Solve sin cos: Find cos6x - 4cos4x + 8cos2x =?

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Discussion Overview

The discussion revolves around solving the equation cos6x - 4cos4x + 8cos2x, with a specific condition involving the sum of sine functions. Participants explore various methods to approach the problem, including numerical solutions and algebraic manipulations.

Discussion Character

  • Mathematical reasoning, Homework-related, Technical explanation

Main Points Raised

  • One participant requests hints for solving the equation involving cosines and a condition on sine functions.
  • Another participant suggests solving the cubic equation numerically to find sin(x) and subsequently using the identity cos^2(x) = 1 - sin^2(x) to evaluate the expression, asserting the answer is 4.
  • A different approach is proposed, rewriting the equation in terms of sine and using identities to derive the answer.
  • Subsequent posts shift focus to a new problem involving sine and tangent functions, with participants discussing potential methods to solve it.
  • One participant suggests testing values for variables a and b to find solutions.
  • Another participant recommends using double angle identities to simplify the expressions in the new problem.
  • One participant expresses gratitude for the assistance received and indicates they have found the answer.

Areas of Agreement / Disagreement

There is no consensus on the solution to the original equation, as participants propose different methods and approaches. The later problem introduces additional complexity, with various strategies discussed but no agreement on a singular approach.

Contextual Notes

Participants do not clarify the assumptions behind their methods, and the discussion includes multiple approaches that may depend on specific interpretations of the equations involved.

johncena
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Can anyone give any hint to solve this ?

If sinx + sin2x + sin3x = 1,
then , cos6x - 4cos4x + 8cos2x =

(a) 1
(b) 4
(c) 2
(d) 3
 
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It's probably not in the spirit of the question but you can just solve that cubic numerically and get sin(x) = 0.5437 (to 4 dp). Then using cos^2(x) = 1- sin^2(x) you can easily evaluate it.

BTW. The answer is (b) 4.
 
Rewrite as: sinx + (sinx)^3 = (cos)^2. Then square both sides and use the identity (sin)^2=1-(cos)^2 and the answer will follow
 
Thanks for the hint . I got the answer .Now can you help me to solve this ?

If [tex]\frac{sin(2a + b)}{sin b}[/tex] = [tex]\frac{n}{m}[/tex] , then tan (a + b)cot a = ?

(A)[tex]\frac{n-m}{n+m}[/tex] (B) [tex]\frac{m-n}{m+n}[/tex]

(C)[tex]\frac{n+m}{n-m}[/tex] (D)[tex]\frac{m+n}{m-n}[/tex]
 
Last edited:
I suppose the easiest way would be to choose values for a and b and see which fail.
 
Write sin(2a+b) as sin [a+(a+b)] and sin(b) as sin[a+(b-a)]. Then use the double angle identities and you should obtain the answer easily.
 
Thank you very much for your help sir...I got the answer easily
 

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