Question 1045772
<pre>
1.   cos(3x) = cos(2x+x) = cos(2x)cos(x)-sin(2x)sin(x) =
(2cos˛(x)-1)cos(x)-2sin(x)cos(x)sin(x) =
2cosł(x)-cos(x)-2sin˛(x)cos(x) =
2cosł(x)-cos(x)-2[1-cos˛(x)]cos(x) =
2cosł(x)-cos(x)-2cos(x)+2cosł(x)
4cosł(x)-3cos(x)

So cos(3x)=4cosł(x)-3cos(x) and cos(2x)=2cos˛(x)-1. 
So cos(6x)=2[cos(3x)]˛-1 or 2[4cosł(x)-3cos(x)]˛-1.
We can expand it to become 
2[16cos<sup>6</sup>(x)-24cos<sup>4</sup>(x)+9cos˛(x)]-1 

Final result: cos(6x) = 

32cos<sup>6</sup>(x) - 48cos<sup>4</sup>(x) + 18cos<sup>2</sup>(x) - 1

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2.   {{{((1^""+tan(y))(1^""+cot(x)))/(cot(x)+tan(y))}}}

{{{(1^""+tan(y))(1^""+cot(x))}}}{{{"÷"}}}{{{(cot(x)^""+tan(y))}}}


{{{((1^""+sin(y)/cos(y))^""(1^""+cos(x)/sin(x)))}}}{{{"÷"}}}{{{(cos(x)/sin(x)+sin(y)/cos(y))}}}

{{{((cos(y)+sin(y))/cos(y))((sin(x)+cos(x))/sin(x))}}}{{{"÷"}}}{{{(cos(x)cos(y)+sin(y)sin(x))/(sin(x)cos(y))}}}

{{{(cos(y)sin(x)+cos(y)cos(x)+sin(y)sin(x)+sin(y)cos(x))/(cos(y)sin(x)))}}}{{{"÷"}}}{{{(cos(x)cos(y)+sin(x)sin(y))/(sin(x)cos(y))}}}

{{{((cos(y)sin(x)^""+sin(y)cos(x))+(cos(y)cos(x)^""+sin(y)sin(x)))/(cos(y)sin(x)))}}}{{{"÷"}}}{{{cos(x-y)/(sin(x)cos(y))}}}

{{{((sin(x)cos(y)^""+cos(x)sin(y))+(cos(x)cos(y)^""+sin(x)sin(y)))/(cos(y)sin(x)))}}}{{{""*""}}}{{{(sin(x)cos(y))/cos(x-y)}}}

{{{(sin(x+y)+cos(x-y))/(cos(y)sin(x)))}}}{{{""*""}}}{{{(sin(x)cos(y))/cos(x-y)}}}

{{{(sin(x+y)+cos(x-y))/(cross(cos(y))cross(sin(x))))}}}{{{""*""}}}{{{(cross(sin(x))cross(cos(y)))/cos(x-y)}}}

{{{(sin(x+y)+cos(x-y))/1}}}{{{""*""}}}{{{1/cos(x-y)}}}

{{{sin(x+y)/cos(x-y)+""^1cross(cos(x-y))/cross(cos(x-y))}}}

{{{sin(x+y)/cos(x-y)+1}}}

{{{1+sin(x+y)/cos(x-y)}}}

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{{{2cos^2(3x) -sin(7x)sin(x) = 1+cos(7x)cos(x)}}}

This one requires some tricks of subtracting and adding the
same quantity to create a use for some double-angle identities. 
We work with the left side:

Subtract 1 and add 1 to create a use for the identity
                {{{cos(2theta)=2cos^2(theta)-1}}}

{{{2cos^2(3x)-1+1-sin(7x)sin(x)}}}

Using that identity, the first two terms become cos(6x) 

{{{cos(6x)+1-sin(7x)sin(x)}}}

Subtract and add cos(7x)cos(x) to create a
use for the identity {{{cos(alpha-beta)=cos(alpha)cos(beta)+sin(alpha)sin(beta)}}}
 
{{{cos(6x)+1-sin(7x)sin(x)-cos(7x)cos(x)+cos(7x)cos(x)}}}

Factor a "-" out of 3rd and 4th terms:

{{{cos(6x)+1-(sin(7x)sin(x)^""+cos(7x)cos(x))+cos(7x)cos(x)}}}

Swap the terms in the parentheses to recognize the identity:

{{{cos(6x)+1-(cos(7x)cos(x)^""+sin(7x)sin(x))+cos(7x)cos(x)}}}

Use the identity

{{{cos(6x)+1-cos(7x-x)+cos(7x)cos(x)}}}

7x-x = 6x

{{{cos(6x)+1-cos(6x)+cos(7x)cos(x)}}}

Cancel the two opposite terms

{{{cross(cos(6x))+1-cross(cos(6x))+cos(7x)cos(x)}}}

{{{1+cos(7x)cos(x)}}}

Tricky, huh?

Edwin</pre>