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العربية

prove tan(2A)-2tan(2A)sin^2(A)=sin(2A)

الحلّ

prove

tan (2 A) - 2 tan (2 A) sin^{2} (A) = sin (2 A)

صحيح

خطوات الحلّ

tan (2 A) - 2 tan (2 A) sin^{2} (A) = sin (2 A)

قم بالعمل على الطرف الأيسر

tan (2 A) - 2 tan (2 A) sin^{2} (A)

sin, cos :

عبّر بواسطة

tan (2 A) - 2 sin^{2} (A) tan (2 A)

tan (x) = \frac{sin ( x )}{cos ( x )}

:Use the basic trigonometric identity

= \frac{sin ( 2 A )}{cos ( 2 A )} - 2 sin^{2} (A) \frac{sin ( 2 A )}{cos ( 2 A )}

\frac{sin ( 2 A )}{cos ( 2 A )} - 2 sin^{2} (A) \frac{sin ( 2 A )}{cos ( 2 A )}

بسّط:

\frac{sin ( 2 A ) - 2 sin ^{2} ( A ) sin ( 2 A )}{cos ( 2 A )}

\frac{sin ( 2 A )}{cos ( 2 A )} - 2 sin^{2} (A) \frac{sin ( 2 A )}{cos ( 2 A )}

2 sin^{2} (A) \frac{sin ( 2 A )}{cos ( 2 A )}

اضرب بـ:

\frac{2 sin ^{2} ( A ) sin ( 2 A )}{cos ( 2 A )}

2 sin^{2} (A) \frac{sin ( 2 A )}{cos ( 2 A )}

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

:اضرب كسور

= \frac{sin ( 2 A ) \cdot 2 sin ^{2} ( A )}{cos ( 2 A )}

= \frac{sin ( 2 A )}{cos ( 2 A )} - \frac{2 sin ^{2} ( A ) sin ( 2 A )}{cos ( 2 A )}

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

فعّل القانون

= \frac{sin ( 2 A ) - 2 sin ^{2} ( A ) sin ( 2 A )}{cos ( 2 A )}

= \frac{sin ( 2 A ) - 2 sin ^{2} ( A ) sin ( 2 A )}{cos ( 2 A )}

= \frac{sin ( 2 A ) - 2 sin ( 2 A ) sin ^{2} ( A )}{cos ( 2 A )}

Rewrite using trig identities

\frac{sin ( 2 A ) - 2 sin ( 2 A ) sin ^{2} ( A )}{cos ( 2 A )}

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

:فعّل متطابقة الزاوية المضاعفة

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

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

\frac{sin ( 2 A ) + ( cos ( 2 A ) - 1 ) sin ( 2 A )}{cos ( 2 A )}

بسّط:

sin (2 A)

\frac{sin ( 2 A ) + ( cos ( 2 A ) - 1 ) sin ( 2 A )}{cos ( 2 A )}

sin (2 A) + (cos (2 A) - 1) sin (2 A)

حلل إلى عوامل:

sin (2 A) cos (2 A)

sin (2 A) + (cos (2 A) - 1) sin (2 A)

sin (2 A)

قم باخراج العامل المشترك

= sin (2 A) (1 - 1 + cos (2 A))

بسّط

= sin (2 A) cos (2 A)

= \frac{sin ( 2 A ) cos ( 2 A )}{cos ( 2 A )}

cos (2 A) :

إلغ العوامل المشتركة

= sin (2 A)

= sin (2 A)

= sin (2 A)

أظهرنا بأنّ الطرفين من الممكن أن تكون لهما نفس الصورة

\Rightarrow صحيح

p ro v e sin^{2} (x) = \frac{1}{csc ^{2} ( x )}

p ro v e \frac{cot ( 4 u ) - 1}{cot ( 4 u ) + 1} = \frac{1 - tan ( 4 u )}{1 + tan ( 4 u )}

p ro v e sin (- x) = sin (x - π)

p ro v e sin (4 a) = 4 sin (a) cos (a) cos (2 a)

p ro v e 3 cos^{2} (x) - 2 = 1 - 3 sin^{2} (x)