In order to solve this problem, we first square both sides of the given equation in order to find the value of 2sin(θ)cos(θ). Using this we will be able to find the required answer.
Suppose that we are given that,
sin(θ)+cos(θ) = 1
Then on squaring both sides we get that,
(sin(θ)+cos(θ))2 = 1
⇒ sin(θ)2+cos(θ)2 + 2sin(θ)cos(θ)= 1
We know that, sin(θ)2+cos(θ)2 = 1. Substituting this identity in the above equation we get that,
⇒ 1 + 2sin(θ)cos(θ)= 1
⇒ 2sin(θ)cos(θ)= 0 , that is, sin(θ)cos(θ)= 0 (Equation A)
What is the value of sin(θ) – cos(θ)?
We consider the square of sin(θ) – cos(θ) as follows,
(sin(θ) – cos(θ))2 = sin(θ)2+cos(θ)2 – 2sin(θ)cos(θ)
We now use the formula, sin(θ)2+cos(θ)2 = 1, and also the fact in bold (Equation A) above that, sin(θ)cos(θ)= 0. Substituting all this in the above equation we get,
(sin(θ) – cos(θ))2 = 1 – 2 x 0 = 1- 0 = 1
Taking square roots on both sides we get that,
sin(θ) – cos(θ) = 1
What is the value of sin(2θ)?
Here we use the following duplication formula,
sin(2θ) = 2sin(θ)cos(θ)
Now, from (Equation A) we know that, sin(θ)cos(θ)= 0. Substituting this in the above duplication formula we get that,
sin(2θ) = 0
Conclusion:
If sin(θ)+cos(θ) = 1, then
sin(θ) – cos(θ) = 1 and,
sin(2θ) = 0
