How do you solve sinx+cosx = sqrt2?

1 Answer
Jun 16, 2015

x=45^@

Explanation:

![https://community-plus.s3.amazonaws.com/AE/35/551.jpg](useruploads.socratic.org)

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/4radians

or

x=45^@