Question

(4cos^2 a)/(clg * a/2 - tg a/2) = sin 2a

Answer 1

Ответ:  Доказано.

Пошаговое объяснение:

Докажите тождество :

$\displaystyle \frac{4\cos ^2 a }{\mathrm{ctg}\frac{a}{2} - \mathrm {tg }\frac{a}{2} }=\sin 2a$

Воспользуемся тем , что :

$\mathrm {ctg }\frac{a}{2} = \dfrac{\cos \frac{a}{2} }{\sin \frac{a}{2} } \\\\\\ \mathrm {tg }\frac{a}{2} = \dfrac{ \sin \frac{a}{2} }{ \cos \frac{a}{2}}$

Таким образом :

$\displaystyle \frac{4\cos ^2 a }{ \dfrac{\cos \frac{a}{2} }{\sin \frac{a}{2} } - \dfrac{ \sin \frac{a}{2} }{ \cos \frac{a}{2}} }=\sin 2a \\\\\\ \frac{4\cos ^2a}{\dfrac{\cos ^2 \frac{a}{2}- \sin ^2\frac{a}{2}}{\frac{1}{2} \cdot 2\sin \frac{a}{2}\cos \frac{a}{2} } } = \sin 2a$

Воспользуемся формулами двойного угла :

$\cos 2a = \cos ^2a- \sin ^2a \Rightarrow \cos a= \cos ^2\frac{a}{2} - \sin ^2 \frac{a}{2} \\\\ \sin 2a = 2\sin a \cos a \Rightarrow \sin a= 2\sin \frac{a}{2}\cos \frac{a}{2}$

И мы получим :

$\displaystyle \frac{4\cos ^{\diagup \!\!\!\!2}a}{\dfrac{\LARGE \diagup \!\!\!\!\!\!\!\!\!\!\!\cos a}{\frac{1}{2} \cdot \sin a } } = \sin 2a \\\\\\\ \frac{1}{2} \cdot 4 \cos a \cdot \sin a = \sin 2a \\\\ 2 \sin a \cos a = \sin 2a \\\\\ \sin 2a = \sin 2a ~\checkmark$

Тождество доказано.

#SPJ1

Answer 2

To prove the given equation, let's simplify the left-hand side (LHS) and right-hand side (RHS) separately and then equate them.

Starting with the LHS:

LHS = (4cos^2(a))/(clg(a/2) - tg(a/2))

Using trigonometric identities, we can rewrite the cosine and tangent functions in terms of sine:

cos^2(a) = 1 - sin^2(a) tg(a/2) = sin(a)/(1 + cos(a))

Substituting these identities into the LHS:

LHS = (4(1 - sin^2(a)))/(clg(a/2) - sin(a)/(1 + cos(a)))

Now, let's simplify the denominator by finding the common denominator:

clg(a/2) - sin(a)/(1 + cos(a)) = [(clg(a/2) * (1 + cos(a))) - sin(a)] / (1 + cos(a))

Now, the LHS becomes:

LHS = (4(1 - sin^2(a))) / [(clg(a/2) * (1 + cos(a))) - sin(a)]

Multiplying the numerator and denominator by (1 - cos(a)), we get:

LHS = [(4(1 - sin^2(a))) * (1 - cos(a))] / [(clg(a/2) * (1 + cos(a))) - sin(a)] * (1 - cos(a))

Expanding and simplifying the numerator:

LHS = [4(1 - sin^2(a) - cos(a) + sin^2(a) * cos(a))] / [(clg(a/2) * (1 + cos(a))) - sin(a)] * (1 - cos(a))

Further simplifying:

LHS = [4 - 4sin^2(a) - 4cos(a) + 4sin^2(a) * cos(a)] / [(clg(a/2) * (1 + cos(a))) - sin(a)] * (1 - cos(a))

Combining like terms:

LHS = [4 - 4cos(a) - 4sin^2(a) + 4sin^2(a) * cos(a)] / [(clg(a/2) * (1 + cos(a))) - sin(a)] * (1 - cos(a))

Factoring out a common factor of 4:

LHS = 4[1 - cos(a) - sin^2(a) + sin^2(a) * cos(a)] / [(clg(a/2) * (1 + cos(a))) - sin(a)] * (1 - cos(a))

Using the identity sin^2(a) = 1 - cos^2(a):

LHS = 4[1 - cos(a) - (1 - cos^2(a)) + (1 - cos^2(a)) * cos(a)] / [(clg(a/2) * (1 + cos(a))) - sin(a)] * (1 - cos(a))

Simplifying further:

LHS = 4[1 - cos(a) - 1 + cos^2(a) + cos(a) - cos^3(a)] / [(clg(a/2) * (1 + cos(a))) - sin(a)] * (1 - cos(a))

Combining like terms:

LHS = 4[cos^2(a) - cos^3(a)] / [(clg(a/2) * (1 + cos(a))) - sin(a)] * (1 - cos(a))

Now, let's simplify the RHS:

RHS = sin(2a)

Using the double-angle identity for sine:

RHS = 2sin(a)cos(a