How Do Double Angle Identities Simplify Trigonometric Equations?

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Homework Help Overview

The discussion revolves around the simplification of trigonometric equations using double angle identities, specifically focusing on the equation sin4x - sin2x / sin2x = cos3x / cosx. Participants are exploring the application of trigonometric identities and the manipulation of expressions to prove or solve the equation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to simplify the left-hand side of the equation but expresses uncertainty about the next steps after reaching a certain point. They question the feasibility of manipulating the expressions further.
  • Some participants clarify the goal of proving the equation rather than solving it, leading to discussions about expanding both sides of the equation and applying relevant identities.
  • Others suggest expanding the right-hand side and exploring the implications of known identities to simplify the expressions involved.

Discussion Status

The discussion is active, with participants providing guidance on how to approach the problem. There is a mix of attempts to clarify the requirements of the problem and suggestions for manipulating the equations. Multiple interpretations of the problem-solving approach are being explored, but no consensus has been reached on a specific method or solution.

Contextual Notes

Participants are working under the constraints of using specific trigonometric identities provided in the thread. There is an emphasis on proving the equation rather than solving it, which shapes the direction of the discussion.

Jatt
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Homework Statement


sin4x-sin2x/sin2x=cos3x/cosx


Homework Equations


sin2x = 2sinxcosx


The Attempt at a Solution



LS = sin(2x + 2x) - sin2x/sin2x
= sin2xcos2x + cos2xsin2x - sin2x/sin2x
= 2sin2xcos2x - sin2x/sin2x
This is where i get stuck...
I don't know what happens if you try to: 2sinx2x - sin2x or 2sin2x/sin2x, is that possible or not? Can you guys help me solve this question? Thanks.
 
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Quick question you are to prove that
\frac{sin4x-sin2x}{sin2x}=\frac{cos3x}{cosx}

?
 
yup.
 
I believe there are two more Relevant equations you need.
 
doesn't matter i was able to solve it, but here's another problem which I'm now stuck with. cosx+cos2x+cos3x=cos2x(1+2cosx).
I've tried many things with this problem, but always seem to get lost.
The only given identites which I'm given to use:
cos2x=cos^2x-sin^2x
cos2x=2cos^2x-1
cos2x=1-2sin^2x
sin2x=2sinxcosx
 
Last edited:
OK, since you have to SOLVE and not prove...
if you expand the RHS you would see that the cos2x cancels out and you are left with

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

then expand out cos(3x) and see if anything gets simpler
 
lol, sorry for not stating this, but i have to prove not solve.
 
well then expand out the LHS
Recall that cos(3x)=cos(2x+x)
 
yea then i get: cos2xcosx + sin2xsinx + cos2x + cosx
 
  • #10
cos2x=2cos^2(x)-1
and sin2x=2sinxcosx
cos^2(x)+sin^2(x)=1

expand out and find all in terms of cosx and hopefully it will work
 

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