Trigonomentry Given that sinx = cos(x+20o) write tanx in terms of sin20o and cos 20o

Question

a) Given that sinx = cos(x+20o) write  tanx in terms of sin20o and cos 20o.        

Hence find the values of tan x.                    

 

    b) Find  the general solution of  3sec²y=5(1+tany)           

Answer 1

 

(A)  sin x = cos (x + 20)

 

      NOTE:  cos (x + 20) = (cos x)(cos 20) - (sin x)(sin 20) ... according to an identity

 

      sin x = cos (x + 20) = (cos x)(cos 20) - (sin x)(sin 20)

 

      sin x + (sin x)(sin 20) = (cos x)(cos 20)

 

      (sin x)(1 + sin 20) = (cos x)(cos 20)

 

      (sin x)(1 + sin 20)/(cos x) = (cos x)(cos 20)/(cos x)

 

      (sin x)(1 + sin 20)/(cos x) = (cos 20)

 

      (sin x)(1 + sin 20)/(cos x)(1 + sin 20) = (cos 20)/(1 + sin 20)

 

      (sin x)/(cos x) = (cos 20)/(1 + sin 20)

 

      tan x = (cos 20)/(1 + sin 20)      

 

 

(B) 3 sec² y = 5 (1 + tan y )

 

      3 [ 1 + tan² y ] = 5 [ 1 + tan y ]

 

      3 + 3•tan² y = 5 + 5•tan y

 

      3•tan² y - 5•tan y + 3 - 5 = 0

 

      3•tan² y - 5•tan y - 2 = 0

 

      NOTE:  substitute ... x = tan y

 

      3x² - 5x - 2 = 0

 

      (3x + 1)(x - 2) = 0

 

       Using the zero product property,

 

       3x + 1 = 0

       3x = -1

       x = -1/3

 

       also

       x - 2 = 0

       x = 2

 

       Substitute x = -1/3 and x = 2 back in for the earlier substitution, and use the calculator inverse trig functions to find the angle measures

 

       tan y = -1/3

       tan^-1(-1/3) = -18.43 degrees, or 360 - 18.43 = 341.57 degrees

 

       also

       tan y = 2

       tan^-1(2) = 63.43 degrees

 

The general form would take those degree measures and add 360n to them, where n is an integer