
 👍 0
 👎 0
posted by Reiny
Respond to this Question
Similar Questions

PreCalculus
Find all solutions to the equation in the interval [0, 2pi) cos4xcos2x=0 So,this is what i've done so far: cos4xcos2x=0 cos2(2x)cos2x (2cos^2(2x)1)(2cos^2(x)1) No idea what to do next.
asked by Maria on March 19, 2012 
Calculus
If √2cos(x)1 = (1+√3)/2 and √2cos(x)+1 = (1√3)/2, find the value of cos4x.
asked by sh on March 5, 2010 
Math  Trig  Double Angles
Prove: cos4x = 8cos^4x  8cos^2x + 1 My Attempt: RS: = 4cos^2x (2cos^2x  1) + 1 = 4 cos^2x (cos2x) + 1 LS: = cos2(2x) = 2cos^2(2x)  1 = (cos^2(2))  cos^2(2x))  1  Prove: 8cos^4x = cos4x + 4cos2x + 3 My Attempt: RS: =
asked by Anonymous on November 17, 2007 
trig
how can you confirm the identity cos^4x = (1/8)(3+ 4cos2x+ cos4x) and sin4x = (4sinxcosx)(2cos^2x1)
asked by Sarah Kim on July 18, 2009 
trig
I don't understand this. can you please tell how you got each answer? describe how the graph of each function is related to a basic trigonometric graph. 1. y= 1/2cos x 2. y= 2cos x+1 3. y= sin x4 4. y= sin(x4) 5. y= sin(3x) 6.
asked by BreAnne on February 10, 2012 
trig
Prove that cos4x=8cos^4x8cos^2x+1 so I'm guessing you begin with the left side cos4x=cos2(2x) then im kinda lost
asked by Natash on November 23, 2008 
Pre Calc
the graph f(x)=cos x Which gives the graph of g(x)=2cos[3(x+(pi/3))]5 write the equation of g(x) in terms of f(x). Is the question basicly asking for me to do g(f(x))?
asked by Rita on June 12, 2011 
math
cosA= 5/9 find cos1/2A are you familiar with the halfangle formulas? the one I would use here is cos A = 2cos^2 (1/2)A  1 5/9 + 1 =2cos^2 (1/2)A 14/9 =2cos^2 (1/2)A cos (1/2)A = √(7)/3
asked by Anonymous on May 3, 2007 
Math
Determine the solution of the equetion: 2cos x1=0? 2cos x=0? and 2cos x+1=0?
asked by Miley on September 8, 2015 
Math
I'm after understanding more than an answer but an answer would be greatly appreciated. The question reads as "On the answer sheet for this question is shown an xy graph of the function [y(x)=x^2cos(3x)] Add to that same picture
asked by Jack on November 17, 2010