if sinx+cosx=2^1/2,find the maximum possible value of x.?

Question

could you please give me a systematic procedure instead of a trial and error method.

Answer 1

sinx + cosx= sqrt(2)
(sinx+cosx)^2 = 2
sin^2(x) + 2sinxcosx + cos^2(x) = 2

Remember that sin^2(x) + cos^2(x) = 1

Therefore 2sinxcosx = 1
sin(2x) = 1
2x = (2n+1).pi/2
x = (2n+1).pi/4
x = pi/4, 3pi/4. 5pi/4 ....

Answer 2

ok by basics sin (45 degrees) = 1 / (2^1/2 )
and cos(450 = 1/(2^1/2)

so the thing is u can get max values only when sinx = cosx (because they are sort of complimentary i.e is if sin increases cos decreases and vice versa ) , that is when x = 45 degrees.

Subsitute it and u will see that sinx + cosx = 1/(2^1/2) + 1/(2^1/2) = 2^1/2 on simplifying

Answer 3

sinx + cosx = sqrt(2)
squaring both sides we get
sin^2 x +cos^2 x +2 sinx cosx = 2
=> 1 + sin2x = 2 (as sin^2 x + cos^2 x = 1)
=> sin2x = 1
=> 2x = (2n +1)pi/2 where n is any integer
=> x = (2n +1)pi/4
as n is any integer we can not find the maximum or minimum
possible values in this case

Answer 4

Yes you should work with tan(x/2)which we call z
2z/(1+z^2)+(1-z^2)/(1+z^2)=sqrt(2)

(-z^2+2z+1)=2^1/2 +2^1/2*z^2
(1+2^1/2)z^2-2z+2^1/2-1 =0
z=((2+-(sqrt(4-4(1)))/2(1+2^1/2)=1/(1+2^1/2) =2^1/2-1

tan x/2= 2^1/2 -1
x/2 = 22.5 +kpi degrees so x= 45 deg +2kpi
No trial and error