# The value of |sin thita + cos thita| is:-a) <=2.b) >=root(2).c) <=root(2).d) none of these.

290 Points
12 years ago

Hi Arnab,

sinx + cosx = √2(sinx/√2 + cosx/√2) = √2 [sin(x+pi/4]

Hence |sinx+cosx| = √2|sin(x+pi/4)|, which is ≤√2.

Option (C).

Hope that helps.

Best Regards,

Swapnil Saxena
102 Points
12 years ago

The equation can take two forms. For sin(thita)+cos(thita)>=0,it is sin(thita)+cos(thita)----(1) , But for sin(thita)+cos(thita)<0 ,it is -(sin(thita)+cos(thita))------(2)

Differentiating the equation with respect to (thita) for finding the the maxima,

(d/d(thita))(sin(thita)+cos(thita))=cos(thita)-sin(thita)

Putting the above equation equal to 0,cos(thita)-sin(thita)=0  ==> cos(thita)=sin(thita) which is only possible for (pi)/4 or 5(pi)/4 .When thita is pi/4 or 5(pie)/4, the value of the equation is (1/root(2))+(1/root(2))=root(2).(Maxima of the equation).The minima will definitely be 0 achieved when thita=3pi/4 or 7pi/4. So option (c) seems to be correct.

Note: The Mod seems to change the minima of the eqaion frm -root(2) to 0, but not the maxima