## anonymous one year ago Simplify: (sin Θ − cos Θ)2 − (sin Θ + cos Θ)2 a. −4sin(Θ)cos(Θ) b. 2 c. sin^2 Θ d. cos^2 Θ

1. anonymous

$(\sin(\theta)-\cos(\theta))^2-(\sin(\theta)+\cos(\theta))^2$ Apply the identity $a^2-b^2=(a+b)(a-b)$ $\implies ((\sin(\theta)-\cos(\theta))+(\sin(\theta)+\cos(\theta)))\times((\sin(\theta)-\cos(\theta))-(\sin(\theta)+\cos(\theta)))$

2. anonymous

Here, we've taken $a=\sin(\theta)-\cos(\theta)$ and $b=\sin(\theta)+\cos(\theta)$

3. anonymous

Wouldnt the answer be 0 tho? since it's cancelled out?

4. anonymous

Nope, open the parenthesis and find out for yourself

5. anonymous

i still dont understand

6. Michele_Laino

hint: you can use these identities: $\Large \begin{gathered} {\left( {\sin \theta - \cos \theta } \right)^2} = {\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2} - 2\sin \theta \cos \theta \hfill \\ \hfill \\ {\left( {\sin \theta + \cos \theta } \right)^2} = {\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2} + 2\sin \theta \cos \theta \hfill \\ \end{gathered}$

7. anonymous

Hm, it seems much more simpler to understand if you apply those identities above, do you understand them @xoamb

8. anonymous

$(a+b)^2=a^2+b^2+2ab$ $(a-b)^2=a^2+b^2-2ab$

9. anonymous

so it would be d?

10. anonymous

Sorry, we just don't give answers, and it seems like all you're doing is just guessing, why not at least show us some work so we can understand where exactly you are having problem understanding the concept