anonymous
  • anonymous
Solve: cos2Ɵ = sin Ɵ
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
myininaya
  • myininaya
\[\cos(2 \theta)-\sin(\theta)=0\] Hint put in terms of sine ------------------------- \[\cos(2 \theta)=\cos^2(\theta)- \sin^2(\theta)=(1-\sin^2(\theta))-\sin^2(\theta)=1-2\sin^2(\theta)\]
anonymous
  • anonymous
is 1−2sin2(θ) the answer ?
myininaya
  • myininaya
No

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
how do i go from there?
myininaya
  • myininaya
\[\cos(2 \theta)=1 - 2 \sin^2(\theta) \text{ is an identity}\] we need to use it in our problem
myininaya
  • myininaya
\[1-2\sin^2(\theta)-\sin(\theta)=0\] you need to solve this for theta
myininaya
  • myininaya
\[-2 \sin^2(\theta)-\sin(\theta)+1=0 \] Can you solve: \[-2u^2-u+1=0\] for u?
myininaya
  • myininaya
if so the relationship btw the equations I put is \[u=\sin(\theta)\] so find what u is then replace u with sin(theta)
anonymous
  • anonymous
the problem above −2u2−u+1=0 you just add u to the other side right?
myininaya
  • myininaya
No solve that for u
myininaya
  • myininaya
\[-2u^2-u+1=0\] Solve this for u
anonymous
  • anonymous
u = √1/2
myininaya
  • myininaya
You should had got two solutions.
anonymous
  • anonymous
Square roots can be both negative and positive. Those would be the two solutions.
anonymous
  • anonymous
oh i see
anonymous
  • anonymous
so it's u= +-√1/2 ?
anonymous
  • anonymous
Yes
myininaya
  • myininaya
\[(-2u+1)(u+1)=0\]
anonymous
  • anonymous
is it 2u + 1 = 0?
myininaya
  • myininaya
You have a*b=0 => either a=0 , b=0 or both=0 so we have -2u+1=0 or u+1=0 Solve both of these linear equations.
anonymous
  • anonymous
-2u+1 i knew it :|
anonymous
  • anonymous
u = 1/2 and u = -1
myininaya
  • myininaya
Write now remember our relationship \[u=\sin(\theta)\]
myininaya
  • myininaya
So we have \[\sin(\theta)=\frac{1}{2} \text{ or } \sin(\theta)=-1\]
myininaya
  • myininaya
Oops right* not write* sorry wrong right/write lol
anonymous
  • anonymous
it's ok
anonymous
  • anonymous
so I can plug in to the calculator : inverse sin (1/2) then
anonymous
  • anonymous
after that multiply by pi/180 to get pi/6 right ?
anonymous
  • anonymous
30 or Pi/6
myininaya
  • myininaya
Well I would take off points if you used a calculator
anonymous
  • anonymous
oh ok
myininaya
  • myininaya
It doesn't give you all the solutions and plus sometimes it gives you an approximation
myininaya
  • myininaya
depending on what calculator you use
myininaya
  • myininaya
There is the unit circle though
anonymous
  • anonymous
it's really confusing :|
anonymous
  • anonymous
but if i use calculator then I'm fine with that :)
myininaya
  • myininaya
when is sine, 1/2?
myininaya
  • myininaya
brainshot gave you one value that satisfies the equation
anonymous
  • anonymous
@myininaya Are you a teacher?
anonymous
  • anonymous
Myininaya is the awsomest teacher of all time :D
anonymous
  • anonymous
Ha yeah, but I meant in reality.
anonymous
  • anonymous
Oh yes in reality.
anonymous
  • anonymous
What do you teach?
anonymous
  • anonymous
Calculus to undegrad students in a university.
anonymous
  • anonymous
Lol, I am talking for her :P :D
myininaya
  • myininaya
@brainshot3 I am. \[\sin(\theta)=\frac{1}{2}\] \[\theta=\frac{\pi}{6}+2 n \pi \text{ this } n 2 \pi \text{ part is the n times we go around the circle}\] But we also have another value in the interval [0,2pi] such that \[\sin(\theta)=\frac{1}{2}\] \[\theta=\frac{5 \pi}{6} +2 n \pi\] And by the way n is an integer
myininaya
  • myininaya
http://www.google.com/imgres?imgurl=http://upload.wikimedia.org/wikipedia/commons/thumb/4/4c/Unit_circle_angles_color.svg/300px-Unit_circle_angles_color.svg.png&imgrefurl=http://en.wikipedia.org/wiki/Unit_circle&h=300&w=300&sz=38&tbnid=zCoPII8QibzEpM:&tbnh=86&tbnw=86&prev=/search%3Fq%3Dunit%2Bcircle%26tbm%3Disch%26tbo%3Du&zoom=1&q=unit+circle&docid=D09vYD7H2hKF-M&sa=X&ei=ja17T7qbDuOh2QW1r7WrAw&ved=0CC8Q9QEwAA&dur=438
anonymous
  • anonymous
and a phd student in mathematics and yet find time to help people here lol
anonymous
  • anonymous
Wow! What do you plan to get a job in?
myininaya
  • myininaya
You really need to familiarize yourself with the unit circle. It is your best friend in trig.
anonymous
  • anonymous
thanks for the link!
myininaya
  • myininaya
You still need to solve the other equation.
myininaya
  • myininaya
\[\sin(\theta)=-1 \]
anonymous
  • anonymous
http://www.khanacademy.org/math/trigonometry/v/the-unit-circle-definition-of-trigonometric-function
anonymous
  • anonymous
i got cos^2 θ = 0 ?
anonymous
  • anonymous
thanks for the link @FoolForMath
anonymous
  • anonymous
\[ \sin(\theta)=-1=\sin (-\frac \pi 2) \implies \theta = n\pi +(-1)^n (-\frac \pi 2) \]
anonymous
  • anonymous
Btw, Myin, I thought that definition of \(2n\pi \) is only for \( \cos \), am I missing something?
anonymous
  • anonymous
This question is really hard for me, I will look over it again tonight thanks guys!
anonymous
  • anonymous
Btw can you guys help me with another one ?
anonymous
  • anonymous
I kinda in a rush :|
anonymous
  • anonymous
\[ \sin(\theta)=-1=\sin (-\frac \pi 2) \implies \theta = n\pi +(-1)^n (-\frac \pi 2) \]\[= n\pi +(-1)^{n+1} \left( \frac \pi 2\right)\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.