IIT JEE , AIEEE Forum - Mathematics , Trignometry

nivedh_89

maximum value =

2......at

=

/4..............
minimum value = -1 .......at

= -

/2................
CHEERS!!!!!!!!!!!!!!!!!!!

-

2 is not possible..........................!!!!!!!!!!!!!!!!!!!!!!!!!

neeraj_agarwal_1990

solution given by flower_ssss is right.....

prathyu

maximum value of sinx & cos x is 1

minimum value of sinx & cosx is -1

tushar.vaidya

max=infinity
min=-1

tushar.vaidya

max=infinity
min=-2

spideyunlimited

sin

+ cos

=

2 ( (1/

2) sin

+ (1/

2) cos

)
=

2 ( sin (

/4 +

) )

now as sin varies from -1 to +1
thus, value of above function varies from -

2 to +

2

*PLZ RATE*

kmmankad

For any form of an eqn, aSin#+bCos#+c,the min is -{(a^2+b^2)}^1/2 +c
and max is {(a^2+b^2)}^1/2 + c....Solving this like so,we get min as -2^1/2 and max as 2^1/2

nigitha_17

reply by kmmankad is absolutly perfect............

somyekathait

max value =

2

min value= -

2

thedumbheadwithnobrain

Find it using the method of method of differentiation

Differentiate and equate it to 0 and find critical points from 0 to 2(pi)

the answer will comeout to be

root2 at pi/4

-root2 at 3(pi)/4

udbarguj

cos

+sin

2 (1/

2cos

+ 1/

2sin

)

2 [sin(pi/4+

)

Max. value

2

Min value -

2

deepamkanjani

max. value = under root 2......at theta= pi/4
min. value = -under root 2 .......at theta= = 3pi/2

nadeemoidu

The answer is -

2,

2
Any function of the asinx+bcosx lies between -

(a²+b²) and

(a²+b²).

It can be derived as follows

asinx+bcosx =

(a²+b²) ( a/

(a²+b²)sinx +b/

(a²+b²)cosx)

Put a/

(a²+b²)= sin y.

=

(a²+b²)( siny sinx+ cosy cosx)
=

(a²+b²)cos(x+y)
cos(x+y) is between -1 and 1.
so asinx+bcosx is between -

(a²+b²) and

(a²+b²).

prathyu

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