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Finding the exact value of an expression involving an inverse trigonometric function

Find the exact value of sin ( cos −1 1 2 + sin −1 3 5 ) . Then check the answer with a graphing calculator.

The pattern displayed in this problem is sin ( α + β ) . Let α = cos −1 1 2 and β = sin −1 3 5 . Then we can write

cos α = 1 2 , 0 α π sin β = 3 5 , π 2 β π 2

We will use the Pythagorean identities to find sin α and cos β .

sin α = 1 cos 2 α = 1 1 4 = 3 4 = 3 2 cos β = 1 sin 2 β = 1 9 25 = 16 25 = 4 5

Using the sum formula for sine,

sin ( cos 1 1 2 + sin 1 3 5 ) = sin ( α + β ) = sin α cos β + cos α sin β = 3 2 4 5 + 1 2 3 5 = 4 3 + 3 10
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Using the sum and difference formulas for tangent

Finding exact values for the tangent of the sum or difference of two angles is a little more complicated, but again, it is a matter of recognizing the pattern.

Finding the sum of two angles formula for tangent involves taking quotient of the sum formulas for sine and cosine and simplifying. Recall, tan x = sin x cos x , cos x 0.

Let’s derive the sum formula for tangent.

tan ( α + β ) = sin ( α + β ) cos ( α + β ) = sin α cos β + cos α sin β cos α cos β sin α sin β = sin α cos β + cos α sin β cos α cos β cos α cos β sin α sin β cos α cos β Divide the numerator and denominator by cos α cos β . = sin α cos β cos α cos β + cos α sin β cos α cos β cos α cos β cos α cos β sin α sin β cos α cos β = sin α cos α + sin β cos β 1 sin α sin β cos α cos β = tan α + tan β 1 tan α tan β

We can derive the difference formula for tangent in a similar way.

Sum and difference formulas for tangent

The sum and difference formulas for tangent are:

tan ( α + β ) = tan α + tan β 1 tan α tan β
tan ( α β ) = tan α tan β 1 + tan α tan β

Given two angles, find the tangent of the sum of the angles.

  1. Write the sum formula for tangent.
  2. Substitute the given angles into the formula.
  3. Simplify.

Finding the exact value of an expression involving tangent

Find the exact value of tan ( π 6 + π 4 ) .

Let’s first write the sum formula for tangent and then substitute the given angles into the formula.

tan ( α + β ) = tan α + tan β 1 tan α tan β tan ( π 6 + π 4 ) = tan ( π 6 ) + tan ( π 4 ) 1 ( tan ( π 6 ) ) ( tan ( π 4 ) )

Next, we determine the individual function values within the formula:

tan ( π 6 ) = 1 3 , tan ( π 4 ) = 1

So we have

tan ( π 6 + π 4 ) = 1 3 + 1 1 ( 1 3 ) ( 1 ) = 1 + 3 3 3 1 3 = 1 + 3 3 ( 3 3 1 ) = 3 + 1 3 1
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Find the exact value of tan ( 2 π 3 + π 4 ) .

1 3 1 + 3

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Finding multiple sums and differences of angles

Given   sin α = 3 5 , 0 < α < π 2 , cos β = 5 13 , π < β < 3 π 2 , find

  1. sin ( α + β )
  2. cos ( α + β )
  3. tan ( α + β )
  4. tan ( α β )

We can use the sum and difference formulas to identify the sum or difference of angles when the ratio of sine, cosine, or tangent is provided for each of the individual angles. To do so, we construct what is called a reference triangle to help find each component of the sum and difference formulas.

  1. To find sin ( α + β ) , we begin with sin α = 3 5 and 0 < α < π 2 . The side opposite α has length 3, the hypotenuse has length 5, and α is in the first quadrant. See [link] . Using the Pythagorean Theorem, we can find the length of side a :
    a 2 + 3 2 = 5 2 a 2 = 16 a = 4
    Diagram of a triangle in the x,y plane. The vertices are at the origin, (4,0), and (4,3). The angle at the origin is alpha degrees, The angle formed by the x-axis and the side from (4,3) to (4,0) is a right angle. The side opposite the right angle has length 5.

    Since cos β = 5 13 and π < β < 3 π 2 , the side adjacent to β is −5 , the hypotenuse is 13, and β is in the third quadrant. See [link] . Again, using the Pythagorean Theorem, we have

    ( −5 ) 2 + a 2 = 13 2 25 + a 2 = 169 a 2 = 144 a = ± 12

    Since β is in the third quadrant, a = –12.

    Diagram of a triangle in the x,y plane. The vertices are at the origin, (-5,0), and (-5, -12). The angle at the origin is Beta degrees. The angle formed by the x axis and the side from (-5, -12) to (-5,0) is a right angle. The side opposite the right angle has length 13.

    The next step is finding the cosine of α and the sine of β . The cosine of α is the adjacent side over the hypotenuse. We can find it from the triangle in [link] : cos α = 4 5 . We can also find the sine of β from the triangle in [link] , as opposite side over the hypotenuse: sin β = 12 13 . Now we are ready to evaluate sin ( α + β ) .

    sin ( α + β ) = sin α cos β + cos α sin β = ( 3 5 ) ( 5 13 ) + ( 4 5 ) ( 12 13 ) = 15 65 48 65 = 63 65
  2. We can find cos ( α + β ) in a similar manner. We substitute the values according to the formula.
    cos ( α + β ) = cos α cos β sin α sin β = ( 4 5 ) ( 5 13 ) ( 3 5 ) ( 12 13 ) = 20 65 + 36 65 = 16 65
  3. For tan ( α + β ) , if sin α = 3 5 and cos α = 4 5 , then
    tan α = 3 5 4 5 = 3 4

    If sin β = 12 13 and cos β = 5 13 , then

    tan β = 12 13 5 13 = 12 5

    Then,

    tan ( α + β ) = tan α + tan β 1 tan α tan β = 3 4 + 12 5 1 3 4 ( 12 5 ) =    63 20 16 20 = 63 16
  4. To find tan ( α β ) , we have the values we need. We can substitute them in and evaluate.
    tan ( α β ) = tan α tan β 1 + tan α tan β = 3 4 12 5 1 + 3 4 ( 12 5 ) = 33 20 56 20 = 33 56
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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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