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

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Homework Help Overview

The discussion revolves around solving a trigonometric equation involving multiple cosine and sine terms. The original poster expresses uncertainty about how to begin, particularly due to the lack of common angles in the equation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants suggest expanding products using trigonometric identities. There is a focus on clarifying the correct use of these identities and the importance of parentheses in mathematical expressions.

Discussion Status

Some participants have provided guidance on using trigonometric identities and the correct formulation of equations. The conversation includes attempts to clarify notation and expressions, with ongoing questions about proper formatting in mathematical writing.

Contextual Notes

There are indications of confusion regarding the application of trigonometric identities and the notation used in the equations. The original poster is seeking assistance with formatting and clarity in their mathematical expressions.

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?
 
Last edited:
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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
 
Last edited:
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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