+0

# Pls help if possible

0
200
14

Calculate sin([arccos(-1/9)]/2) and put the answer as a fraction.

Jan 18, 2022

#1
+2

Hint: Jan 18, 2022
#2
-1

I'm sorry, but I don't really understand the hint.

ImDecentAtMath  Jan 18, 2022
#3
+2

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.

Melody  Jan 18, 2022
#4
0

cos(theta) would be -2/3

ImDecentAtMath  Jan 18, 2022
#5
+2

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.

Melody  Jan 18, 2022
edited by Melody  Jan 18, 2022
edited by Melody  Jan 19, 2022
#6
-1

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.

ImDecentAtMath  Jan 18, 2022
#7
+2

Oh sorry.

I didn't notice the brackets properly.  My fault.

Melody  Jan 19, 2022
edited by Melody  Jan 19, 2022
#8
-1

So basically, it's the arccos of -1/9, then that divided by 2, and then the sin of that.

ImDecentAtMath  Jan 19, 2022
#9
+2

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/

Jan 19, 2022
#10
+2

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.

=^._.^=

catmg  Jan 19, 2022
#11
0

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)?

ImDecentAtMath  Jan 19, 2022
edited by ImDecentAtMath  Jan 19, 2022
edited by ImDecentAtMath  Jan 19, 2022
#12
+1

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!

ImDecentAtMath  Jan 19, 2022
#13
0

Nice job. :DDD

=^._.^=

catmg  Jan 19, 2022
#14
+2

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}\\

Jan 19, 2022
edited by Melody  Jan 19, 2022
edited by Melody  Jan 19, 2022