Respuesta :
Solution
From the giiven information we have.
[tex]\begin{gathered} STEP1)\cos \theta\text{ =- }\frac{3}{5}\text{ } \\ F\text{rom the question we are to find} \\ \sin 2\theta\text{ + cos2}\theta \\ But\colon\text{ sin2}\theta\text{ = 2sin}\theta\cos \theta \\ Also.\text{ cos2}\theta=1-2sin^2\theta \\ \text{Then:} \\ \sin 2\theta\text{ + cos2}\theta \\ (\text{2sin}\theta\cos \theta)\text{ + }(1-2sin^2\theta) \\ \\ \end{gathered}[/tex][tex]\begin{gathered} STEP2\colon\text{But note that} \\ \text{from cos }\theta\text{ = -}\frac{3}{5}\text{ = }\frac{ADJ}{\text{HYP}} \\ \text{Then we will have }to\text{ find the OP}\P \\ By\text{ using pythagoras Theorem} \\ \text{HYp}^2=Opp^2+Adj^2 \\ 5^2=Opp^2+3^2 \\ 25=Opp^2\text{ + 9} \\ 25-9=Opp^2 \\ 16=Opp^2 \\ \text{Opp}^2\text{ = 16} \\ \text{Opp = }\sqrt[]{16} \\ Opp\text{ = 4} \\ \text{Therefore} \\ S\text{in}\theta\text{ =}\frac{\text{Opp}}{\text{HYP}}\text{ =- }\frac{4}{5} \end{gathered}[/tex][tex]\begin{gathered} STEP3\colon\text{Therefore from step 1, we have} \\ s\text{in2}\theta\text{ + cos2}\theta \\ (2\sin \theta\cos \theta)+(1-2sin^2\theta) \\ (2\times-\frac{4}{5}\text{ }\times-\frac{3}{5}\text{ ) + (1 -2}\times(-\frac{4}{5})^2\text{)} \\ (\frac{24}{25})\text{ + (-}\frac{32}{25}\text{)} \\ \frac{24}{25}-\frac{32}{25} \\ -\frac{8}{25} \end{gathered}[/tex]