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prove cos(60)=cos^2(30)-sin^2(30)

الحلّ

prove

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

صحيح

خطوات الحلّ

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

قم بالعمل على الطرف الأيسر

cos (6 0^{\circ})

بسّط

= \frac{1}{2}

قم بالعمل على الطرف الأيمن

cos^{2} (3 0^{\circ}) - sin^{2} (3 0^{\circ})

cos^{2} (3 0^{\circ}) - sin^{2} (3 0^{\circ})

بسّط:

\frac{1}{2}

cos^{2} (3 0^{\circ}) - sin^{2} (3 0^{\circ})

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

cos^{2} (3 0^{\circ})

cos (3 0^{\circ})

بسّط:

\frac{3}{2}

cos (3 0^{\circ})

استخدم المتطابقة الأساسيّة التالية:

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

= (\frac{3}{2})^{2}

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

:فعّل قانون القوى

= \frac{( 3 ) ^{2}}{2 ^{2}}

(3)^{2} : 3

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

:فعْل قانون الجذور

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

(a^{b})^{c} = a^{b c}

:فعّل قانون القوى

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

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

\frac{1}{2} \cdot 2

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

:اضرب كسور

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

2 :

إلغ العوامل المشتركة

= 1

= 3

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

sin^{2} (3 0^{\circ}) = \frac{1}{2 ^{2}}

sin^{2} (3 0^{\circ})

sin (3 0^{\circ})

بسّط:

\frac{1}{2}

sin (3 0^{\circ})

استخدم المتطابقة الأساسيّة التالية:

sin (3 0^{\circ}) = \frac{1}{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{1}{2}

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

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

:فعّل قانون القوى

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

1^{a} = 1

فعّل القانون

1^{2} = 1

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

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

بسّط

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

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

فعّل القانون

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

3 - 1 = 2 :

اطرح الأعداد

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

2 :

إلغ العوامل المشتركة

= \frac{1}{2}

= \frac{1}{2}

= \frac{1}{2}

أظهرنا بأنّ الطرفين من الممكن أن تكون لهما نفس الصورة

\Rightarrow صحيح

p ro v e tan (6 a) sec (3 a) = 2 sin (3 a) sec (6 a)

p ro v e (1 - sin^{2} (a)) (1 + tan^{2} (a)) = 1

p ro v e sin^{2} (x) + sin^{2} (x) cot^{2} (x) = 1

p ro v e tan (x) = \frac{1}{cot ( x )}

p ro v e sin^{2} (\frac{θ}{2}) = \frac{sec ( θ ) - 1}{2 sec ( θ )}