Respuesta :
- A. (sin θ - cos θ) - (sin θ + cos θ)² = sin θ - cos θ - 1 - sin 2θ
- B. (sin θ - cos θ) - (sin θ + cos θ)² = - 2sin 2θ
- C. (sin θ - cos θ) - (sin θ + cos θ)² = 2
Further explanation
We will simplify a form related to trigonometric identity.
For more practice and learning, this time we prepare three cases. Most likely one of them is included in the real question being asked.
Problem A
The Process:
[tex]\boxed{ \ (sin \theta - cos \theta) - (sin \theta + cos \theta)(sin \theta + cos \theta) \ }[/tex]
[tex]\boxed{ \ (sin \theta - cos \theta) - (sin^2 \theta + cos^2 \theta + 2sin \theta cos \theta) \ }[/tex]
Recall that [tex]\boxed{ \ sin^2\theta + cos^2\theta = 1 \ }[/tex]
[tex]\boxed{ \ (sin \theta - cos \theta) - (1 + 2sin \theta cos \theta) \ }[/tex]
Recall that [tex]\boxed{ \ 2sin \theta cos \theta = sin \ 2\theta \ }[/tex]
[tex]\boxed{ \ (sin \theta - cos \theta) - (1 + sin \ 2\theta) \ }[/tex]
Therefore, the result is [tex]\boxed{\boxed{ \ sin \theta - cos \theta - 1 - sin \ 2\theta \ }}[/tex]
Problem B
[tex]\boxed{ \ (sin \theta - cos \theta)^2 - (sin \theta + cos \theta)^2 = \ ? \ }[/tex]
The Process:
[tex]\boxed{ \ (sin \theta - cos \theta)^2 - (sin \theta + cos \theta)^2 = \ ? \ }[/tex]
Let's expand both squares.
[tex]\boxed{ \ (sin \theta - cos \theta)(sin \theta - cos \theta) - (sin \theta + cos \theta)(sin \theta + cos \theta) \ }[/tex]
[tex]\boxed{ \ sin^2 \theta + cos^2 \theta - 2sin \theta cos \theta - (sin^2 \theta + cos^2 \theta + 2sin \theta cos \theta) \ }[/tex]
[tex]\boxed{ \ sin^2 \theta + cos^2 \theta - 2sin \theta cos \theta - sin^2 \theta - cos^2 \theta - 2sin \theta cos \theta \ }[/tex]
[tex]\boxed{ \ - 2sin \theta cos \theta - 2sin \theta cos \theta \ }[/tex]
[tex]\boxed{ \ - 4sin \theta cos \theta \ }[/tex]
Recall that [tex]\boxed{ \ 2sin \theta cos \theta = sin \ 2\theta \ }[/tex]
[tex]\boxed{ \ - 2(2sin \theta cos \theta) \ }[/tex]
Therefore, the result is [tex]\boxed{ \ - 2sin \ 2\theta \ }[/tex]
Problem C
[tex]\boxed{ \ (sin \theta - cos \theta)^2 + (sin \theta + cos \theta)^2 = \ ? \ }[/tex]
The Process:
[tex]\boxed{ \ (sin \theta - cos \theta)^2 + (sin \theta + cos \theta)^2 = \ ? \ }[/tex]
Let's expand both squares.
[tex]\boxed{ \ (sin \theta - cos \theta)(sin \theta - cos \theta) + (sin \theta + cos \theta)(sin \theta + cos \theta) \ }[/tex]
[tex]\boxed{ \ sin^2 \theta + cos^2 \theta - 2sin \theta cos \theta + (sin^2 \theta + cos^2 \theta + 2sin \theta cos \theta) \ }[/tex]
[tex]\boxed{ \ sin^2 \theta + cos^2 \theta - 2sin \theta cos \theta + sin^2 \theta + cos^2 \theta + 2sin \theta cos \theta \ }[/tex]
Recall that [tex]\boxed{ \ sin^2\theta + cos^2\theta = 1 \ }[/tex]
[tex]\boxed{ \ 1 - 2sin \theta cos \theta + 1 + 2sin \theta cos \theta \ }[/tex]
Therefore, the result is [tex]\boxed{\boxed{ \ 2 \ }}[/tex]
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Keywords: simplify, (sin Θ − cos Θ) − (sin Θ + cos Θ)², (sin Θ − cos Θ)² − (sin Θ + cos Θ)², (sin Θ − cos Θ) + (sin Θ + cos Θ)², trigonometric identity,
