Find the exact value of the following under the given conditions a.cos(a-b) b.(a-b) c.tan(a-b)

Question

sin a=3/5, 0<β<pi/2

cos b=-1/6, pi<β<3pi/2

Answer 1

sin a = (3/5)  0 < a < pi/2 (QUADRANT I)

cos b = (-1/6)  pi < b < 3pi/2 (QUADRANT III)

 (1) First draw triangles for each function, and use Pythagorean Theorem to calculate the missing sides.

Triangle with angle a:               Triangle with angle b:

sin a = y/r = 3/5                        cos b = x/r = -1/6

x^2 + 3^2 = 5^2                        (-1)^2 + y^2 = 6^2

x^2 + 9 = 25                               1 + y^2 = 36

x^2 = 16                                        y^2 = 35

x = 4                                              y = sqrt (35)        

(This is a standard

3-4-5 right triangle.)

 (2) Solve the first problem using the Difference Formula for Cosine.

    Using the given information and the x and y values from above:

    sin a = (3/5)       (Quadrant I)

    cos a = (4/5)  

    sin b = - sqrt (35) /6   (Quadrant III)

    cos b = (-1/6)  

    cos (a – b) = cos a cos b + sin a sin b

                      = (4/5)(-1/6) + (3/5)( - sqrt (35) /6)

                      = (-4/30) + ( -3 sqrt (35) /30)

                      = (-4 – 3 sqrt (35) ) / 30

(3) Solve the second problem using inverse trig functions, i.e., arcsin and arccos.

      We need to find the values of angles a and b, and remember to consider

      the quadrant in which each angle is positioned.

     a = arcsin (3/5)

     a = 0.64  (use radians, since domain was given in radians) 

       The easy way to get the measure of angle b, which resides in Quadrant III, is to find the

        measure of the POSITIVE angle (in Quadrant I) and then add pi to get the angle

        measure in Quadrant III.  To find the value of b, 

             b = arcos (1/6)

             b = 1.4

        Now add pi,  b = 1.4 + 3.14

                               b = 4.54

 

(4) Solve the third problem using Difference Formula for Tangent.

            tan (a – b) = (tan a - tan b) / (1 + tan a tan b)

           tan a = sin a / cos a = (3/5) / (4/5) = 3/4

           tan b = sin b / cos b = (- sqrt (35) /6 ) / (-1/6)  = sqrt (35)   

            tan (a – b) =     (3/4) - sqrt (35)  

                                   --------------------------

                                   1 + (3/4)( sqrt (35) )

                               = 4       (3/4) - sqrt (35)                    (Multiply fraction by 4/4

                                  -- x  --------------------------              to simplify)

                                  4      1 + (3/4)( sqrt (35) )

           tan (a – b) =   3 – 4 sqrt (35)   

                                 --------------------------

                                  4 + 3 sqrt (35)       

I hope this helps you understand this problem – and solve others that are similar!