* How To Solve an Equation using Trig Identities *

~ Reciprocal Identities ~ 

(sinx)(cscx)=1                           (cosx)(secx)=1                            (tanx)(cotx)=1  

sinx =    1          cscx =   1   

           cscx                   sinx

 

cosx =   1          secx =   1   

           secx                   cosx

 

tanx =   1          cotx =    1  

           cotx                   tanx                         

 

~ Pythagorean Identities ~

1.)  r2 = x2 + y2

      r2     r2     r2 

   1= cos2x + sin2x

 

2.)  r2 = x2 + y2

      x2    x2    x2

    sec2x = 1 + tan2x

 

3.) r2 = x2 + y2

      y2    y2    y2

    csc2x = cot2x +1

 

~ Quotient Identities ~

1.) sinx  = y/r = (y)(r) = (y)  = tanx

     cosx        x/r  (r)(x)    (x)

    sinx  = tanx

     cosx

 

2.) cosx = x/r = (x)(r) + (x) = cotx

    sinx       y/r   (r)(y)    (y)

    cosx = cotx

    sinx

     

 

Problem A:                                

csc x tan x = secx

  1    x  sinx

sinx     cosx 

  1 

cosx

=secx

   

 

Problem B:  

(secx-1)(secx+1)(cos2x)=sin2x

(sec2x-1)(cos2x)

(tan2x)(cos2x)

sin2x x (cos2x)

cos2x

=sin2x

 

       As you see above in both problem A and problem B there are two equations given to you and following them are the answers belonging to each of the questions asked.  The answers for each problem were derived from the eight basic trig identities, all eight identities are shown in the blocks above, also shown in the block above is a diagram that showing how both the Pythagorean and the Quotient identities are formed.   

      When solving each problem  keep in you mind all identities and try to look for and substitution that you can make to replace and identity with what was given to you to make the problem easier for you to solve.     

~As you can see in problem A my first substitution was removing cscx and putting in 1 over sinx, then talking tanx and putting in sinx over cosx.  Then you will see that both sinx cancel out and leaving you with 1 over cosx which is one of you reciprocal identities that gives you secx and that is the answer you are looking for. 

~To solve problem B I multiplied  (secx-1) and (secx+1) together to get sec2x and if you notice it can be replaced with tanx, then again if you take out the tanx and put in sinx over cosx the bottom cosx and the other cosx will cancel out leaving you with sin2x which is the answer you are wanting to end up with.