Question

The period of 2sin2x-5cos2x/7cosx-8sinx
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Answer 1

Let f(x)=2sin2x−5cosx7cosx−8sinxf(x)=2sin⁡2x−5cos⁡x7cos⁡x−8sin⁡x

Property of periodic function of period T.

f(x+T)=f(x)f(x+T)=f(x)

⇒sinx,cosx⇒sin⁡x,cos⁡x are periodic functions of period 2π2π

by putting T=2πT=2π

We conclude that,

2sin2(x+2π)−5cos(x+2π)7cos(x+2π)−8sin(x+2π)=2sin2x−5cosx7cosx−8sinx2sin⁡2(x+2π)−5cos⁡(x+2π)7cos⁡(x+2π)−8sin⁡(x+2π)=2sin⁡2x−5cos⁡x7cos⁡x−8sin⁡x

2sin2x−5cosx7cos(x)−8sinx=f(x)2sin⁡2x−5cos⁡x7cos⁡(x)−8sin⁡x=f(x)

So, that period of f(x)is2π

Answer 2

$\displaystyle\sf{We\:have\:to\:find\:the\:Period\:of\:\:\:\frac{2sin2x-5cos2x}{7cosx-8sinx}$

• $\displaystyle\sf{2sin2x=\frac{2\pi}{2}=\frac{\pi}{1}$

• $\displaystyle\sf{5cos2x=\frac{2\pi}{2}=\frac{\pi}{1}$

• $\displaystyle\sf{7cosx=\frac{2\pi}{1}$

• $\displaystyle\sf{8sinx=\frac{2\pi}{1}$

$\displaystyle\sf{Period=\frac{LCM\:of\:(\pi,\pi,2\pi,2\pi)}{HCF\:of\:(1,1,1,1)}$

• $\displaystyle\sf{LCM\:of\:(\pi,\pi,2\pi,2\pi)=2\pi$

• $\displaystyle\sf{HCF\:of\:(1,1,1,1)=1}$

$\displaystyle\sf{Period=\frac{2\pi}{1}=2\pi$

$\boxed{\bold{\displaystyle\sf{Period\:of\:\:\:\frac{2sin2x-5cos2x}{7cosx-8sinx}=2\pi}}}$