IIT JEE , AIEEE Forum - Mathematics , Trignometry - Trigonometry IIT-JEE question

sahilgupta_iit

Prove that cos(sin

) >₌ sin(cos

).

JEE - 1981

ankitagg

we know
sin

+cos

<=root2<pie/2
therefore sin

+cos

<pie/2
pie/2-sin

>cos

sin(pie/2-sin

)>sin

cos

cos(sin

)>sin

(cos

)

sahilgupta_iit

please answer it guys

Milo_Breakin_free

Use ur knowledge of functions.....i.e.
here now sinx & cos x have values wihin the range [-1,1]...........
Now if we look at it.....cos 0=1 so,cos x is actually a decreasing funcion in the domain [0,pi]
While sin x is an increasing function since its value starts from sin 0=0..........
So,it is clear that for as x tends to zero cos x tends to be greater than sin x........
clearly from x belonging to [0,1],cos x should have greater values than sin x
so,cos(sin x)>=sin(cos x)

ankitagg

did u find something wrong in my solution.please tell

sahilgupta_iit

but how would you prove this in a subjective pattern?

sahilgupta_iit

ya, you are using both the inverse relation and the general one.... in the first line you have written pi as well as root 2

nknikhilesh1

sin X + cos X=root2 sin(pi/4+theta)

sin X + cos

root 2 <pi/2 as root 2=1.414 and pi/2=1.57

sinX + cos X<pi/2

cos X<pi/2-sin X.........................(1)

takin sin on both sides of equation 1 we get the required result

Please rate me if it helped u,

sahilgupta_iit

i have rated you but in the question it greater than or EQUAL to.

ashish_banga

sin

<

sin(sin

) < sin

sin(cos

) < cos

as cos

is decreasing in 0, pi/2 cos(sin

) >cos

> sin(cos

)

Ask & Discuss Questions with Community & Experts