Question #d76e9

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Question #d76e9

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Answer 1 · 2017-07-24T18:38:34

$0$ $0$

Explanation:

$sin 2 x = 1$ $sin2x=1$ , $2 x = \frac{π}{2}$ $2x=pi/2$ or $x = \frac{π}{4}$ $x=pi/4$

Hence

$u = cos (8 \cdot \frac{π}{4}) + 2 cos (6 \cdot \frac{π}{4}) + cos (4 \cdot \frac{π}{4})$ $u=cos(8*pi/4)+2cos(6*pi/4)+cos(4*pi/4)$

$= cos (2 π) + 2 cos (\frac{3 π}{2}) + cos (π)$ $=cos(2pi)+2cos((3pi)/2)+cos(pi)$

$= 1 + 2 \cdot 0 - 1$ $=1+2*0-1$

$= 0$ $=0$

1) I found $x$ $x$ for $sin 2 x = 1$ $sin2x=1$ condition.

2) I substituted to $x = \frac{π}{4}$ $x=pi/4$ .

Answer 2 · 2017-07-24T23:27:24

$0$ $0$ .

Explanation:

$sin (2 x) = 1$ $sin(2x)=1$ .

In general, equations of the form $sin (θ) = K$ $sin(theta)=K$ and solved by $θ = arcsin (K) + 2 n π$ $theta = arcsin(K)+2npi$ and $π - θ = arcsin (K) + 2 n π$ $pi-theta=arcsin(K)+2npi$ , $n in ZZ$ , because of the periodicity of the sine function#.

Note that $arcsin(1)=pi/2$ . This means that if $sin(2x)=1$ ,

$2x=pi/2+2npi$ ,
$x=pi/4+npi$ .

$pi-2x=pi/2+2npi$ ,
$2x=pi/2-2npi$ ,
$x=pi/4-npi$ .

As $n in ZZ$ these solution sets are identical.
So if $sin(2x)=1$ then $x=pi/4+npi$ , $n in ZZ$ .

Then

$cos(8(pi/4+npi)+2cos(6(pi/4+npi))+cos(4(pi/4+npi))$ ,
$cos(2pi+4(2npi))+2cos((3pi)/2+3(2npi))+cos(pi+2(2npi))$ .

Multiplies of the periodicity can be removed and the values substituted to give,

$1+2*0-1=0$ .

In this case, it was safe to assume $sin(2x)=1 => x=pi/4$ . This will not always be the case! So it is good to solve more generally.

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Question #d76e9

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