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Using the power-reducing formulas to prove an identity

Use the power-reducing formulas to prove

sin 3 ( 2 x ) = [ 1 2 sin ( 2 x ) ] [ 1 cos ( 4 x ) ]

We will work on simplifying the left side of the equation:

sin 3 ( 2 x ) = [ sin ( 2 x ) ] [ sin 2 ( 2 x ) ] = sin ( 2 x ) [ 1 cos ( 4 x ) 2 ] Substitute the power-reduction formula . = sin ( 2 x ) ( 1 2 ) [ 1 cos ( 4 x ) ] = 1 2 [ sin ( 2 x ) ] [ 1 cos ( 4 x ) ]
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Use the power-reducing formulas to prove that 10 cos 4 x = 15 4 + 5 cos ( 2 x ) + 5 4 cos ( 4 x ) .

10 cos 4 x = 10 ( cos 2 x ) 2 = 10 [ 1 + cos ( 2 x ) 2 ] 2 Substitute reduction formula for cos 2 x . = 10 4 [ 1 + 2 cos ( 2 x ) + cos 2 ( 2 x ) ] = 10 4 + 10 2 cos ( 2 x ) + 10 4 ( 1 + cos 2 ( 2 x ) 2 ) Substitute reduction formula for cos 2 x . = 10 4 + 10 2 cos ( 2 x ) + 10 8 + 10 8 cos ( 4 x ) = 30 8 + 5 cos ( 2 x ) + 10 8 cos ( 4 x ) = 15 4 + 5 cos ( 2 x ) + 5 4 cos ( 4 x )

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Using half-angle formulas to find exact values

The next set of identities is the set of half-angle formulas    , which can be derived from the reduction formulas and we can use when we have an angle that is half the size of a special angle. If we replace θ with α 2 , the half-angle formula for sine is found by simplifying the equation and solving for sin ( α 2 ) . Note that the half-angle formulas are preceded by a ± sign. This does not mean that both the positive and negative expressions are valid. Rather, it depends on the quadrant in which α 2 terminates.

The half-angle formula for sine is derived as follows:

sin 2 θ = 1 cos ( 2 θ ) 2 sin 2 ( α 2 ) = 1 ( cos 2 α 2 ) 2 = 1 cos α 2 sin ( α 2 ) = ± 1 cos α 2

To derive the half-angle formula for cosine, we have

cos 2 θ = 1 + cos ( 2 θ ) 2 cos 2 ( α 2 ) = 1 + cos ( 2 α 2 ) 2 = 1 + cos α 2 cos ( α 2 ) = ± 1 + cos α 2

For the tangent identity, we have

tan 2 θ = 1 cos ( 2 θ ) 1 + cos ( 2 θ ) tan 2 ( α 2 ) = 1 cos ( 2 α 2 ) 1 + cos ( 2 α 2 ) = 1 cos α 1 + cos α tan ( α 2 ) = ± 1 cos α 1 + cos α

Half-angle formulas

The half-angle formulas    are as follows:

sin ( α 2 ) = ± 1 cos α 2
cos ( α 2 ) = ± 1 + cos α 2
tan ( α 2 ) = ± 1 cos α 1 + cos α = sin α 1 + cos α = 1 cos α sin α

Using a half-angle formula to find the exact value of a sine function

Find sin ( 15° ) using a half-angle formula.

Since 15° = 30° 2 , we use the half-angle formula for sine:

sin 30° 2 = 1 cos 30° 2 = 1 3 2 2 = 2 3 2 2 = 2 3 4 = 2 3 2

Remember that we can check the answer with a graphing calculator.

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Given the tangent of an angle and the quadrant in which the angle lies, find the exact values of trigonometric functions of half of the angle.

  1. Draw a triangle to represent the given information.
  2. Determine the correct half-angle formula.
  3. Substitute values into the formula based on the triangle.
  4. Simplify.

Finding exact values using half-angle identities

Given that tan α = 8 15 and α lies in quadrant III, find the exact value of the following:

  1. sin ( α 2 )
  2. cos ( α 2 )
  3. tan ( α 2 )

Using the given information, we can draw the triangle shown in [link] . Using the Pythagorean Theorem, we find the hypotenuse to be 17. Therefore, we can calculate sin α = 8 17 and cos α = 15 17 .

Diagram of a triangle in the x,y-plane. The vertices are at the origin, (-15,0), and (-15,-8). The angle at the origin is alpha. The angle formed by the side (-15,-8) to (-15,0) forms a right angle with the x axis. The hypotenuse across from the right angle is length 17.
  1. Before we start, we must remember that if α is in quadrant III, then 180° < α < 270° , so 180° 2 < α 2 < 270° 2 . This means that the terminal side of α 2 is in quadrant II, since 90° < α 2 < 135° .

    To find sin α 2 , we begin by writing the half-angle formula for sine. Then we substitute the value of the cosine we found from the triangle in [link] and simplify.

    sin α 2 = ± 1 cos α 2 = ± 1 ( 15 17 ) 2 = ± 32 17 2 = ± 32 17 1 2 = ± 16 17 = ± 4 17 = 4 17 17

    We choose the positive value of sin α 2 because the angle terminates in quadrant II and sine is positive in quadrant II.

  2. To find cos α 2 , we will write the half-angle formula for cosine, substitute the value of the cosine we found from the triangle in [link] , and simplify.
    cos α 2 = ± 1 + cos α 2 = ± 1 + ( 15 17 ) 2 = ± 2 17 2 = ± 2 17 1 2 = ± 1 17 = 17 17

    We choose the negative value of cos α 2 because the angle is in quadrant II because cosine is negative in quadrant II.

  3. To find tan α 2 , we write the half-angle formula for tangent. Again, we substitute the value of the cosine we found from the triangle in [link] and simplify.
    tan α 2 = ± 1 cos α 1 + cos α = ± 1 ( 15 17 ) 1 + ( 15 17 ) = ± 32 17 2 17 = ± 32 2 = 16 = −4

    We choose the negative value of tan α 2 because α 2 lies in quadrant II, and tangent is negative in quadrant II.

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Questions & Answers

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Practice Key Terms 3

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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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