Question

найти sin4a если известно, что tg(a+pi/4)=-√2

Answer 1

$mathop{mathrm{tg}}left(alpha+dfracpi4right)=dfrac{mathop{mathrm{tg}}alpha+mathop{mathrm{tg}}fracpi4}{1-mathop{mathrm{tg}}alphamathop{mathrm{tg}}fracpi4}=dfrac{1+mathop{mathrm{tg}}alpha}{1-mathop{mathrm{tg}}alpha}=-sqrt2\ 1+mathop{mathrm{tg}}alpha=sqrt{2}(mathop{mathrm{tg}}alpha-1)\ mathop{mathrm{tg}}alpha=dfrac{sqrt2+1}{sqrt2-1}=3+2sqrt2=t$

Формулы тангенса половинного угла:
$cos2alpha=dfrac{1-t^2}{1+t^2}qquadsin2alpha=dfrac{2t}{1+t^2}$

Тогда
$sin4alpha=2sin2alphacos2alpha=dfrac{4t(1-t^2)}{(1+t^2)^2}$

$t^2=(3+2sqrt2)^2=17+12sqrt2\ (1+t^2)^2=(18+12sqrt2)^2=36(3+2sqrt2)^2\ sin4alpha=dfrac{-4(3+2sqrt2)(16+12sqrt2)}{36(3+2sqrt2)^2}=-dfrac{4(4+3sqrt2)}{9(3+2sqrt2)}=\=-dfrac{4(4+3sqrt2)(3-2sqrt2)}{9}=-dfrac{4sqrt2}9$

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Другой способ. Найдём sin и cos, потом удвоим угол.
$begin{cases}cos^2left(alpha+dfracpi4right)=dfrac1{1+mathop{mathrm{tg}}^2(alpha+fracpi4)}=dfrac13\ sin^2left(alpha+dfracpi4right)=1-cos^2left(alpha+dfracpi4right)=dfrac23end{cases}\ begin{cases}-sin2alpha=cosleft(2alpha+dfracpi2right)=2cos^2left(alpha+dfracpi4right)-1=-dfrac13\ cos^22alpha=sin^2left(2alpha+dfracpi2right)=4sin^2left(alpha+dfracpi4right)cos^2left(alpha+dfracpi4right)=dfrac89end{cases}$

Если tg(α + π/4) < 0, то πn + π/2 < α + π/4 < πn + π, πn + π/4 < α < πn + 3π4. Значит, 2πn + π/2 < 2α < 2πn + 3π/2, и cos 2α < 0.

$begin{cases}sin2alpha=dfrac13\ cos2alpha=-dfrac{2sqrt2}3end{cases}\ sin4alpha=2sin2alphacos2alpha=-dfrac{4sqrt2}9$