- #1

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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

- Thread starter TonyC
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- #1

- 86

- 0

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

- #2

VietDao29

Homework Helper

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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,

- #3

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How would I express that in radical form?

- #4

VietDao29

Homework Helper

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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,

- #5

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I came up with:

sq rt 10 - sq rt 26 all over 6

sq rt 10 - sq rt 26 all over 6

- #6

VietDao29

Homework Helper

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Yup, that's correct.

Viet Dao,

Viet Dao,

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