B=a+y т.к. углы a и B, B и y накрест лежащие
Log8(log4(log2(16)))=log8(log4(4))=log8(1)=0
Решение
<span>sin²x/4-cos²x/4=1/2
- (cos</span>²x/4 - sin²x/4) = 1/2
cos<span>²x/4 - sin²x/4 = - 1/2
</span>cos[2*(x/4)] = - 1/2
cosx/2 = - 1/2
x/2 = +-arccos(-1/2) + 2πk, k ∈ Z
x/2 = +- [π - arccos(1/2)] + 2πk, k ∈ Z
x/2 = +- [π - π/3<span>)] + 2πk, k ∈ Z
</span>x/2 = +- [2π/3<span>)] + 2πk, k ∈ Z
</span>x = +- [4π/3<span>)] + 4πk, k ∈ Z</span>
Если условие выглядит так: tg((pi/4)+x))=(1+tgx)/(1-tgx), то решение:
tg((pi/4)+x)) = (tg(pi/4)+tgx)/(1-tg(pi/4)tgx) = /tg(pi/4) = 1/ = (1+tgx)/(1-tgx) тождество доказано.