Solve Trig Equation: cosθ=4/9, 3π/2≤θ≤2π

[TonyC]

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

 

[VietDao29]

Nope, I don't think it's right.
You have:
\frac{3\pi}{2} \leq \theta \leq 2\pi
\Leftrightarrow \frac{3\pi}{4} \leq \frac{\theta}{2} \leq \pi
So \sin \frac{\theta}{2} > 0, and \cos \frac{\theta}{2} < 0
You also have:
\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta
Therefore:
\cos 2 \theta = \cos ^ 2 \theta - \sin ^ 2 \theta = 2\cos ^ 2 \theta - 1
In other word:
\cos \theta = \cos ^ 2 \frac{\theta}{2} - \sin ^ 2 \frac{\theta}{2} = 2\cos ^ 2 \frac{\theta}{2} - 1
From the equation, you will work out cos(theta / 2). Remember that cos(theta / 2) < 0.
You can then use
\cos ^ 2 \alpha + \sin ^ 2 \alpha = 1
to find out sin(theta / 2). Remember sin(theta / 2) > 0.
Viet Dao,

 

[TonyC]

How would I express that in radical form?

 

[VietDao29]

What do you mean?
Have you covered:
\cos 2 \theta = \cos ^ 2 \theta - \sin ^ 2 \theta = 2\cos ^ 2 \theta - 1 = 1 - 2\sin ^ 2 \theta yet?
Viet Dao,

 

[TonyC]

I came up with:
sq rt 10 - sq rt 26 all over 6

 

[VietDao29]

Yup, that's correct.
Viet Dao,