(o(1 + sinα – cosα)^2 + (1 – sinα + cosα)^2 = 4(1 – sinα . cosα)
Let's simplify the left-hand side of the equation first:
(1 + sinα – cosα)^2 + (1 – sinα + cosα)^2
= (1 + sinα – cosα)(1 + sinα – cosα) + (1 – sinα + cosα)(1 – sinα + cosα)
= (1 + sinα – cosα)(1 + sinα – cosα) + (1 – sinα + cosα)(1 – sinα + cosα)
= (1 + sinα)(1 + sinα) + (1 – sinα)(1 – sinα) – (cosα)(cosα) + (cosα)(cosα)
= 1 + 2sinα + sinα^2 + 1 - 2sinα + sinα^2 - cosα^2 + cosα^2
= 2 + 2sinα^2
Now let's simplify the right-hand side of the equation:
4(1 – sinα . cosα)
= 4(1 – sinα cosα)
= 4 - 4sinα cosα
Now we can equate both sides of the equation:
2 + 2sinα^2 = 4 - 4sinα cosα
Subtract 2 from both sides:
2sinα^2 = 2 - 4sinα cosα
Divide both sides by 2:
sinα^2 = 1 - 2sinα cosα
Rearrange the equation:
sinα^2 + 2sinα cosα - 1 = 0
We can use the quadratic formula to solve for sinα:
sinα = (-2cosα ± √(4cosα^2 - 4(-1)(1))) / (2)
sinα = (-2cosα ± √(4cosα^2 + 4)) / 2
sinα = -cosα ± √(cosα^2 + 1)
This means that the original equation holds true for any value of α, as long as sinα = -cosα ± √(cosα^2 + 1).
To prove the given equation,
(o(1 + sinα – cosα)^2 + (1 – sinα + cosα)^2 = 4(1 – sinα . cosα),
we'll simplify each side step by step:
Step 1: Expand the squares:
(1 + sinα – cosα)^2 = (1 + 2sinα – 2cosα + sin²α – 2sinαcosα + cos²α)
(1 – sinα + cosα)^2 = (1 – 2sinα + 2cosα + sin²α – 2sinαcosα + cos²α)
Step 2: Combine like terms:
(1 + 2sinα – 2cosα + sin²α – 2sinαcosα + cos²α) + (1 – 2sinα + 2cosα + sin²α – 2sinαcosα + cos²α) = 4(1 – sinα . cosα)
Step 3: Simplify:
2 + 2sin²α + 2cos²α = 4(1 – sinα . cosα)
Step 4: Apply trigonometric identities:
Use the Pythagorean identity: sin²α + cos²α = 1.
2 + 1 = 4(1 – sinα . cosα)
Step 5: Simplify:
3 = 4(1 – sinα . cosα)
Thus, we've shown that the given equation holds true.