+189 Calculate sin([arccos(-1/9)]/2) and put the answer as a fraction.
+118687 ok fair enough.
theta is in the second quadrant so cos(theta) will be negative
Now tell me what cos(theta) is.
Answer quickly if you can.
+118687 No it wouldn't...
cos theta = adj / hyp = ( -1/9)/2 [which would simplify to -1/18]
If \(cos\theta = (-1/9)/2\)
then what is
\(acos ((-1/9)/2)\) ?
Maybe it might help to think of it as an equation
\(\qquad cos \theta = ((-1/9)/2)\\ so\\ acos[cos \theta] = acos ((-1/9)/2)\\ \qquad \qquad\theta= acos ((-1/9)/2)\\ \qquad \text{[since acos and cos cancel each other out.]}\)
You have my right angle triangle diagram with theta on it.
Work out the third side using pythagoras.
Then read sin(theta) off it.
+189 Sorry, I got confused. I will explain why. The context of this problem is cos(theta)=-1/9 and cos(theta/2)<0. And I am trying to find the sin(theta/2). When I was finding cos(theta/2) I did arccos(-1/9) then divided that by two, and did the cosine of that.
+189 So basically, it's the arccos of -1/9, then that divided by 2, and then the sin of that.
+118687 I don't have time to answer now. (even for myself) but I would probably start by letting
theta = acos (-1/9)
I will get back to it later/
+2407 Adding on to what Melody said. :))
Let theta = arccos(-1/9)
cos(theta) = -1/9
Note that the range of arccos is [-pi/2, pi2].
Can you figure out what sin(theta) is?
Hint: look at Melody's hint with the triangle and see if you can construct another one.
=^._.^=
+189 I could use a unit circle and then use the Pythagorean Theorem to say sin(theta)^2+(1/9)^2=1^2 or sin(theta)^2=80/81. This leads to sin(theta)=(sqrt80)/9.
Am I correct because I feel like I did something wrong?
And if I got sin(theta) right, how can I evaluate sin(theta/2)?
+189 I figured it out! Thanks for your help guys. I figured out that I could use the trig identity that sin(theta/2)=+-sqrt((1-cos(theta))/2). Then, I substitued cos(theta) with -1/9 and got sin(theta/2)=+-sqrt(5/9).
P.S. I am so relieved I got this right because I spent so long trying to figure it out!
+118687 I know what formula you are after but I can never remember them of the top of my head so...
Let
\(\theta = acos(\frac{-1}{9})\\ \quad note: cos \theta = \frac{-1}{9}\\ find \;\;sin\frac{\theta }{2}\)
--------------------------------------------
\(cos(\frac{\theta}{2}+\frac{\theta}{2})=cos^2\frac{\theta}{2}-sin^2\frac{\theta}{2}\\ cos(\frac{\theta}{2}+\frac{\theta}{2})=1-sin^2\frac{\theta}{2}-sin^2\frac{\theta}{2}\\ cos\theta=1-2sin^2\frac{\theta}{2}\\ \frac{1-cos\theta}{2}=sin^2\frac{\theta}{2}\\ sin\frac{\theta}{2}=\pm \sqrt{\frac{1-cos\theta}{2}}\\ sin\frac{\theta}{2}=\pm \sqrt{\frac{1-\frac{-1}{9}}{2}}\\ sin\frac{\theta}{2}=\pm \sqrt{\frac{\frac{10}{9}}{2}}\\ sin\frac{\theta}{2}=\pm \sqrt{\frac{10}{18}}\\ sin\frac{\theta}{2}=\pm\sqrt{\frac{5}{9}}\\ sin\frac{\theta}{2}=\pm \frac{\sqrt5}{3}\\ sin\left[\frac{acos(\frac{-1}{9})}{2}\right]=\pm\frac{\sqrt5}{3}\\\)
Theta has to be in the second quadrant but theta/2 can be in the first or second quadrent .
Either way. sin(theta) is positive!
\(\boxed{sin\left[\frac{acos(\frac{-1}{9})}{2}\right]=\frac{\sqrt5}{3}}\)
LaTex:
cos(\frac{\theta}{2}+\frac{\theta}{2})=cos^2\frac{\theta}{2}-sin^2\frac{\theta}{2}\\
cos(\frac{\theta}{2}+\frac{\theta}{2})=1-sin^2\frac{\theta}{2}-sin^2\frac{\theta}{2}\\
cos\theta=1-2sin^2\frac{\theta}{2}\\
\frac{1-cos\theta}{2}=sin^2\frac{\theta}{2}\\
sin\frac{\theta}{2}=\pm \sqrt{\frac{1-cos\theta}{2}}\\
sin\frac{\theta}{2}=\pm \sqrt{\frac{1-\frac{-1}{9}}{2}}\\
sin\frac{\theta}{2}=\pm \sqrt{\frac{\frac{10}{9}}{2}}\\
sin\frac{\theta}{2}=\pm \sqrt{\frac{10}{18}}\\
sin\frac{\theta}{2}=\pm\sqrt{\frac{5}{9}}\\
sin\frac{\theta}{2}=\pm \frac{\sqrt5}{3}\\
sin\left[\frac{acos(\frac{-1}{9})}{2}\right]=\frac{\sqrt5}{3}\\
