Question

Только 4 и 5 ⬆️⬆️⬆️⬆️⬆️⬆️⬆️

Answer 1

Ответ:

$4)\ \ \displaystyle \frac{sina}{1+cosa}=\frac{2\, sin\frac{a}{2}\cdot cos\frac{a}{2}}{2cos^2\frac{a}{2}}=\frac{sin\frac{a}{2}}{cos\frac{a}{2}}=tg\frac{a}{2}\\\\\\\frac{1-cosa}{sina}=\frac{2sin^2\frac{a}{2}}{2\, sin\frac{a}{2}\cdot cos\frac{a}{2}}=\frac{sin\frac{a}{2}}{cos\frac{a}{2}}=tg\frac{a}{2}\\\\\\tg\frac{a}{2}=tg\frac{a}{2}$

$5)\ \ sin^6a+cos^6a+3sin^2a\cdot cos^2a=\\\\=(\underbrace {sin^2a+cos^2a}_{1})(sin^4a-sin^2a\cdot cos^2a+cos^4a)+3sin^2a\cdot cos^2a=\\\\=sin^4a+2sin^2a\cdot cos^2a+cos^4a=(\underbrace{sin^2a+cos^2a}_{1})^2=1^2=1\\\\1=1$

Answer 2

$4)\dfrac{Sin\alpha}{1+Cos\alpha } =\dfrac{1-Cos\alpha }{Sin\alpha}\\\\\dfrac{Sin\alpha}{1+Cos\alpha } -\dfrac{1-Cos\alpha }{Sin\alpha }=\dfrac{Sin\alpha\cdot Sin\alpha-(1+Cos\alpha)\cdot(1-Cos\alpha)}{Sin\alpha\cdot(1+Cos\alpha)} =\\\\=\dfrac{Sin^{2}\alpha-1+Cos^{2}\alpha}{Sin\alpha\cdot(1+Cos\alpha)}=\dfrac{1-1}{Sin\alpha\cdot(1+Cos\alpha)}=\boxed0$

Если разность левой и провой частей тождества равна нулю, то они равны . Тождество доказано .

$5)Sin^{6}\alpha +Cos^{6}\alpha+3Sin^{2}\alpha Cos^{2}\alpha=1\\\\(Sin^{2}\alpha )^{3}+(Cos^{2}\alpha )^{3}+3Sin^{2}\alpha Cos^{2}\alpha=\\\\=\underbrace{(Sin^{2}\alpha +Cos^{2}\alpha)}_{1}(Sin^{4} \alpha-Sin^{2}\alpha Cos^{2}\alpha +Cos^{4}\alpha)+3Sin^{2}\alpha Cos^{2}\alpha=\\\\=Sin^{4} \alpha-Sin^{2}\alpha Cos^{2}\alpha +Cos^{4}\alpha+3Sin^{2}\alpha Cos^{2}\alpha=\\\\=Sin^{4} \alpha+2Sin^{2}\alpha Cos^{2}\alpha +Cos^{4}\alpha=(Sin^{2}\alpha+Cos^{2}\alpha)^{2}=1\\\\1=1$

Тождество доказано