Mathematics, 03.02.2020 00:02, joycewingate919

Idont understand this.. evaluate the trig functions using its period as an aid. sin5π. what is this \"period\"
4ff597f9e4b01c7be8c88410,anonymous, "thu jul 05 2012 15: 34: 49 gmt+0200 (cest)

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Basically to convert 5pi to degree you multiply by 180/pi and this converts it so pi cancels out and you get 900 degree than 900-360 = 540 than 540 -360 = 180 and sin of 180 is zero!,

but you can use the unit circle chart to do this basically all even pi will fall at 360 or 0 degree and odd pi will fall at 180 degree so in the next question 7 pi will give you 180 degree (if you do it the long way 7pi * 180/pi = 1260 - 360 = 900 -360 = 540- 360 = 180) and cosine of 180 is 1, so cos 7pi is 1

for 8pi/3 long way will give you 120 degree but if you use unit circle it will help you even more just go around the circle and count pi/3, basically pi/3 means 180/3 parts so 60 degree for each so 6pi/3 is 360 and so remaining is 2pi/3 means 60+60 = 120 degree. 30, 60, and 90 degree special triangle tells you the value of hypotenuse is 2, opposite side is radical 3 and adjacent is 1, since its in second quadrant the 1 or x value is gonna be negative so cosine would be -1/2. try in calculator it should give you -0.5

do the same for the rest!

look up also 30, 60, and 90 degree special triangle and use the values to cosine and sin or goggle it and they will show you how to solve for it ... faster than calculator.

I hope my answer has come to your help. Thank you for posting your question here in . We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

answered: Guest

step-by-step explanation:

answered: Guest

x= -3 and $x=\frac{-3+/-3i\sqrt{3} }{2}$

step-by-step explanation:

to find the roots of a cubic, use division or factoring to find the factors then set equal to 0. as this is a difference of cubes, it factors into

$(x^3-a^3)=$ $(x-a)(x^2+ax+a^2)$

where a is the cube root of $a^3$ .

here this is 27 and the cube root of 27 is 3. so we write using the form:

$(x-3)(x^2+(3)x+3^2)\\(x-3)(x^2+3x+9)$ .

set each factor to 0 and solve for x.

x-3=0

x=3

the quadratic expression does not factor and must be solved using the quadratic formula.

$x=\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

where a = 1, b=3, and c=9

$x=\frac{-3+/-\sqrt{3^2-4(1)(9)} }{2(1)} \\x=\frac{-3+/-\sqrt{9-36)} }{2} \\x=\frac{-3+/-\sqrt{-27)} }{2}\\x=\frac{-3+/-3i\sqrt{3} }{2}$

this cubic has two complex roots.

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Idont understand this.. evaluate the trig functions using its period as an aid. sin5π. what is thi...