The distributive property of division over addition and substraction is idk
very simple, we use the formula sin(a+b)=sinacosb +
sinbcosa and sin(20)=2sinacosa
5pi = 2pi/3+3pi/3,
First, we use sin(a+b)=sinacosb + sinbcosa
sin(5pi/3)=sin(2pi/3+3pi/3)=
sin(2pi/3+pi)=
sin(2pi/3)cos(pi) +sin(pi)cos(2pi/3)
but we know that sin(pi)=
0, and cos (pi) = -1, so sin(5pi/3)=
- sin(2pi/3)
now, use sin(2a)=2sinacosa,
sin(5pi/3)= - sin(2pi/3)= -2sin(pi/3)cos(pi/3)
sin<span>(5pi/3)=
-2sin(pi/3)cos(pi/3)</span>
<span>sin(pi/3)= 0.86,
cos(pi/3)=0.5, finally we have </span>sin<span>(5pi/3)= -0.86 x 0.5= -0.43</span>
Easiest method solving 2 unknowns is Substitution.
So there will normally be 2(ormore)equations given.
Say
X+Y= a
bX=cY + d
a b c d are numbers they give
So first we rearrange the first equation
X = a - Y
No we substitute this into the 2nd equation so that we can get Y
since we know X = a-Y,
b(a-Y) = cY + d
we rearrange and we can find Y for this, it is just Y and numbers
hence finding Y, we sub the value for Y into X = a-Y and we get X also .
