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Evaluating trigonometric expressions...

I am having trouble in evaluating the following expression:


sec⁡(1/2arcsin⁡2/3)\sec (1/2 \arcsin 2/3)


Here is my working out:
Let's forget about the reciprocal for a moment.

cos⁡(1/2arcsin⁡2/3)\cos (1/2 \arcsin 2/3)

let:u=arcsin⁡2/3u = \arcsin 2/3
so: sin⁡u=2/3\sin u = 2/3

Therefore (using Pythagorean Identities):
cos⁡u=5/3\cos u = \sqrt 5/3

So:
1/(cos⁡(1/2u))1 / (\cos (1/2 u))
(From cos⁡2+sin⁡2=1\cos^2 + \sin^2 = 1 )



Substituting the half angle formula:



((1+cos⁡u)/2)\sqrt((1+ \cos u) / 2)

Substituting back cos u:

((1+5/3)/2) \sqrt((1+ \sqrt 5/3) / 2)


((3+53/2) \sqrt((\dfrac{3 + \sqrt 5}{3}/ 2)

(3+53∗1/2) \sqrt(\dfrac{3 + \sqrt 5}{3}* 1/2)

(3+56) \sqrt(\dfrac{3 + \sqrt 5}{6})

(3+5)6 \dfrac{\sqrt (3 + \sqrt 5)}{\sqrt6}

Putting the original reciprocal back:

6(3+5) \dfrac{\sqrt6}{\sqrt (3 + \sqrt 5)}


I am unsure how the book gives the following answer:

(18−6(5))2 \dfrac{\sqrt(18 - 6 \sqrt(5))}{2}




Can someone please tell me where I have gone wrong?
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B_9710
18
Original post by Incongruous
I am having trouble in evaluating the following expression:


sec⁡(1/2arcsin⁡2/3)\sec (1/2 \arcsin 2/3)


Here is my working out:
Let's forget about the reciprocal for a moment.

cos⁡(1/2arcsin⁡2/3)\cos (1/2 \arcsin 2/3)

let:u=arcsin⁡2/3u = \arcsin 2/3
so: sin⁡u=2/3\sin u = 2/3

Therefore (using Pythagorean Identities):
cos⁡u=5/3\cos u = \sqrt 5/3

So:
1/(cos⁡(1/2u))1 / (\cos (1/2 u))
(From cos⁡2+sin⁡2=1\cos^2 + \sin^2 = 1 )



Substituting the half angle formula:



((1+cos⁡u)/2)\sqrt((1+ \cos u) / 2)

Substituting back cos u:

((1+5/3)/2) \sqrt((1+ \sqrt 5/3) / 2)


((3+53/2) \sqrt((\dfrac{3 + \sqrt 5}{3}/ 2)

(3+53∗1/2) \sqrt(\dfrac{3 + \sqrt 5}{3}* 1/2)

(3+56) \sqrt(\dfrac{3 + \sqrt 5}{6})

(3+5)6 \dfrac{\sqrt (3 + \sqrt 5)}{\sqrt6}

Putting the original reciprocal back:

6(3+5) \dfrac{\sqrt6}{\sqrt (3 + \sqrt 5)}


I am unsure how the book gives the following answer:

(18−6(5))2 \dfrac{\sqrt(18 - 6 \sqrt(5))}{2}




Can someone please tell me where I have gone wrong?


Multiply by 3−53−5 \displaystyle \frac{\sqrt{3-\sqrt 5 }}{\sqrt{3-\sqrt 5}} .
Similar to how you would normally rationalise the denominator.
(edited 7 years ago)
Original post by B_9710
Multiply by 3−53−5 \displaystyle \frac{\sqrt{3-\sqrt 5 }}{\sqrt{3-\sqrt 5}} .
Similar to how you would normally rationalise the denominator.


Thank you

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