How do I prove that sin(a) + cos(a) ≤ √2 for all angles (a)?

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Answer 1 · 2018-04-06T14:20:53

Please see below.

We know that for any angle #A#, the maximum value of #sinA# is #1# i.e. #sinA<1#.

Now #sina+cosa#

= #sqrt2(sina*1/sqrt2+cosa*1/sqrt2)#

= #sqrt2(sina*cos(pi/4)+cosa*sin(pi/4))#

= #sqrt2*sin(a+pi/4)#

Now maximum value of sine of any angle is #1#, hence maximum value of #sin(a+pi/4)# too is #1#

therefore maximum value of #sqrt2*sin(a+pi/4)# is#sqrt2#

i.e. maximum value of #sina+cosa# is #sqrt2# or #sina+cosa<=sqrt2# for all angles #a#.

Answer 2 · 2018-04-06T14:23:17

Shown below

This can be approached with the " r - alpha method"

# color(red)(1 ) cos(a) + color(blue)(1)sin(a) = r sin ( a+ beta ) #

#=> = r = sqrt(color(red)(1)^2 +color(blue)(1)^2 ) = sqrt(2) #

#=> cos(a) + sin(a) = sqrt(2) sin(a)cos(beta) + sqrt(2) sin(beta)cos(a) #

Equating...

#1 = sqrt(2) sin(beta ) #

#1 = sqrt(2) cos(beta) #

#=> tan(beta ) =1 => beta = pi/4 #

#=> cos(a) + sin(a) = sqrt(2) sin(a+pi/4) #

We know #sin(theta) <= 1 #

#=> sin(a + pi/4) <= 1 #

#=> sqrt(2) sin(a+pi/4) <= sqrt(2) #

#therefore cos(a) + sin(a) <= sqrt(2) #

Answer 3 · 2018-04-06T15:59:39

We have:

#(sina + cosa)^2 ≤ 2#

#sin^2a + 2sinacosa + cos^2a ≤ 2#

#1 + 2sinacosa ≤ 2#

#2sinacosa ≤ 1#

#sin(2a) ≤ 1#

The function #sin(2x)# has a maximum at #y = 1#. So by definition #sin(2a) ≤ 1#.

Hopefully this helps!