What is the period of #f(theta) = tan ( ( 15 theta)/7 )- sec ( ( 5 theta)/ 6 ) #?

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Answer 1 · 2016-07-10T18:03:14

Period #P=(84pi)/5=52.77875658#

The given #f(theta)=tan((15theta)/7)-sec ((5theta)/6)#

For #tan ((15theta)/7)#, period #P_t=pi/(15/7)=(7pi)/15#

For #sec ((5theta)/6)#, period #P_s=(2pi)/(5/6)=(12pi)/5#

To get the period of #f(theta)=tan((15theta)/7)-sec ((5theta)/6)#,
We need to obtain the LCM of the #P_t# and #P_s#

The solution

Let #P# be the required period
Let #k# be an integer such that #P=k*P_t#
Let #m# be an integer such that #P=m*P_s#

#P=P#
#k*P_t=m*P_s#
#k*(7pi)/15=m*(12pi)/5#

Solving for #k/m#

#k/m=(15(12)pi)/(5(7)pi)#

#k/m=36/7#

We use #k=36# and #m=7#
so that
#P=k*P_t=36*(7pi)/15=(84pi)/5#

also

#P=m*P_s=7*(12pi)/5=(84pi)/5#

Period #P=(84pi)/5=52.77875658#

Kindly see the graph and observe two points to verify for the period

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