How to solve (1+cosA)^2 - (1-cosA)^2 - sin^2A = ctgA*sinA*cosA

  • Thread starter Thread starter makarov1901
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around the equation (1+cosA)^2 - (1-cosA)^2 - sin^2A = ctgA*sinA*cosA, which involves trigonometric identities and simplifications. Participants are exploring the validity of the equation and its components.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to simplify the equation by moving sin^2A to the right side and expanding the left side. Some participants question the validity of the simplification and whether the equation represents an identity.

Discussion Status

Participants are actively discussing the nature of the equation, with some suggesting that it may not be an identity due to its failure to hold for all values of A. There is a recognition of potential typos in the problem statement, and differing interpretations are being explored.

Contextual Notes

There is a mention of specific values for A, such as A=pi/2, which highlight the discrepancies between the left and right sides of the equation. Participants are also considering the implications of the equation not being an identity.

makarov1901
Messages
3
Reaction score
0

Homework Statement


(1+cosA)^2-(1-cosA)^2-sin^2A=ctgA*sinA*cosA

Homework Equations


The Attempt at a Solution



I moved sin^2 to the right side, then expanded the left side and got to:
1+2cosA+cos^2A-1+2cosA-cos^2A=1
When I cancel the left side I get:
4cosA=1

I'd be very grateful if anyone help me.
 
Last edited:
Physics news on Phys.org


Looks good so far. What's the problem?

(I assume when you wrote sin^2, you meant sin2 a.)
 


You assumed right.
The problem is that both parts of the identity should be equal.
As far as I know, 4cosA is not equal to 1.
 


It's not an identity. For example, when a=pi/2, the lefthand side is -1 while the righthand side equals 0.

Perhaps there's a typo in the problem.
 


My task is to transform both parts so that they're equal. I don't know why you say it's not an identity.
 


If it were an identity, it would hold for all values of a. It clearly doesn't; therefore, it's not an identity.
 

Similar threads

  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 9 ·
Replies
9
Views
12K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K