<em>1) (6sin(π/12))*(cos(π/12))=3sinπ/6=3*0.5=</em><em>1.5</em>
<em>2) sin²(5α)-cos²(5α)= </em><em>-cos10α</em>
<em>3)(2tg2α)/(1-tg²(2α))=</em><em>tg4α</em>
<em>4)sin( A/6)cos(A/6)=</em><em>0.5sin (A/3)</em>
<em>5) 20sin²(4α)cos²(4α)=(5*(2sin(4α))*cos(4α))²=</em><em>5sin²8α</em>
<em>6)(sin3α)(cos3α)=</em><em>0.5sin6α</em>
Cosx-2sinxcosx=0
cosx(1-2sinx)=0
cosx=0⇒x=π/2+πn,n∈z
sinx=1/2⇒x=(-1)^k*π/6+πk,k∈z
<span>1)2Cos</span>²<span>3</span>α <span>- Cos6</span>α = 2Cos²3α - 2Cos²α +1 = 1<span>
2) 2Cos</span>²α<span>/ 1+ Cos2</span>α= 2Cos²α/(1 + 2Cos²α -1) = 2Cos²α/2Cos²α = 1<span>
3) 1 - Cos (</span>π <span>- 2</span>α<span>)/ 1 - Sin</span>²α = 1+Cos2α/Сos²α= (1 +2Cos²α -1)/Cos²α =2<span>
4) Cos2 </span>α<span> + Sin</span>²α<span>/ 1 - Sin</span>²α= (Cos²α - Sin²α + Sin²α)/Сos²α = 1
{x+y-xy=-14
{x+y+xy=2
Пусть х+у=а
ху=b
{a-b=-14
{a+b=2
Складываем уравнения системы:
2а=-12
а=-6
-6+в=2
в=8
{x+y=-6
{xy=8
x=-6-y
(-6-y)y=8
-6y-y²=8
-y²-6y-8=0
y²+6y+8=0
D=36-32=4=2²
y₁=(-6-2)/2=-8/2=-4 x₁=-6-(-4)=-6+4=-2
y₂=-4/2=-2 x₂=-6+2=-4
Ответ: х₁=-2 у₁=-4
х₂=-4 у₂=-2
<em>1) Все производные- табличные, находим по формулам, первая равна </em><em>6х-5.</em>
<em>2) </em><em>3х²+1/(2√х)</em>
<em>3) </em><em>4х³+cosx-sinx</em>