Question #fdeca

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Answer 1 · 2014-10-26T22:03:31

$cos (x)$ $cos(x)$

Recall our trig identities:

$sin (2 x) = 2 sin (x) cos (x)$ $sin(2x) = 2sin(x)cos(x)$

$cos(2x) = cos^2(x) – sin^2(x)$

$sin^2(x)+cos^2(x)=1$

Looking at our problem:

$cos(2x) cos(x) + sin(2x) sin(x)$

Let's substitute for $sin(2x)$ . This will give us:

$cos(2x) cos(x) + (2sin(x)cos(x))* sin(x)$

$=cos(2x)cos(x)+2sin^2(x)cos(x)$

Now, let's substitute $sin^2(x)$ using the identity $sin^2(x)+cos^2(x)=1$

This gives us:

$cos(2x)cos(x)+2sin^2(x)cos(x)$

$= cos(2x)cos(x)+2(1-cos^2(x))*cos(x)$

Now, let's substitute $cos(2x)$ using the identity $cos(2x) = cos^2(x) – sin^2(x)$

This gives us:

$cos(2x)cos(x)+2(1-cos^2(x))*cos(x)$

$=(cos^2(x)-sin^2(x))*cos(x)+2(1-cos^2(x))*cos(x)$

Now, let's substitute the sin^2(x) inside the parenthesis using the identity, $sin^2(x)+cos^2(x)=1$

This gives us:

$(cos^2(x)-sin^2(x))*cos(x)+2(1-cos^2(x))*cos(x)$

$=(cos^2(x)-(1-cos^2(x)))*cos(x)+2(1-cos^2(x))*cos(x)$

$=(cos^2(x)-1+cos^2(x))*cos(x)+2(1-cos^2(x))*cos(x)$

$=(2cos^2(x)-1)*cos(x)+2(1-cos^2(x))*cos(x)$

Now, multiplying through with our $cos(x)$ 's:

$(2cos^2(x)-1)*cos(x)+2(1-cos^2(x))*cos(x)$

$=2cos^3(x)-cos(x)+2(cos(x)-cos^3(x))$

$=2cos^3(x)-cos(x)+2cos(x)-2cos^3(x)$

The $2cos^3(x)$ and the $-2cos^3(x)$ cancel out,

and we are left with:

$-cos(x)+2cos(x)$

which equals:

$cos(x)$