If #Sin(theta) - Cos(theta) = -sqrt(2) sin(theta)#, what is #Sin(theta) + Cos (theta)#?

#sintheta + sqrt(2)sin theta = costheta#

#sintheta(1 + sqrt(2)) = costheta#

#1 + sqrt(2) = costheta/sintheta#

By the quotient identity #costheta/sintheta = cot theta#

#1 + sqrt(2) = cot theta#

We must now determine the values of #sintheta# and #costheta#.

First, we must define #cottheta#: #cottheta= "adjacent"/"opposite"#.

Therefore, the side adjacent #theta# measures #1 + sqrt(2)# and the side opposite measures #1#. We will need to find the hypotenuse for #sintheta# and #costheta#.

Let y be the hypotenuse. By pythagorean theorem we have

#(1 + sqrt(2))^2 + 1^2 = y^2#

#1 + 2sqrt(2) + 2 + 1 = y^2#

#4 + 2sqrt(2) = y^2#

#sqrt(4 + 2sqrt(2)) = y#

The definition of #sintheta = "opposite"/"hypotenuse"# and the definition of #costheta = "adjacent"/"hypotenuse"#

Therefore, #sintheta = 1/(sqrt(4 + 2(sqrt(2)))# and #costheta = (1 + sqrt(2))/(sqrt(4 + 2(sqrt(2))#

Since we have been asked to evaluate #sintheta + costheta# we add our two expressions. Luckily, they are on equal denominators!

#costheta + sin theta = +-(2 + sqrt(2))/(sqrt(4 + 2(sqrt(2))))#

#= +-sqrt(1 + sqrt(2)/2#

Hopefully this helps!

General discussion to determine the sign of #=>sintheta+costheta#

From the given condition

#sintheta+sqrt2sintheta=costheta#

#=>(sqrt2+1)sintheta=costheta#

#:.sintheta/costheta=1/(sqrt2+1)....(1)#

#tantheta=1/(sqrt2+1)>0#

#tantheta# being positve #theta# should be in 1st or 3rd quadrant.when #theta# is in the 1st quadrant is positive but in 3rd quadrant #sintheta and costheta# both negative and the the value #sintheta+costheta#should be negative.

So # color(red)(sintheta+costheta" will be either +ve or -ve")#

Method -I (A tricky approach)

From equation (1) have

#sintheta/costheta=1/(sqrt2+1)....(1)#

and

Inverting and ratiolising numerator of RHS

#costheta/sintheta=sqrt2+1=1/(sqrt2-1)....(2)#

Adding equation(1) and equation (2)we get

#sintheta/costheta+costheta/sintheta=1/(sqrt2+1)+1/(sqrt2-1)#

#(sin^2theta+cos^2theta)/(sinthetacostheta)=(2sqrt2)/((sqrt2-1)(sqrt2+1))=(2sqrt2)/1#

Inverting and rearranging we get

#(2sinthetacostheta)/(sin^2theta+cos^2theta)=1/(sqrt2)#

Adding 1 on both sides we get

#1+(2sinthetacostheta)/(sin^2theta+cos^2theta)=1+1/(sqrt2)#

#(sintheta+costheta)^2/(sin^2theta+cos^2theta)=(sqrt2+1)/sqrt2#

#=>(sintheta+costheta)^2=(sqrt2+1)/sqrt2#

#=>sintheta+costheta=+-sqrt((sqrt2+1)/sqrt2)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Method - II

Given
#sintheta-costheta=-sqrt2sintheta#

Dividing both sides by# sqrt2 " "#we have

#=>1/sqrt2sintheta-1/sqrt2costheta=-sintheta#

(Considering unit of angle in degree.)

#=>cos(45)sintheta-sin(45)costheta=-sintheta...(1)#

#=>sin(theta-45)=sin(-theta)#

#=>theta-45=-theta#

#=>2theta=45#

#=>theta=22.5#

Another solution is posible satisfying positive value of #tantheta# when #theta# is in third quadrant

Then from eq (1)

#sin(theta-45)=sin(360-theta)#
#=>theta=405/2=180+22.5#

Now

#sintheta+costheta#

#=sqrt2(1/sqrt2sintheta+1/sqrt2costheta)#
#=sqrt2sin(theta+45)....(2)#

#color (blue)("when " " theta=22.5#

Inserting #theta=22.5#

#=sqrt2sin(22.5+45)#

#=sqrt2sin(90-22.5)#

#=sqrt2cos(22.5)#

#=sqrt2sqrt(1/2(1+cos45))#

#=sqrt(1+cos45)#

#=sqrt(1+1/sqrt2)#

#=sqrt((sqrt2+1)/sqrt2)#

#color (green)("Again when " " theta=180+22.5#

we put #theta=180+22.5#

in eq(2)

we get

#=>sintheta+costheta =sqrt2sin(theta+45)#

#=sqrt2sin(180+22.5+45)#

#=sqrt2sin(180+22.5+45+22.5-22.5)#

#=sqrt2sin(270-22.5)#

#=-sqrt2cos(22.5)#

#=-sqrt2(sqrt(1/2(1+cos45)))#

#=-sqrt2(sqrt(1/2(1+1/sqrt2)))#

#=-sqrt(1+1/sqrt2)#

#=-sqrt((sqrt2+1)/sqrt2)#

So combining these two we get

#sintheta+costheta=+-sqrt((sqrt2+1)/sqrt2)#