18 I’m pretty sure that is the answer that
U want
A trig identity is <span>asinucosu=<span>a/2</span>sin(2u)</span>So you can write your equation as<span>y=sin(x)cos(x)=<span>1/2</span>sin(2x)</span>Use the crain rule here<span><span>y′</span>=<span>d/<span>dx</span></span><span>1/2</span>sin(2x)=<span>1/2</span>cos(2x)<span>d/<span>dx</span></span>2x=cos(2x)</span>The curve will have horizontal tangents when y' = 0.<span><span>y′</span>=0=cos(2x)</span>On the interval [-pi, pi], solution to that is<span><span>x=±<span>π4</span>,±<span><span>3π</span>4</span></span></span>
They are both equal because the zeros don't count after the number 5.
Answer:
1. 13
2. 11
3. 10
4. 9
5. 12
6. 16
7. 14
8. 40
Step-by-step explanation:
