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شائع علم المثلثات >

prove sqrt(2)sin(x+45)=sin(x)+cos(x)

الحلّ

prove

2 sin (x + 4 5^{\circ}) = sin (x) + cos (x)

الحلّ

صحيح

خطوات الحلّ

2 sin (x + 4 5^{\circ}) = sin (x) + cos (x)

قم بالعمل على الطرف الأيسر

2 sin (x + 4 5^{\circ})

Rewrite using trig identities

sin (x + 4 5^{\circ})

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

:فعّل متطابقة الجمع لزوايا

= sin (x) cos (4 5^{\circ}) + cos (x) sin (4 5^{\circ})

sin (x) cos (4 5^{\circ}) + cos (x) sin (4 5^{\circ})

بسّط:

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

sin (x) cos (4 5^{\circ}) + cos (x) sin (4 5^{\circ})

sin (x) cos (4 5^{\circ}) = \frac{2 sin ( x )}{2}

sin (x) cos (4 5^{\circ})

cos (4 5^{\circ})

بسّط:

\frac{2}{2}

cos (4 5^{\circ})

استخدم المتطابقة الأساسيّة التالية:

cos (4 5^{\circ}) = \frac{2}{2}

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{2}{2}

= \frac{2}{2} sin (x)

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

:اضرب كسور

= \frac{2 sin ( x )}{2}

cos (x) sin (4 5^{\circ}) = \frac{2 cos ( x )}{2}

cos (x) sin (4 5^{\circ})

sin (4 5^{\circ})

بسّط:

\frac{2}{2}

sin (4 5^{\circ})

استخدم المتطابقة الأساسيّة التالية:

sin (4 5^{\circ}) = \frac{2}{2}

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{2}{2}

= \frac{2}{2} cos (x)

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

:اضرب كسور

= \frac{2 cos ( x )}{2}

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

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

فعّل القانون

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

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

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

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

بسّط:

sin (x) + cos (x)

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

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

:اضرب كسور

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

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

اختزل:

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

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

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

:فعْل قانون الجذور

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

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

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

:فعّل قانون القوى

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

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

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

اطرح الأعداد

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

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

:فعْل قانون الجذور

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

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

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

2

قم باخراج العامل المشترك

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

2 :

إلغ العوامل المشتركة

= sin (x) + cos (x)

= sin (x) + cos (x)

أظهرنا بأنّ الطرفين من الممكن أن تكون لهما نفس الصورة

\Rightarrow صحيح

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