Question

If 0 is in Quadrant III and cos 0 = = }, what is sin 20 + cos 20? - 국 25 SK - 27 다 10

Answer 1

From the giiven information we have.

`\begin{gathered} STEP1)\cos \theta\text{ =- }(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}`
`\begin{gathered} STEP2\colon\text{But note that} \\ \text{from cos }\theta\text{ = -}(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{ =- }(4)/(5) \end{gathered}`
`\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*-(4)/(5)\text{ }*-(3)/(5)\text{ ) + (1 -2}*(-(4)/(5))^2\text{)} \\ ((24)/(25))\text{ + (-}(32)/(25)\text{)} \\ (24)/(25)-(32)/(25) \\ -(8)/(25) \end{gathered}`