How can I solve this trigonometric equation using the given equations?

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Suraj M
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Homework Statement



$$ \cos(2x) × \cos(\frac x2) - \cos(3x)× \cos (\frac{9}{2} x) = \sin(5x)× \sin \frac {5}{2}x $$

Homework Equations


$$ \cos(x) + \cos(y) = 2 \cos{\frac{x+y}{2}} \cos{\frac{x-y}{2}} $$
$$ \cos(x) - \cos (y) = -2 \sin \frac{x+y}{2} \sin \frac{x-y}{2} $$

The Attempt at a Solution


I don't know how to start, as there are no common angles(2x,x/2,3x...).
How do i start off?
 
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First, start by using parentheses when necessary. It's not true that
$$\cos x + \cos y = 2\cos\left(x+\frac y2\right)\cos\left(x-\frac y2\right)$$ as you wrote.

It might help to write the identities using variables other than x and y to avoid confusion, e.g.
$$\cos a + \cos b = 2\cos \frac{a+b}2 \cos \frac {a-b}2.$$ To deal with the term ##\cos 2x \cos \frac x2##, you want ##\frac{a+b}2 = 2x## and ##\frac{a-b}2 = \frac x2##. Solve for ##a## and ##b##. What do you get?
 
vela said:
First, start by using parentheses when necessary. It's not true that
$$\cos x + \cos y = 2\cos\left(x+\frac y2\right)\cos\left(x-\frac y2\right)$$ as you wrote.

It might help to write the identities using variables other than x and y to avoid confusion, e.g.
$$\cos a + \cos b = 2\cos \frac{a+b}2 \cos \frac {a-b}2.$$ To deal with the term ##\cos 2x \cos \frac x2##, you want ##\frac{a+b}2 = 2x## and ##\frac{a-b}2 = \frac x2##. Solve for ##a## and ##b##. What do you get?
I actually don't know how you right those equations using parentheses, so pardon this:
##a = {\frac{5}{2}} x ## and ##b = {\frac{3}{2}} x ##
and then if i do the same for the second term... ##c = {\frac{15}{2}} x## and ##d = {\frac{-3}{2}} x##
so then my LHS becomes:
$${\frac{1}{2}}\left({\cos \left({\frac{5}{2}} x \right) +\cos{\frac{3}{2}}x } \right)- {\frac{1}{2}} \left( [\cos {\frac{15}{2}} x + \cos{\frac{3}{2}} x ] \right) $$
$$ = {\frac{1}{2}}\left[(-2) × \sin(-5x) × \sin{\frac{5}{2}}x \right] $$
= RHS
, Thanks a lot:smile:
now could you please tell me how to right those equations using parentheses,please
 
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