The Algebra of Trigonometric Functions: Solutions

Here are some solutions to the Exercises to accompany the section The Algebra of Trigonometric Functions.  

Basic Trigonometric Identities

1. Use the geometry of the unit circle:

    

   to match each of  the following function evaluations on the left with the corresponding exact value on the right.  Note: The same values may be used more than once.

   cos(p) sin(p) cos(-p/2) sin(-p/2) cos(2p/3)  sin(2p/3) cos(-p/4)  sin(-p/4) cos(5p/6) sin(5p/6) sin(7p/3) cos(7p/3) sin(-4p/5) cos(-4p/5)

   1 » 0.9 » 0.8 » 0.7 1/2 = 0.5 » 0.3 0 - » -0.3 -1/2 = -0.5 » -0.7 - » -0.8 - » -0.9 -1

   Hint: First find the correct point on the unit circle, estimate the correct value to within 0.1, then use the approximate values to match up with the exact value.  Compare each answer with the values the table of exact values from the text 

   Solution

   We can see that cos(p) = -1, sin(p) = 0, cos(-p/2) = 0, and sin(-p/2) = -1 directly from the unit circle.  We can see, from the symmetry of the circle, that cos(2p/3) = -cos(p/3), which visually and from the table is exactly -0.5.  Similarly, sin(2p/3) = sin(p/3) = » 0.9, cos(-p/4) = cos(p/4) = » 0.7, sin(-p/4) = -sin(p/4) = - » -0.7, cos(5p/6) = -cos(p/6) = - » -0.9, sin(5p/6) = sin(p/6) = 0.5 (i.e., our estimate is exactly right), sin(7p/3) = sin(p/3) = , cos(7p/3) = cos(p/3) = 0.5, sin(-4p/5) = -sin(p/5) = » -0.3, cos(-4p/5) = -cos(p/5) = - » -0.8.

   Back to Exercises.

2. Use your answers to the previous Exercise and the definitions of the trigonometric functions to give exact values for the following:
   1. tan(p)
   2. sec(p)
   3. sec(-p/2)
   4. csc(-p/2)
   5. sec(2p/3) 
   6. cot(2p/3)
   7. csc(-p/4) 
   8. tan(-p/4)
   9. sec(5p/6)
   10. csc(5p/6)
   11. cot(7p/3)
   12. csc(7p/3)
   13. sec(-4p/5)
   14. tan(-4p/5)

   Solution

   tan(p) = sin(p)/cos(p) = 0/(-1) = 0.
   sec(p) = 1/cos(p) = 1/(-1) = -1.
   sec(-p/2) = 1/cos(-p/2) = 1/0 = undefined.
   csc(-p/2) = 1/sin(-p/2) = 1/(-1) = -1.
   sec(2p/3) = 1/cos(2p/3) = 1/(-0.5) = -2.

   cot(2p/3) = cos(2p/3)/sin(2p/3) = .
   csc(-p/4) = 1/sin(-p/4) = .
   tan(-p/4) = sin(-p/4)/cos(-p/4) = .
   sec(5p/6) = 1/cos(5p/6) = . 
   csc(5p/6) = 1/sin(5p/6) = 1/(0.5) = 2.
   cot(7p/3) = cos(7p/3)/sin(7p/3) = .
   csc(7p/3) = 1/sin(7p/3) = .
   sec(-4p/5) = 1/cos(-4p/5) = .
   tan(-4p/5) = sin(-4p/5)/cos(-4p/5) = .

   Back to Exercises.

3. Use the indicated identity to solve for the indicated value:
   1. Use the Pythagorean identity to solve for sin(2p/5), starting from cos(2p/5) = .

      Solution

      Solving the Pythagorean identity for sin, plugging in t = 2p/5, substituting the value for cos(2p/5), and simplifying gives:

      .

   2. Use the double angle formula for sin to solve for sin(2p/5), starting from cos(p/5) = and sin(p/5) = .  Compare your answer to that from part a).

      Solution

      This gives .  Plugging this and , from part a), into a calculator gives the same approximate value of 0.951057, so they seem to be equal.  Squaring this expression and multiplying out, we obtain:

      .

      Since both expressions are positive, and their squares are equal, they must be equal.

   3. Use the difference identity for cos to solve for cos(p/12) = cos(p/4 - p/6).

      Solution

      cos(p/12) = cos(p/4 - p/6) = cos(p/4)cos(p/6) + sin(p/4)sin(p/6) = + (0.5) = .

   4. Use the Pythagorean identity to solve for cos(p/12), starting from the value of sin(p/12) = from the text.  Compare your answer to that from part c).

      Solution

      Solving the Pythagorean identity for cos, plugging in t = p/12, substituting the value for sin(p/12), and simplifying gives:

      .

      Plugging this and , from part c), into a calculator gives the same approximate value of 0.965926.  As before, if we square this expression and multiply out, we obtain:

      ,

      so they must be equal.

   Back to Exercises.

4. Follow the given instructions to derive even more accurate approximation formulas for sin and cos.
   1. Use the double angle formula for sin and the approximation formulas from the text for sin and cos (i.e., sin(t) » t and cos(t) » 1 - t²/2) to determine an approximation formula for sin(2t).

      Solution

      sin(2t) = 2sin(t)cos(t) » 2t(1 - t²/2) = 2t - t³.

   2. Replace t by t/2 in your approximation formula from part a) to give a more elaborate approximation formula for sin(t).  By comparing the graphs of your approximation formula, t, and sin(t):

      explain whether or not your formula is a more accurate approximation than sin(t) » t.

