Question

Use your understanding of how to identify the parametersr of the sinusoidal function y=asin(bx+c)+d or y =acos(bx+c) to write the equation of a simpler function p(x) that describes the graph

Answer 1

\[(sinx +\sin3x)/(2\sin2x) which is = \to \cos x\]

Answer 2

umm I think the cos one would be better

Answer 3

So you have to prove the identity
\[\frac{\sin x+\sin3x}{2\sin2x}=\cos x~~?\]

Answer 4

yes

Answer 5

well actuallyyy i need to fund write an new equation to describe the graph but i'm lost

Answer 6

Recall the angle sum identity for sine:
\[\sin(a+b)=\sin a\cos b+\sin b\cos a\]
Double angle identities:
\[\sin2x=2\sin x\cos x\\
\cos2x=\cos^2x-\sin^2x\]
Pythagorean identity:
\[\sin^2x+\cos^2x=1\]
Setting \(a=x\) and \(b=2x\), you have \(x+y=3x\) and
\[\sin3x=\sin x\cos 2x+\sin 2x\cos x\]
Substitute in the initial exression and simplify:
\[\begin{align*}\frac{\sin x+\sin3x}{2\sin2x}&=\frac{\sin x+\sin x\cos2x+\sin2x\cos x}{2\sin2x}\\\\
&=\frac{\sin x+\sin x\cos2x}{2\sin2x}+\frac{\sin2x\cos x}{2\sin2x}\\\\
&=\frac{\sin x+\sin x\cos2x}{4\sin x\cos x}+\frac{\cos x}{2}\\\\
&=\frac{1+\cos2x}{4\cos x}+\frac{\cos x}{2}\\\\
&=\frac{1+\cos^2x-\sin^2x}{4\cos x}+\frac{\cos x}{2}\\\\
&=\frac{\cos^2x+\cos^2x}{4\cos x}+\frac{\cos x}{2}\\\\
&=\frac{2\cos^2x}{4\cos x}+\frac{\cos x}{2}\\\\
&=\frac{\cos x}{2}+\frac{\cos x}{2}\\\\
&=\cos x
\end{align*}\]

Answer 7

omg I am so sorry I already did that part! I just was wondering how to write the equation of a simpler function with either y=asin(bx+c)+d or y=acos(bx+c)

Answer 8

I hope you're not mad >.>

Answer 9

No worries. The "simpler" function would be the completely reduced version, so I would say it's just \(p(x)=\cos x\). I'm not sure what kind of answer you're expected to give...

Answer 10

i think it's just cos too but they want me to pick of of those two formulas (y=asin(bx+c) & y=acos(bx+c) I'm not sure how to approach that.

Answer 11

I would think you have to indicate the values of \(a,b,c,d\). Since \(p(x)=\cos x\), you have \(a=b=1\) and \(c=d=0\).

Answer 12

sooo do you think I should pick the equation y=acos(x+c) and say 1cos(x+0)?

Answer 13

I would leave it as \(\cos x\). The 1s and 0s are implied.

Answer 14

ok thanks so much