If sin∅+cos∅ = 1 , find sin∅.cos∅.

Explanation:

The original equation is in the form a+b = 1, where

a = sin(theta)

b = cos(theta)

Square both sides of a+b = 1 to get

(a+b)^2 = 1^2

a^2+2ab+b^2 = 1

(a^2+b^2)+2ab = 1

From here notice that a^2+b^2 is sin^2+cos^2 = 1, which is the pythagorean trig identity. So we go from (a^2+b^2)+2ab = 1 to 1+2ab = 1 to 2ab = 0 to ab = 0

Therefore,

sin(theta)*cos(theta) = 0

sin ∅ cos ∅ = 0.

Step-by-step explanation:

(sin∅+cos∅)^2 = 1^2 = 1

(sin∅+cos∅)^2  = sin^2∅ + cos^2∅ + 2sin ∅ cos ∅ = 1

But  sin^2∅ + cos^2∅ = 1, so:  

2sin ∅ cos ∅ + 1 = 1

2 sin ∅ cos ∅ = 1 - 1 = 0

sin ∅ cos ∅ = 0.

hi! i think the answer is 1/2 : )