How do you solve #sinx+cosx = sqrt2#?

Question

14281 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-16T11:11:04

x=#45^@#

#cosx=b/c##" "color(blue)((1))#

#sinx=a/c##" "color(blue)((2))#

Given that,

#sinx+cosx=sqrt2##" "color(blue)((3))#

Substitute #" "color(blue)((1))# and #color(blue)((2))# in #" "color(blue)((3))#

#a/c+b/c=sqrt2##=>(a+b)/c=sqrt2#

#a+b=csqrt2#
squaring on both sides

#(a+b)^2=(csqrt(2))^2##=>(a+b)^2=2c^2#

#a^2+b^2+2ab=2c^2#

#c^2+2ab=2c^2#[since from Pythagoras]

#c^2=2ab#

#a^2+b^2=2ab#[since from Pythagoras]

#a^2+b^2-2ab=0#

#(a-b)^2=0#

#a-b=0#

#a=b##" "color(blue)((4))#

equation#" "color(blue)((4))# substitute in#" "color(blue)((1))#

#cos x=a/c##" "color(blue)((5))#

From#" "color(blue)((5))#&#" "color(blue)((2))#

#cosx=sinx#

#sin(pi/2-x)=sinx#

#cancel(sin)(pi/2-x)=cancel(sin)x#

#pi/2-x=x##=>2x=pi/2#

#x=pi/4#radians

or

#x=45^@#