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How would I solve this?

How would I solve this?-example-1
User Yexo
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Explanation:

In these types of problems, you need to use various trigonometric identities to solve this.

First, note what are you trying to solve.

We are trying to find


\cos( ( \alpha )/(2) )

Using our trig identities, we know that


\cos( ( \alpha )/(2) ) = \sqrt{ (1 + \cos( \alpha ) )/(2) }

Next, in order to solve this problem, we need to find cos(alpha).

So using our given info, of tan(alpha), we must find a way to manipulate that information to get cos(alpha)

Notice that


\tan {}^(2) ( \alpha ) + 1 = \sec {}^(2) ( \alpha )


\sqrt{ \tan {}^(2) ( \alpha ) + 1 } = \sec( \alpha )

And that


\cos( \alpha ) = (1)/( \sec( \alpha ) )

So,


\cos( \alpha ) = \frac{1}{ \sqrt{ \tan {}^(2) ( \alpha ) + 1 } }

Using the interval, since cos is negative over the interval (pi,3pi/2),

our answer will be negative


\cos( \alpha ) = - \frac{1}{ \sqrt{ \tan {}^(2) ( \alpha ) + 1 } }

We know that


\tan( \alpha ) = (11)/(60)

So


\cos( \alpha ) = - \frac{1}{ \sqrt{( (11)/(60) ) {}^(2) + 1 } }


\cos( \alpha ) = - \frac{1}{ \sqrt{ (121)/(3600) + 1} }


\cos( \alpha ) = - (1)/( (61)/(60) )


\cos( \alpha ) = - (60)/(61)

Since we know what cos(alpha) is, We can solve cos(alpha/2), likewise, cos will be negative since the former is negative over pi and 3pi/2


\cos( ( \alpha )/(2) ) = - \sqrt{ (1 + ( - (60)/(61) ))/(2) }


\cos( ( \alpha )/(2) ) = - \sqrt{ ( (1)/(61) )/(2) }


\cos( ( \alpha )/(2) ) = - \sqrt{ (1)/(122) }


\cos( ( \alpha )/(2) ) = - (1)/( √(122) )


\cos( ( \alpha )/(2) ) = - ( √(122) )/(122)

User Typo Johnson
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