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شائع علم المثلثات >

sin(piy)=cos(1)sin(pi/6)-1/2

الحلّ

sin (π y) = cos (1) sin (\frac{π}{6}) - \frac{1}{2}

الحلّ

y = \frac{- 0.23192 \dots}{π} + 2 n, y = 1 + \frac{0.23192 \dots}{π} + 2 n

+1

درجات

y = - 4.22975 \dots^{\circ} + 114.59155 \dots^{\circ} n, y = 61.52553 \dots^{\circ} + 114.59155 \dots^{\circ} n

خطوات الحلّ

sin (π y) = cos (1) sin (\frac{π}{6}) - \frac{1}{2}

sin (\frac{π}{6}) = \frac{1}{2}

sin (\frac{π}{6})

استخدم المتطابقة الأساسيّة التالية:

sin (\frac{π}{6}) = \frac{1}{2}

sin (\frac{π}{6})

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{1}{2}

= \frac{1}{2}

sin (π y) = cos (1) \frac{1}{2} - \frac{1}{2}

Apply trig inverse properties

sin (π y) = cos (1) \frac{1}{2} - \frac{1}{2}

sin (π y) = cos (1) \frac{1}{2} - \frac{1}{2} :

حلول عامّة لـ

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

π y = arcsin (cos (1) \frac{1}{2} - \frac{1}{2}) + 2 πn, π y = π + arcsin (- cos (1) \frac{1}{2} + \frac{1}{2}) + 2 πn

π y = arcsin (cos (1) \frac{1}{2} - \frac{1}{2}) + 2 πn, π y = π + arcsin (- cos (1) \frac{1}{2} + \frac{1}{2}) + 2 πn

π y = arcsin (cos (1) \frac{1}{2} - \frac{1}{2}) + 2 πn

حلّ:

y = \frac{arcsin ( \frac{1}{2} cos ( 1 ) - \frac{1}{2} )}{π} + 2 n

π y = arcsin (cos (1) \frac{1}{2} - \frac{1}{2}) + 2 πn

π

اقسم الطرفين على

π y = arcsin (cos (1) \frac{1}{2} - \frac{1}{2}) + 2 πn

π

اقسم الطرفين على

\frac{π y}{π} = \frac{arcsin ( cos ( 1 ) \frac{1}{2} - \frac{1}{2} )}{π} + \frac{2 πn}{π}

بسّط

y = \frac{arcsin ( \frac{1}{2} cos ( 1 ) - \frac{1}{2} )}{π} + 2 n

y = \frac{arcsin ( \frac{1}{2} cos ( 1 ) - \frac{1}{2} )}{π} + 2 n

π y = π + arcsin (- cos (1) \frac{1}{2} + \frac{1}{2}) + 2 πn

حلّ:

y = 1 + \frac{arcsin ( - \frac{1}{2} cos ( 1 ) + \frac{1}{2} )}{π} + 2 n

π y = π + arcsin (- cos (1) \frac{1}{2} + \frac{1}{2}) + 2 πn

π

اقسم الطرفين على

π y = π + arcsin (- cos (1) \frac{1}{2} + \frac{1}{2}) + 2 πn

π

اقسم الطرفين على

\frac{π y}{π} = \frac{π}{π} + \frac{arcsin ( - cos ( 1 ) \frac{1}{2} + \frac{1}{2} )}{π} + \frac{2 πn}{π}

بسّط

\frac{π y}{π} = \frac{π}{π} + \frac{arcsin ( - cos ( 1 ) \frac{1}{2} + \frac{1}{2} )}{π} + \frac{2 πn}{π}

\frac{π y}{π}

بسّط:

y

\frac{π y}{π}

π :

إلغ العوامل المشتركة

= y

\frac{π}{π} + \frac{arcsin ( - cos ( 1 ) \frac{1}{2} + \frac{1}{2} )}{π} + \frac{2 πn}{π}

بسّط:

1 + \frac{arcsin ( - \frac{1}{2} cos ( 1 ) + \frac{1}{2} )}{π} + 2 n

\frac{π}{π} + \frac{arcsin ( - cos ( 1 ) \frac{1}{2} + \frac{1}{2} )}{π} + \frac{2 πn}{π}

\frac{a}{a} = 1

فعّل القانون

\frac{π}{π} = 1

= 1 + \frac{arcsin ( \frac{1}{2} - \frac{1}{2} cos ( 1 ) )}{π} + \frac{2 πn}{π}

\frac{2 πn}{π}

اختزل:

2 n

\frac{2 πn}{π}

π :

إلغ العوامل المشتركة

= 2 n

= 1 + \frac{arcsin ( \frac{1}{2} - \frac{1}{2} cos ( 1 ) )}{π} + 2 n

y = 1 + \frac{arcsin ( - \frac{1}{2} cos ( 1 ) + \frac{1}{2} )}{π} + 2 n

y = 1 + \frac{arcsin ( - \frac{1}{2} cos ( 1 ) + \frac{1}{2} )}{π} + 2 n

y = 1 + \frac{arcsin ( - \frac{1}{2} cos ( 1 ) + \frac{1}{2} )}{π} + 2 n

y = \frac{arcsin ( \frac{1}{2} cos ( 1 ) - \frac{1}{2} )}{π} + 2 n, y = 1 + \frac{arcsin ( - \frac{1}{2} cos ( 1 ) + \frac{1}{2} )}{π} + 2 n

أظهر الحلّ بالتمثيل العشريّ

y = \frac{- 0.23192 \dots}{π} + 2 n, y = 1 + \frac{0.23192 \dots}{π} + 2 n

رسم

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