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

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The discussion focuses on solving a complex trigonometric equation involving multiple cosine and sine terms. Participants suggest using trigonometric identities to simplify the equation, emphasizing the importance of clarity in notation, particularly with parentheses. One user expresses confusion about applying these identities and seeks guidance on proper formatting for equations. The conversation highlights the need for careful manipulation of terms to find a solution. Ultimately, the thread illustrates the collaborative effort in tackling challenging trigonometric problems.
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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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Why do you not simply start expanding your products using the equations you have quoted and see what you get?
 
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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Suraj M said:
now could you please tell me how to write those equations using parentheses, please
cos ((x+y)/2) instead of cos (x+y/2)
 
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