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شائع علم المثلثات >

prove sin^2(90-b)+cos^2(b-450)=1

الحلّ

prove

sin^{2} (9 0^{\circ} - b) + cos^{2} (b - 45 0^{\circ}) = 1

الحلّ

صحيح

خطوات الحلّ

sin^{2} (9 0^{\circ} - b) + cos^{2} (b - 45 0^{\circ}) = 1

قم بالعمل على الطرف الأيسر

sin^{2} (9 0^{\circ} - b) + cos^{2} (b - 45 0^{\circ})

Rewrite using trig identities

cos (b - 45 0^{\circ})

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

:فعّل متطابقة الطرح لزوايا

= cos (b) cos (45 0^{\circ}) + sin (b) sin (45 0^{\circ})

cos (b) cos (45 0^{\circ}) + sin (b) sin (45 0^{\circ})

بسّط:

sin (b)

cos (b) cos (45 0^{\circ}) + sin (b) sin (45 0^{\circ})

cos (b) cos (45 0^{\circ}) = 0

cos (b) cos (45 0^{\circ})

cos (45 0^{\circ}) = 0

cos (45 0^{\circ})

cos (45 0^{\circ}) = cos (9 0^{\circ})

cos (45 0^{\circ})

36 0^{\circ} + 9 0^{\circ}

كـ

45 0^{\circ}

اكتب مجددًا

= cos (36 0^{\circ} + 9 0^{\circ})

cos (x + 36 0^{\circ}) = cos (x)

cos :

استخدم دوريّة الـ

cos (36 0^{\circ} + 9 0^{\circ}) = cos (9 0^{\circ})

= cos (9 0^{\circ})

= cos (9 0^{\circ})

استخدم المتطابقة الأساسيّة التالية:

cos (9 0^{\circ}) = 0

cos (9 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}

= 0

= 0

= 0 \cdot cos (b)

0 \cdot a = 0

فعّل القانون

= 0

sin (b) sin (45 0^{\circ}) = sin (b)

sin (b) sin (45 0^{\circ})

sin (45 0^{\circ}) = 1

sin (45 0^{\circ})

sin (45 0^{\circ}) = sin (9 0^{\circ})

sin (45 0^{\circ})

36 0^{\circ} + 9 0^{\circ}

كـ

45 0^{\circ}

اكتب مجددًا

= sin (36 0^{\circ} + 9 0^{\circ})

sin (x + 36 0^{\circ}) = sin (x)

sin :

استخدم دوريّة الـ

sin (36 0^{\circ} + 9 0^{\circ}) = sin (9 0^{\circ})

= sin (9 0^{\circ})

= sin (9 0^{\circ})

استخدم المتطابقة الأساسيّة التالية:

sin (9 0^{\circ}) = 1

sin (9 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}

= 1

= 1

= 1 \cdot sin (b)

sin (b) \cdot 1 = sin (b) :

اضرب

= sin (b)

= 0 + sin (b)

0 + sin (b) = sin (b)

= sin (b)

= sin (b)

= sin^{2} (9 0^{\circ} - b) + (sin (b))^{2}

Rewrite using trig identities

sin (9 0^{\circ} - b)

sin (s - t) = sin (s) cos (t) - cos (s) sin (t)

:فعّل متطابقة الطرح لزوايا

= sin (9 0^{\circ}) cos (b) - cos (9 0^{\circ}) sin (b)

sin (9 0^{\circ}) cos (b) - cos (9 0^{\circ}) sin (b)

بسّط:

cos (b)

sin (9 0^{\circ}) cos (b) - cos (9 0^{\circ}) sin (b)

sin (9 0^{\circ}) cos (b) = cos (b)

sin (9 0^{\circ}) cos (b)

sin (9 0^{\circ})

بسّط:

1

sin (9 0^{\circ})

استخدم المتطابقة الأساسيّة التالية:

sin (9 0^{\circ}) = 1

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}

= 1

= 1 \cdot cos (b)

1 \cdot cos (b) = cos (b) :

اضرب

= cos (b)

cos (9 0^{\circ}) sin (b) = 0

cos (9 0^{\circ}) sin (b)

cos (9 0^{\circ})

بسّط:

0

cos (9 0^{\circ})

استخدم المتطابقة الأساسيّة التالية:

cos (9 0^{\circ}) = 0

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}

= 0

= 0 \cdot sin (b)

0 \cdot a = 0

فعّل القانون

= 0

= cos (b) - 0

cos (b) - 0 = cos (b)

= cos (b)

= cos (b)

= (cos (b))^{2} + (sin (b))^{2}

(a) = a

:احذف الأقواس

= cos^{2} (b) + sin^{2} (b)

Rewrite using trig identities

cos^{2} (b) + sin^{2} (b)

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

:فعّل نطريّة فيتاغوروس

= 1

= 1

أظهرنا بأنّ الطرفين من الممكن أن تكون لهما نفس الصورة

\Rightarrow صحيح

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cos^2(2x)+sin^2(x)=1

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sin^2(x)-3cos(x)-4=0

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tan^2(x)+sec(x)=5

tan^{2} (x) + sec (x) = 5

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