Question

Докажите тождество: sin2t/1+cos2t*cost/1+cost=tgt/2 sin2t/1+cos2t*cost/1+cost*cos1/2/1+cost t/2=tg1/4

Answer 1

$1)~\boldsymbol{\dfrac{\sin(2t)}{1+\cos(2t)}\cdot\dfrac{\cos t}{1+\cos t}=}\dfrac{2\sin t\cos t}{2\cos^2t}\cdot\dfrac{\cos t}{1+\cos t}=\\\\\\~~~=\dfrac{\sin t}{1+\cos t}=\dfrac{2\sin \frac t2\cdot\cos \frac t2}{2\cos^2 \frac t2}=\dfrac{\sin \frac t2}{\cos \frac t2}=\bold{ \text{tg }\frac t2}~~~\blacksquare$

2) В условии в третьей и четвёртой дроби исправлены опечатки

$\boldsymbol{\dfrac{\sin(2t)}{1+\cos(2t)}\cdot\dfrac{\cos t}{1+\cos t}\cdot\dfrac{\cos \frac t2}{1+\cos \frac t2}=}\\\\\\=\dfrac{2\sin t\cos t}{2\cos^2t}\cdot\dfrac{\cos t}{1+\cos t}\cdot\dfrac{\cos \frac t2}{1+\cos \frac t2}=$

$=\dfrac{\sin t}{1+\cos t}\cdot\dfrac{\cos \frac t2}{1+\cos \frac t2}=\dfrac{2\sin \frac t2\cos \frac t2}{2\cos^2 \frac t2}\cdot\dfrac{\cos \frac t2}{1+\cos \frac t2}=\\\\\\=\dfrac{\sin \frac t2}{1+\cos \frac t2}=\dfrac{2\sin \frac t4\cos \frac t4}{2\cos^2 \frac t4}=\dfrac{\sin \frac t4}{\cos \frac t4}=\bold{ \text{tg }\frac t4}~~~\blacksquare$

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Использованы формулы двойного аргумента

$\sin (2\alpha )=2\sin \alpha \cos \alpha \\\cos(2\alpha )=2\cos^2\alpha -1\\1+\cos(2\alpha )=2\cos^2\alpha$

Answer 2

1) (sin(2t))/(1+cos(2t)) *((сost)/(1+cos(t)) =

(((2sint)*(cost))/(2cos²t))*(cost/(2cos²(t/2)))=(tgt)*cost/(2cos²(t/2))=

(sint)/(2cos²(t/2))=(2sin(t/2))*cos(t/2)/(2cos²(t/2))=tg(t/2)

Bоспользовался дважды формулой (1+cosα)=2cos²α  ; формулой синуса двойного аргумента sin2α=2(sinα)*(cosα)   и tgα=sinα/cosα.

2) Докажем второе  тождество, используя те же формулы.

((sin(2t))/(1+cos(2t)))*(cost/(1+cost))*(cos(t/2))/(1+cos(t/2))=tg(t/4)

1) упростим ((sin(2t))/(1+cos(2t)))=(2sint)(сost)/(2cos²t)=sint/(cost)=tgt

2) умножим (tgt)*(cost/(1+cost))=(sint)/(2cos²(t/2))=

(2sin(t/2))*(cos(t/2))/(2cos²(t/2))=tg(t/2)

3) умножим (tg(t/2))*((cos(t/2))/(1+cos(t/2))=sin(t/2)/(2cos²(t/4)=

(2sin(t/4)*(cos(t/4))/(2cos²(t/4))=tg(t/4)

Требуемое доказано.