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شائع علم المثلثات >

prove cos(pi/3+x)=sin(30-x)

الحلّ

prove

cos (\frac{π}{3} + x) = sin (3 0^{\circ} - x)

الحلّ

صحيح

خطوات الحلّ

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

قم بالعمل على الطرف الأيسر

cos (6 0^{\circ} + x)

Rewrite using trig identities

cos (6 0^{\circ} + x)

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

:فعّل متطابقة الجمع لزوايا

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

cos (6 0^{\circ}) cos (x) - sin (6 0^{\circ}) sin (x)

بسّط:

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

cos (6 0^{\circ}) cos (x) - sin (6 0^{\circ}) sin (x)

cos (6 0^{\circ})

بسّط:

\frac{1}{2}

cos (6 0^{\circ})

استخدم المتطابقة الأساسيّة التالية:

cos (6 0^{\circ}) = \frac{1}{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{1}{2}

= \frac{1}{2} cos (x) - sin (6 0^{\circ}) sin (x)

sin (6 0^{\circ})

بسّط:

\frac{3}{2}

sin (6 0^{\circ})

استخدم المتطابقة الأساسيّة التالية:

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

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

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

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

قم بالعمل على الطرف الأيمن

sin (3 0^{\circ} - x)

Rewrite using trig identities

sin (3 0^{\circ} - x)

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

:فعّل متطابقة الطرح لزوايا

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

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

بسّط:

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

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

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

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

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

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

أظهرنا بأنّ الطرفين من الممكن أن تكون لهما نفس الصورة

\Rightarrow صحيح

أمثلة شائعة

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

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

cos(θ)= 2/5 ,cot(θ)<0

cos (θ) = \frac{2}{5}, cot (θ) < 0

solvefor b,s(t)=tan(bt^2)

so l v e f or b, s (t) = tan (b t^{2})

tan(x)= 80/90

tan (x) = \frac{80}{90}

4cos^2(x)-11cos(x)+6=0

4 cos^{2} (x) - 11 cos (x) + 6 = 0