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More trig homework

  • Thread starter TonyC
  • Start date
86
0
given cos theta=4/9, where 3pi/2 is less than/equal theta greater than/equal 2pi.
Find exact value of sin1/2theta+cos1/2theta

I have come up with sq rt10 + sq rt26 / 6

Just don't have a warm fuzzy
 

Answers and Replies

VietDao29
Homework Helper
1,417
1
Nope, I don't think it's right.
You have:
[tex]\frac{3\pi}{2} \leq \theta \leq 2\pi[/tex]
[tex]\Leftrightarrow \frac{3\pi}{4} \leq \frac{\theta}{2} \leq \pi[/tex]
So [tex]\sin \frac{\theta}{2} > 0[/tex], and [tex]\cos \frac{\theta}{2} < 0[/tex]
You also have:
[tex]\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta[/tex]
Therefore:
[tex]\cos 2 \theta = \cos ^ 2 \theta - \sin ^ 2 \theta = 2\cos ^ 2 \theta - 1[/tex]
In other word:
[tex]\cos \theta = \cos ^ 2 \frac{\theta}{2} - \sin ^ 2 \frac{\theta}{2} = 2\cos ^ 2 \frac{\theta}{2} - 1[/tex]
From the equation, you will work out cos(theta / 2). Remember that cos(theta / 2) < 0.
You can then use
[tex]\cos ^ 2 \alpha + \sin ^ 2 \alpha = 1[/tex]
to find out sin(theta / 2). Remember sin(theta / 2) > 0.
Viet Dao,
 
86
0
How would I express that in radical form?
 
VietDao29
Homework Helper
1,417
1
What do you mean?
Have you covered:
[tex]\cos 2 \theta = \cos ^ 2 \theta - \sin ^ 2 \theta = 2\cos ^ 2 \theta - 1 = 1 - 2\sin ^ 2 \theta[/tex] yet?
Viet Dao,
 
86
0
I came up with:
sq rt 10 - sq rt 26 all over 6
 
VietDao29
Homework Helper
1,417
1
Yup, that's correct.
Viet Dao,
 

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