<span>we have that
the cube roots of 27(cos 330° + i sin 330°) will be
</span>∛[27(cos 330° + i sin 330°)]
we know that
e<span>^(ix)=cos x + isinx
therefore
</span>∛[27(cos 330° + i sin 330°)]------> ∛[27(e^(i330°))]-----> 3∛[(e^(i110°)³)]
3∛[(e^(i110°)³)]--------> 3e^(i110°)-------------> 3[cos 110° + i sin 110°]
z1=3[cos 110° + i sin 110°]
cube root in complex number, divide angle by 3
360nº/3 = 120nº --> add 120º for z2 angle, again for z3
<span>therefore
</span>
z2=3[cos ((110°+120°) + i sin (110°+120°)]------ > 3[cos 230° + i sin 230°]
z3=3[cos (230°+120°) + i sin (230°+120°)]--------> 3[cos 350° + i sin 350°]
<span>
the answer is
</span>z1=3[cos 110° + i sin 110°]<span>
</span>z2=3[cos 230° + i sin 230°]
z3=3[cos 350° + i sin 350°]<span>
</span>
Use the trig identity
2*sin(A)*cos(A) = sin(2*A)
to get
sin(A)*cos(A) = (1/2)*sin(2*A)
So to find the max of sin(A)*cos(A), we can find the max of (1/2)*sin(2*A)
It turns out that sin(x) maxes out at 1 where x can be any expression you want. In this case, x = 2*A.
So (1/2)*sin(2*A) maxes out at (1/2)*1 = 1/2 = 0.5
The greatest value of sin(A)*cos(A) is 1/2 = 0.5
Answer:
The shadow is 11 Ft. away from the pole base.
Step-by-step explanation:
I just usually do this trick in situations like these where if <em>x</em> equals 1, and <em>y</em> equals 2, then <em>z </em>is gonna be 3. 2 goes after 1, and 3 goes after 2. This trick can work in other counting-up situations like this, as long as the number go up, and are right next to each other.
