Question

I already asked this but i didn't understand the answer? Solve for a exactly: cos(2a)= -sin(a) with 0≤a≤2π

Answer 1

It's multiple choice but i dont even know how you would plug them in?
A. a=π
B. a=π/3
C. a=π/4
D. a=π/2

Answer 2

cos(2a) = 1 - 2sin^2(a) this is a standard trig identity
so substituting this in the original equation
1 - 2sin^2(a) = -sin ( a)
2sin^2(a) - sin (a) - 1 = 0
this a quadratic equation which can be solved by factoring
(2sin (a) + 1 )(sin (a) - 1) = 0
sin (a) = -1/2 or sin (a) = 1
angle whose sine is 1/2 = pi/6 and angle whose sine is 1 is pi/2

negative 1/2 - angle is in 3rd and 4th quadrant ie 7pi/6 and 11pi/6

there are 3 roots 7pi/6 , 11pi/6 and pi/2

in degrees this is 210,330 and 90 degrees.

Answer 3

Whoops... guess jimmy got it for us. lol

Answer 4

relation betewwen degrees and radians

pi radians = 180 degrees
2pi radians = 360 degrees

rads to degrees : - multiply by 180 / pi
degrees to rads : multiply by pi / 180

Answer 5

Is there anyway i can do it without trigg identities because id really like not having to try and quickly relearn them:/

Answer 6

the only other way is to plug in the 4 values and see which one fits
the pi/2 will satisfy the equation
cos (2 * pi/2) = cos pi or cos 180 degrees; use calculator to get -1
-sin pi/2 : use calculator to find sin pi/2 or 90 degrees = 1
so -sin pi/2 = -1
so pi/2 satisfies the equation
d0 is the correct answer