Help With Double Angle And Half Angle Formulas(Algebra 2)!

[CarribeanKid]

Hey, i just joined this site, but i am having a very hard time with this stuff. I was wondering if you guys could help me with the following problems, i would really appreciate it. I will Post the 3 problems and then tell you how i am confused. These are homework questions, but i really need help.

Find the exact values of Sin2θ,Cosθ,Sin θ/2 and Cosθ/2.

Cosθ= 1/5; 270%<θ<360%

Cosθ= -1/4; 90%<θ<180%

Cosθ=-12/13; 180%<θ<270%

As you can see all the problems i need help with have to do with cosine. This is because i somewhat understand what to do with Sine, but i am unclear about cosine. I know the formulas for the double and half angles, and i understand how to use them. But as far as getting started, i am very confused. So if anyone could help, or even walk me through it, i would really appreciate it. I am a pretty good student, but i require specific details for many things, as i am not as quick to apply as others. Anyways,thanks for the help.[/img][/list][/code]

 

[arthur ohlsten]

1)
cosB=1/5
cosB=.2 answer

from the right triangle in 4th quadrant
sinB= - [sqrt24]/ 5

sin2B= 2sinB cos B
sin 2B= -2[sqrt24]/25 answer

sin [B/2 +B/2] = 2 sin B/2 cos B/2
sin B= 2 sin B/2 cos B/2

cos[B/2+B/2]= cos^2 B/2 -sin^2 B/2
cosB= 1-sin^2B/2 -sin^2B/2
2 sin ^2 B/2 = 1- cosB
sin B/2 = sqrt {[1-cosB] / 2}
sinB/2= sqrt {1- 1/5]/2 }
sinB/2 = sqrt[ 4/10] answer

cosB=cos^2 B/2 -sin^2 B/2
1/5 = cos^2 B/2 - [1-1/5]/2
1/5 = cos^2 B/2 -4/10
6/10 =cos^2 B/2
cos B/2 = sqrt3 /5 answer

The remaining two you should be able to do .
Arthur

 

[soroban]

Re: Help With Double Angle And Half Angle Formulas(Algebra 2

Hello, CarribeanKid!

Welcome aboard!

I'll show you my appoach to #2 . . .

\(\displaystyle \cos\theta\,=\,-\frac{1}{4},\;90^o\,<\,\theta\,<\,180^o\)

Find the exact values of: \(\displaystyle \,\sin2\theta,\;\cos2\theta,\;\sin\frac{\theta}{2},\;\cos\frac{\theta}{2}\)

You will need these formulas/identities:

\(\displaystyle \sin2x\;=\;2\cdot\sin x\cdot\cos x\) . . . . . \(\displaystyle \cos2x\;=\;\cos^2x\,-\,\sin^2x\)

\(\displaystyle \sin\frac{x}{2}\;=\;\pm\sqrt{\frac{1\,-\,\cos x}{2}}\). . . . . . \(\displaystyle \cos\frac{x}{2}\;=\;\pm\sqrt{\frac{1\,+\,\cos x}{2}}\)

Also: \(\displaystyle \sin^2x\,+\,\cos^2x\:=\:1\;\;\Rightarrow\;\;\sin x\:=\:\pm\sqrt{1\,-\,\cos^2x}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

(a) \(\displaystyle \,\sin2\theta\;\;\) Identity: \(\displaystyle \:\sin2\theta\:=\:2\cdot\sin\theta\cdot\cos\theta\)

We know that: \(\displaystyle \cos\theta\,=\,-\frac{1}{4}\)
We need the value of \(\displaystyle \sin\theta\)

Identity: \(\displaystyle \sin\theta\:=\:\pm\sqrt{1\,-\,\cos^2\theta}\)

We have: \(\displaystyle \:\sin\theta\;=\;\pm\sqrt{1\,-\,\left(-\frac{1}{4}\right)^2} \;=\;\pm\sqrt{\frac{15}{16}}\;=\;\pm\frac{\sqrt{15}}{4}\)

Since \(\displaystyle \theta\) is in Quadrant 2, sine is positive: \(\displaystyle \:\sin\theta \,=\,\frac{\sqrt{15}}{4}\)

Therefore: \(\displaystyle \L\:\sin2\theta \;=\;2\left(\frac{\sqrt{15}}{4}\right)\left(-\frac{1}{4}\right)\;=\;\fbox{-\frac{\sqrt{15}}{8}}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

(b) \(\displaystyle \cos2\theta\;\;\) Identity: \(\displaystyle \:\cos2\theta\:=\:\cos^2\theta\,-\,\sin^2\theta\)

Hence: \(\displaystyle \L\cos2\theta\;=\;\left(-\frac{1}{4}\right)^2\,-\,\left(\frac{\sqrt{15}}{4}\right)^2\;=\;\frac{1}{16}\,-\,\frac{15}{16}\;=\;-\frac{14}{16}\;=\;\fbox{-\frac{7}{8}}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

(c) \(\displaystyle \sin\frac{\theta}{2}\;\;\) Identity: \(\displaystyle \:\sin\frac{\theta}{2} \:=\:\pm\sqrt{\frac{1\,-\,\cos\theta}{2}}\)

Hence: \(\displaystyle \:\sin\frac{\theta}{2}\;=\;\pm\sqrt{\frac{1\,-\,(-\frac{1}{4})}{2}} \;=\;\pm\sqrt{\frac{(\frac{5}{4})}{2}} \;=\;\pm\sqrt{\frac{5}{8}} \;=\;\pm\frac{\sqrt{10}}{4}\)

Since \(\displaystyle \theta\) is in Quadrant 2, \(\displaystyle \frac{\theta}{2}\) is in Quadrant 1, where sine is positive.

Therefore: \(\displaystyle \L\:\sin\frac{\theta}{2}\;=\:\fbox{\frac{\sqrt{10}}{4}}\)

=~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

(d) \(\displaystyle \cos\frac{\theta}{2}\;\;\) Identity: \(\displaystyle \,\cos\frac{\theta}{2}\:=\:\pm\sqrt{\frac{1\,+\,\cos\theta}{2}}\)

Hence: \(\displaystyle \:\cos\frac{\theta}{2}\;=\;\pm\sqrt{\frac{1\,-\,\(-\frac{1}{4})}{2}} \;=\;\pm\sqrt{\frac{(\frac{3}{4})}{2}} \:=\:\pm\sqrt{\frac{3}{8}}\:=\:\pm\frac{\sqrt{6}}{4}\)

Since \(\displaystyle \frac{\theta}{2}\) is in Quadrant 1, cosine is positive.

Therefore: \(\displaystyle \L\:\cos\frac{\theta}{2}\;=\;\fbox{\frac{\sqrt{6}}{4}}\)

 

[CarribeanKid]

Hey just wanted to say thanks to you guys. Once again, i am by no means just going to copy these answers, but my book just has examples for SINE and it expects us to figure out Cosine, which is obviously easy, but like i said, i am not as quick to catch on. Anyways, i really appreciate the help, I have a test today so wish me luck!

 

[arthur ohlsten]

Good luck, and glad to have helped
Arthur