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Using the power-reducing formulas to prove an identity

Use the power-reducing formulas to prove

sin3(2x)=[12sin(2x)][1cos(4x)]

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

sin3(2x)=[sin(2x)][sin2(2x)]=sin(2x)[1cos(4x)2]Substitute the power-reduction formula.=sin(2x)(12)[1cos(4x)]=12[sin(2x)][1cos(4x)]
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Use the power-reducing formulas to prove that 10cos4x=154+5cos(2x)+54cos(4x).

10cos4x=10(cos2x)2=10[1+cos(2x)2]2Substitute reduction formula for cos2x.=104[1+2cos(2x)+cos2(2x)]=104+102cos(2x)+104(1+cos2(2x)2)Substitute reduction formula for cos2x.=104+102cos(2x)+108+108cos(4x)=308+5cos(2x)+108cos(4x)=154+5cos(2x)+54cos(4x)

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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 α2, the half-angle formula for sine is found by simplifying the equation and solving for sin(α2). 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 α2 terminates.

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

sin2θ=1cos(2θ)2sin2(α2)=1(cos2α2)2=1cosα2sin(α2)=±1cosα2

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

cos2θ=1+cos(2θ)2cos2(α2)=1+cos(2α2)2=1+cosα2cos(α2)=±1+cosα2

For the tangent identity, we have

tan2θ=1cos(2θ)1+cos(2θ)tan2(α2)=1cos(2α2)1+cos(2α2)=1cosα1+cosαtan(α2)=±1cosα1+cosα

Half-angle formulas

The half-angle formulas    are as follows:

sin(α2)=±1cosα2
cos(α2)=±1+cosα2
tan(α2)=±1cosα1+cosα=sinα1+cosα=1cosαsinα

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:

sin30°2=1cos30°2=1322=2322=234=232

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 tanα=815 and α lies in quadrant III, find the exact value of the following:

  1. sin(α2)
  2. cos(α2)
  3. tan(α2)

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 sinα=817 and cosα=1517.

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<α2<270°2. This means that the terminal side of α2 is in quadrant II, since 90°<α2<135°.

    To find sinα2, 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.

    sinα2=±1cosα2=±1(1517)2=±32172=±321712=±1617=±417=41717

    We choose the positive value of sinα2 because the angle terminates in quadrant II and sine is positive in quadrant II.

  2. To find cosα2, we will write the half-angle formula for cosine, substitute the value of the cosine we found from the triangle in [link] , and simplify.
    cosα2=±1+cosα2=±1+(1517)2=±2172=±21712=±117=1717

    We choose the negative value of cosα2 because the angle is in quadrant II because cosine is negative in quadrant II.

  3. To find tanα2, 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.
    tanα2=±1cosα1+cosα=±1(1517)1+(1517)=±3217217=±322=16=−4

    We choose the negative value of tanα2 because α2 lies in quadrant II, and tangent is negative in quadrant II.

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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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