If #sinA - cosA = 1#, then what is the value of #cosA + sinA#?

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Answer 1 · 2018-02-09T16:43:41

Squaring both sides:

#(sinA- cosA)^2 = 1^2#

#sin^2A - 2sinAcosA + cos^2A = 1#

#1 - 2sinAcosA = 1#

#0 = 2sinAcosA#

#0 = sin(2A)#

Now if we repeat the same thing with #sinA + cosA = B#, we see that

#(sinA + cosA)^2 = B^2#

#sin^2A + cos^2A + 2sinAcosA= B^2#

#sin(2A) = B^2 - 1#

From the above step, recall that #sin(2a)# must equal #0#. Therefore,

#0 = B^2 - 1#

#0 = (B + 1)(B - 1)#

#B = 1 or -1#

Therefore, #cosA + sinA =+-1#, as required.

Hopefully this helps!

Answer 2 · 2018-02-09T17:47:37

Using formula
#(a+b)^2+(a-b)^2=2(a^2+b^2)# we get

#(sinA+cosA)^2+(sinA-cosA)^2=2(sin^2A+cos^2A)#

#=>(sinA+cosA)^2+1^2=2*1=2#

#=>sinA+cosA=pmsqrt(2-1)=pm1#