Respuesta :
[tex]\bf \begin{array}{llll} [sin(\theta )-cos(\theta )] &-& [sin(\theta )+cos(\theta )]^2\\\\ &&sin^2(\theta )+2sin(\theta )cos(\theta )+cos^2(\theta )\\\\ &&sin^2(\theta )+cos^2(\theta )+2sin(\theta )cos(\theta )\\\\ &&1+2cos(\theta )sin(\theta )\\\\ &&1+sin(2\theta ) \end{array} \\\\\\ sin(\theta )-cos(\theta )-[1+sin(2\theta )]\implies sin(\theta )-cos(\theta )-1-sin(2\theta )[/tex]
I don't see it simplifying further.
The value of
[tex]$(\sin (\theta)-\cos (\theta))-(\sin (\theta)+\cos (\theta))^{2}$[/tex] [tex]${data-answer}amp;=\sin (\theta)-\cos (\theta)-\sin (2 \theta)-1[/tex].
What are Trigonometric identities?
Trigonometric identities are equalities that involve trigonometric functions and are true for every value of the occurring variables for which both sides of the equality are defined.
Let the equation be
[tex]$(\sin (\theta)-\cos (\theta))-(\sin (\theta)+\cos (\theta))^{2}$[/tex]
[tex]${data-answer}amp;(\sin (\theta)-\cos (\theta))=\sin (\theta)-\cos (\theta) \\[/tex]
[tex]${data-answer}amp;=\sin (\theta)-\cos (\theta)-(\sin (\theta)+\cos (\theta))^{2}[/tex]
Simplifying the above equation as
[tex]$(\sin (\theta)+\cos (\theta))^{2}: \quad \sin (2 \theta)+1$[/tex]
[tex]$=\sin (\theta)-\cos (\theta)-(\sin (2 \theta)+1)$[/tex]
By using distributive law:
- (a + b) = - a - b
[tex]${data-answer}amp;-(\sin (2 \theta)+1)=-\sin (2 \theta)-1 \\[/tex]
[tex]${data-answer}amp;=\sin (\theta)-\cos (\theta)-\sin (2 \theta)-1[/tex]
To learn more about trigonometric identities, refer:
https://brainly.com/question/3785172
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