      Solution

      sin(2(t/2)) » 2(t/2) - (t/2)³ = t - t³/8, so that sin(t) » t - t³/8.  Graphing sin(t), f(t) = t, and g(t) = t - t³/8, as shown in the Example in the following applet:

      we can see that, while both  f and  g are good approximations for small values of t, g tends to stay closer to the graph of sin(t) for even larger values of t.

   3. Use the double angle formula for cos and the approximation formula from the text for cos (i.e., cos(t) » 1 - t²/2) to determine another approximation formula for cos(2t).

      Solution

      cos(2t) = 2cos²(t) - 1 » 2(1 - t²/2)² - 1 = 2( 1 - 2t²/2 + t⁴/4) - 1 = 1 - 2t² + t⁴/2.
      

   4. Replace t by /2 in your approximation formula from part c) to give a more elaborate approximation formula for cos(t).  By comparing the graphs of your approximation formula, 1 - t²/2, and cos(t):

      explain whether or not your formula is a more accurate approximation than cos(t) » 1 - t²/2.

      Solution

      cos(2(t/2)) » 1 - 2(t/2)² + (t/2)⁴/2 = 1 - t²/2 + t⁴/32, so that cos(t) » 1 - t²/2 + t⁴/32.  Graphing cos(t), f(t) = 1 - t²/2, and g(t) = 1 - t²/2 + t⁴/32, as shown in the Example in the following applet::

      As before, we can see that, although both  f and  g are good approximations for small values of t, g tends to stay closer to the graph of cos(t) for even larger values of t.
      

   Back to Exercises

More Advanced Identities

1. Use the indicated identity to solve for the indicated value:
   1. Use the half angle formula for cos to solve for cos(p/12), starting from cos(p/6) = .  Compare your answer with the value from the previous Exercise. 

      Solution

      Using the half angle formula for cos with t = p/6:

      ,

      plugging in cos(p/6) = , and simplifying gives .  This is the same expression which we obtained in part d) of the previous Exercise.

   2. Use the half angle formula for tan solve for tan(p/12), from cos(p/6) = and sin(p/6) = 0.5.  Compare your answer with the value of tan(p/12) = sin(p/12)/cos(p/12), using the value of sin(p/12) = from the text and cos(p/12) from the previous Exercise.

      Solution

      Using the half angle formula for tan with t = p/6:

      ,  

      plugging in cos(p/6) = and sin(p/6) = 0.5, and simplifying gives .  On the other hand, tan(p/12) = sin(p/12)/cos(p/12).  Plugging in the values, sin(p/12) = and cos(p/12) = gives:

      tan(p/12) = ,

      which is the same.

   3. Use the half angle formula for cos to solve for cos(p/5), starting from cos(2p/5) = .  Compare your answer with the value given in the table from the text. 

      Solution

      Since we can see from the unit circle that cos(p/5) > 0, we can use the half angle formula for cos with t = 2p/5 to obtain:

       

      Plugging in cos(2p/5) = and simplifying gives .  Plugging this expression or the expression from the text into a calculator gives the same approximate value of 0.809017.  Squaring and multiplying out gives:

      ,

      so the two different expressions must be equal.

   4. Use the half angle formula for tan to solve for tan(p/5), from cos(2p/5) = and sin(2p/5) = from a previous Exercise.  Compare your answer with the value of tan(p/5) = sin(p/5)/cos(p/5) = .

      Solution

      Using the half angle formula for tan with t = 2p/5:

      ,  

      plugging in cos(2p/5) = and sin(2p/5) = , and simplifying gives .  Plugging this expression or the expression into a calculator gives the same approximate value of 0.726543.  Since:

      ,

      and:

      ,

      we have:

      or .

   Back to Exercises.

2. Follow the given instructions to derive more trigonometric identities for tan, cot, sec, and csc.
   1. Divide both sides of the Pythagorean identity by cos² and simplify to show that sec²(t) = 1 + tan²(t).

      Solution

      Starting with 1 = cos²(t) + sin²(t) and dividing both sides by cos²(t) gives:

      Simplifying each side and applying the definitions, sec(t) = 1/cos(t) and tan(t) = sin(t)/cos(t), yields:

      so that sec²(t) = 1 + tan²(t).

   2. Starting with the equation tan(t - s) = sin(t - s)/cos(t - s), use the difference formulas for sin and cos to give a difference formula for tan in terms of sin and cos.

      Solution

      Using the difference formulas for sin and cos in the numerator and denominator gives:

   3. Divide the numerator and denominator of your formula from part b) by cos(t)cos(s), and simplify to give a difference formula for tan(t - s) in terms of tan(t) and tan(s).

      Solution

      Dividing the numerator and denominator of by cos(t)cos(s), and simplifying gives:

      .

      Using the definition, tan(t) = sin(t)/cos(t), we can express this as:

      .

   4. Divide the numerator and denominator of your formula from part b) by sin(t)sin(s), and simplify to give a difference formula for tan(t - s) in terms of cot(t) and cot(s).

      Solution

      Dividing the numerator and denominator of by sin(t)sin(s), and simplifying gives:

      .

      Using the definition, cot(t) = cos(t)/sin(t), we can express this as:

      .

   5. Use your formula from part d) to verify the cofunction identity tan(p/2 - t) = cot(t).  

      Solution

      Plugging in p/2 for t and t for s in and simplifying gives: