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شائع علم المثلثات >

(cos(α-30))/1 =(sin(60))/3

الحلّ

\frac{cos ( α - 3 0 ^{\circ} )}{1} = \frac{sin ( 6 0 ^{\circ} )}{3}

الحلّ

α = 1.27795 \dots + 36 0^{\circ} n + 3 0^{\circ}, α = 36 0^{\circ} - 1.27795 \dots + 36 0^{\circ} n + 3 0^{\circ}

+1

راديان

α = 1.27795 \dots + \frac{π}{6} + 2 πn, α = 2 π - 1.27795 \dots + \frac{π}{6} + 2 πn

خطوات الحلّ

\frac{cos ( α - 3 0 ^{\circ} )}{1} = \frac{sin ( 6 0 ^{\circ} )}{3}

sin (6 0^{\circ}) = \frac{3}{2}

sin (6 0^{\circ})

استخدم المتطابقة الأساسيّة التالية:

sin (6 0^{\circ}) = \frac{3}{2}

sin (6 0^{\circ})

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

= \frac{3}{2}

= \frac{3}{2}

\frac{cos ( α - 3 0 ^{\circ} )}{1} = \frac{\frac{3}{2}}{3}

\frac{a}{1} = a

فعّل القانون

cos (α - 3 0^{\circ}) = \frac{\frac{3}{2}}{3}

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

: استخدم ميزات الكسور التالية

cos (α - 3 0^{\circ}) = \frac{3}{2 \cdot 3}

2 \cdot 3 = 6 :

اضرب الأعداد

cos (α - 3 0^{\circ}) = \frac{3}{6}

Apply trig inverse properties

cos (α - 3 0^{\circ}) = \frac{3}{6}

cos (α - 3 0^{\circ}) = \frac{3}{6} :

حلول عامّة لـ

cos (x) = a \Rightarrow x = arccos (a) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (a) + 36 0^{\circ} n

α - 3 0^{\circ} = arccos (\frac{3}{6}) + 36 0^{\circ} n, α - 3 0^{\circ} = 36 0^{\circ} - arccos (\frac{3}{6}) + 36 0^{\circ} n

α - 3 0^{\circ} = arccos (\frac{3}{6}) + 36 0^{\circ} n, α - 3 0^{\circ} = 36 0^{\circ} - arccos (\frac{3}{6}) + 36 0^{\circ} n

α - 3 0^{\circ} = arccos (\frac{3}{6}) + 36 0^{\circ} n

حلّ:

α = arccos (\frac{1}{2 3}) + 36 0^{\circ} n + 3 0^{\circ}

α - 3 0^{\circ} = arccos (\frac{3}{6}) + 36 0^{\circ} n

arccos (\frac{3}{6}) + 36 0^{\circ} n

بسّط:

arccos (\frac{1}{2 3}) + 36 0^{\circ} n

arccos (\frac{3}{6}) + 36 0^{\circ} n

\frac{3}{6} = \frac{1}{2 3}

\frac{3}{6}

6

حلل إلى عوامل:

2 \cdot 3

6 = 2 \cdot 3

حلّل إلى عوامل

= \frac{3}{2 \cdot 3}

\frac{3}{2 \cdot 3}

اختزل:

\frac{1}{2 3}

\frac{3}{2 \cdot 3}

n a = a^{\frac{1}{n}}

:فعْل قانون الجذور

3 = 3^{\frac{1}{2}}

= \frac{3 ^{\frac{1}{2}}}{2 \cdot 3}

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

:فعّل قانون القوى

\frac{3 ^{\frac{1}{2}}}{3 ^{1}} = \frac{1}{3 ^{1 - \frac{1}{2}}}

= \frac{1}{2 \cdot 3 ^{- \frac{1}{2} + 1}}

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

اطرح الأعداد

= \frac{1}{2 \cdot 3 ^{\frac{1}{2}}}

a^{\frac{1}{n}} = n a

:فعْل قانون الجذور

3^{\frac{1}{2}} = 3

= \frac{1}{2 3}

= \frac{1}{2 3}

= arccos (\frac{1}{2 3}) + 36 0^{\circ} n

α - 3 0^{\circ} = arccos (\frac{1}{2 3}) + 36 0^{\circ} n

انقل

3 0^{\circ}

إلى الجانب الأيمن

α - 3 0^{\circ} = arccos (\frac{1}{2 3}) + 36 0^{\circ} n

للطرفين

3 0^{\circ}

أضف

α - 3 0^{\circ} + 3 0^{\circ} = arccos (\frac{1}{2 3}) + 36 0^{\circ} n + 3 0^{\circ}

بسّط

α = arccos (\frac{1}{2 3}) + 36 0^{\circ} n + 3 0^{\circ}

α = arccos (\frac{1}{2 3}) + 36 0^{\circ} n + 3 0^{\circ}

α - 3 0^{\circ} = 36 0^{\circ} - arccos (\frac{3}{6}) + 36 0^{\circ} n

حلّ:

α = 36 0^{\circ} - arccos (\frac{1}{2 3}) + 36 0^{\circ} n + 3 0^{\circ}

α - 3 0^{\circ} = 36 0^{\circ} - arccos (\frac{3}{6}) + 36 0^{\circ} n

36 0^{\circ} - arccos (\frac{3}{6}) + 36 0^{\circ} n

بسّط:

36 0^{\circ} - arccos (\frac{1}{2 3}) + 36 0^{\circ} n

36 0^{\circ} - arccos (\frac{3}{6}) + 36 0^{\circ} n

\frac{3}{6} = \frac{1}{2 3}

\frac{3}{6}

6

حلل إلى عوامل:

2 \cdot 3

6 = 2 \cdot 3

حلّل إلى عوامل

= \frac{3}{2 \cdot 3}

\frac{3}{2 \cdot 3}

اختزل:

\frac{1}{2 3}

\frac{3}{2 \cdot 3}

n a = a^{\frac{1}{n}}

:فعْل قانون الجذور

3 = 3^{\frac{1}{2}}

= \frac{3 ^{\frac{1}{2}}}{2 \cdot 3}

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

:فعّل قانون القوى

\frac{3 ^{\frac{1}{2}}}{3 ^{1}} = \frac{1}{3 ^{1 - \frac{1}{2}}}

= \frac{1}{2 \cdot 3 ^{- \frac{1}{2} + 1}}

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

اطرح الأعداد

= \frac{1}{2 \cdot 3 ^{\frac{1}{2}}}

a^{\frac{1}{n}} = n a

:فعْل قانون الجذور

3^{\frac{1}{2}} = 3

= \frac{1}{2 3}

= \frac{1}{2 3}

= 36 0^{\circ} - arccos (\frac{1}{2 3}) + 36 0^{\circ} n

α - 3 0^{\circ} = 36 0^{\circ} - arccos (\frac{1}{2 3}) + 36 0^{\circ} n

انقل

3 0^{\circ}

إلى الجانب الأيمن

α - 3 0^{\circ} = 36 0^{\circ} - arccos (\frac{1}{2 3}) + 36 0^{\circ} n

للطرفين

3 0^{\circ}

أضف

α - 3 0^{\circ} + 3 0^{\circ} = 36 0^{\circ} - arccos (\frac{1}{2 3}) + 36 0^{\circ} n + 3 0^{\circ}

بسّط

α = 36 0^{\circ} - arccos (\frac{1}{2 3}) + 36 0^{\circ} n + 3 0^{\circ}

α = 36 0^{\circ} - arccos (\frac{1}{2 3}) + 36 0^{\circ} n + 3 0^{\circ}

α = arccos (\frac{1}{2 3}) + 36 0^{\circ} n + 3 0^{\circ}, α = 36 0^{\circ} - arccos (\frac{1}{2 3}) + 36 0^{\circ} n + 3 0^{\circ}

أظهر الحلّ بالتمثيل العشريّ

α = 1.27795 \dots + 36 0^{\circ} n + 3 0^{\circ}, α = 36 0^{\circ} - 1.27795 \dots + 36 0^{\circ} n + 3 0^{\circ}

رسم

أمثلة شائعة

tan(x)tan(7x)=1,x

tan (x) tan (7 x) = 1, x

3sec^2(x)+tan(x)-6=0

3 sec^{2} (x) + tan (x) - 6 = 0

2sin(x)+cot(x)=csc(x)

2 sin (x) + cot (x) = csc (x)

8cos^2(x)+2cos(x)-3=0

8 cos^{2} (x) + 2 cos (x) - 3 = 0

cos(A)= 1/(sqrt(65)),cos(B)= 1/(sqrt(5))

cos (A) = \frac{1}{65}, cos (B) = \frac{1}{5}