Question

what is the period of cos(3x/5) - sin(2x/7)...frnds pls answer tis question...

Answer 1

Basically, period of any function $h(x)=f_1(x)\pm f_2(x)$ is the LCM of period of f₁(x) and f₂(x)

Here, h(x) = cos(3x/5) - sin(2x/7)
f₁(x) = cos(3x/5) , period of f₁(x) is 2π/(3/5) = 10π/3
f₂(x) = sin(2x/7), period of f₂(x) is 2π/(2/7) = 7π/1

Now, period of h(x) = LCM { 10π/3, 7π/1}
we know, LCM of a/b and c/d is LCM of {a, c}/HCF of {b, d}
so, period of h(x) = LCM { 10π, 7π}/HCF of { 3,1} = 70π

Hence, period of cos(3x/5) - sin(2x/7) is 70π

Answer 2

Step-by-step explanation:

if period of f(X)=T

period 9f f(ax)=T/a

period of cos 3x/5=2π/(3/5)=10π/3

period of sin(2x/7)= 2π/(2/7)=7π

T=LCM of 10π/3,7π

T=HCF(10π,7π//)/LCM(1,3)

T=70 π

Hope it helps you