If #sin(picosa)=cos(pisina)# then #a=#?

1 Answer
Feb 22, 2018

#a=2npi+-pi/4+-cos^(-1)(1/(2sqrt2))# where #m# is an integer.

Explanation:

We use the identity #sinx=cos(pi/2-x)=cos(x-pi/2)#

Hence #sin(picosa)=cos(pisina)# can be written as

#cos(picosa-pi/2)=cos(pisina)#

Hence #picosa-pi/2=2npi+-pisina#, where #n# is an integer.

or #cosa=2n+1/2+-sina=(4n+1)/2+-sina#

This leads to two situations

(1) #cosa=(4n+1)/2+sina#

or #cosa-sina=(4n+1)/2#

or #sqrt2cos(a+pi/4)=(4n+1)/2#

or #cos(a+pi/4)=(4n+1)/(2sqrt2)#

Now as #|cosa|<=1#, we can have #n=0# only and then

#cos(a+pi/4)=1/(2sqrt2)#

and #a+pi/4=2mpi+-cos^(-1)(1/(2sqrt2))#

and #a=2mpi-pi/4+-cos^(-1)(1/(2sqrt2))#

(2) #cosa=(4n+1)/2-sina#

or #cosa+sina=(4n+1)/2#

or #sqrt2cos(a-pi/4)=(4n+1)/2#

or #cos(a-pi/4)=1/(2sqrt2)# - because of similar restrictions as mentioned above.

and #a-pi/4=2mpi+-cos^(-1)(1/(2sqrt2))#

or #a=2mpi+pi/4+-cos^(-1)(1/(2sqrt2))#

Here #m# is an integer.