"If sec α=419,0<α<π2, then find the exact value..." was answered by students like you

College
Calculus and Analysis
Integral Calculus
Rational exponents and radicals

Ramsey

Answered question

2020-11-07

If sec $α = \frac{41}{9}, 0 < α < \frac{π}{2},$ then find the exact value of each of the following.
a) $\sin, \frac{α}{2}$
b) $\cos, \frac{α}{2}$
c) $\tan, \frac{α}{2}$

Answer & Explanation

cyhuddwyr9

Skilled2020-11-08Added 90 answers

Step 1
Given that,

\sec α = \frac{41}{9} and α

is in furst quadrant.
In first quandrant, all the trigonometric angles are positive.

\sec α = \frac{h y p}{a d j} = \frac{41}{9}

Using Pythagorean Theorem,

{(o p p)}^{2} = {(h y p)}^{2} - {(a d j)}^{2}

= {(41)}^{2} - {(9)}^{2}

= 1681 - 81

= 1600

o p p = \pm \sqrt{1600}

o p p = \pm 40

Since, all the angles area positive we take

o p p = 40

Step 2
a)

\sin, \frac{α}{2} :

Using the formula,

\sin, \frac{α}{2} = \frac{\sqrt{1 - \cos α}}{2}

... (1)

\cos α = \frac{1}{\sec α} = \frac{1}{\frac{41}{9}} = \frac{9}{41}

Step 3
Substitute the above value in equation (1), we get

\sin, \frac{α}{2} = \sqrt{\frac{1 - \cos α}{2}}

\sin, \frac{α}{2} = \sqrt{\frac{1 - \frac{9}{41}}{2}}

= \sqrt{\frac{\frac{41 - 9}{41}}{2}}

= \sqrt{\frac{32}{41 \times 2}}

= \sqrt{\frac{16}{41}}

= \frac{\sqrt{16}}{\sqrt{41}}

= \frac{4}{\sqrt{41}}

= \frac{4}{\sqrt{41}} \times \frac{\sqrt{41}}{\sqrt{41}}

\sin, \frac{α}{2} = \frac{4 \sqrt{41}}{41}

Step 4
b)

\cos, \frac{α}{2}

Using the formula,

\cos, \frac{α}{2} = \sqrt{\frac{1 + \cos α}{2}}

..(2)
Substitute

\cos α = \frac{9}{41}

in the above formula, we get

\cos, \frac{α}{2} = \sqrt{\frac{1 + \cos α}{2}}

= \sqrt{\frac{1 + \frac{9}{41}}{2}}

= \sqrt{\frac{50}{41 \times 2}}

= \sqrt{\frac{25}{41}}

= \frac{\sqrt{25}}{\sqrt{41}}

= \frac{5}{\sqrt{41}}

= \frac{5}{\sqrt{41}} \times \frac{\sqrt{41}}{\sqrt{41}}

\cos, \frac{α}{2} = \frac{5 \sqrt{41}}{41}

Step 5
c)

\tan, \frac{α}{2}

Using the formula,

\tan, \frac{α}{2} =

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