 36 0^{\circ} n

رسم

أمثلة شائعة

cos(x)=(11)/(sqrt(17*38))

cos (x) = \frac{11}{17 \cdot 38}

(sin(90)}{5.8}=\frac{sin(x))/5

\frac{sin ( 9 0 ^{\circ} )}{5.8} = \frac{sin ( x )}{5}

6cos^2(x)-sin(x)-5=0,0<= x<= 2pi

6 cos^{2} (x) - sin (x) - 5 = 0, 0 \leq x \leq 2 π

sin(4x)=sin(3x)

sin (4 x) = sin (3 x)

4cos(x)+3=5

4 cos (x) + 3 = 5