Math (Trigonometry [Polar Form])
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TRIG!
Posted by hayden on Monday, February 23, 2009 at 4:05pm. sin^6 x + cos^6 x=1  (3/4)sin^2 2x work on one side only! Responses Trig please help!  Reiny, Monday, February 23, 2009 at 4:27pm LS looks like the sum of cubes sin^6 x +
asked by hayden on February 23, 2009 
precal
Simplify the given expression........? (2sin2x)(cos6x) sin 2x and cos 6x can be expressed as a series of terms that involve sin x or cos x only, but the end result is not a simplification. sin 2x = 2 sinx cosx cos 6x = 32 cos^6 x
asked by ethan on May 17, 2007 
math
Determine exact value of cos(cos^1(19 pi)). is this the cos (a+b)= cos a cos b sina sin b? or is it something different. When plugging it in the calculator, do we enter it with cos and then the (cos^1(19 pi)).
asked by Anonymous on April 14, 2012 
math
Find the exact value of cos 300 degrees. thanks guys cos 300 = 1/2 = 0.500 how do you know? I am supposed to show my work. You ought to know the rule on 306090 triangles. If the hyp is 2, the shorter side is 1, and the longer
asked by Bill on June 2, 2007 
Precalculus
Solve Cos^2(x)+cos(x)=cos(2x). Give exact answers within the interval [0,2π) Ive got the equation down to cos^2(x)+cos(x)+1=0 or and it can be simplified too sin^2(x)+cos(x)=0 If you could tell me where to go from either of
asked by Josh on February 23, 2012 
maths
Find the roots of z^6 + 1 and hence resolve z^6 + 1into read quadratic factors; deduce that cos3x = 4[cos(x) cos(pi/6)][(cos(x) cos(pi/2)][(cos(x) cos(5pi/6)]
asked by jake on November 10, 2008 
maths
Find the roots of z^6 + 1 and hence resolve z^6 + 1into read quadratic factors; deduce that cos3x = 4[cos(x) cos(pi/6)][(cos(x) cos(pi/2)][(cos(x) cos(5pi/6)]
asked by jake on November 10, 2008 
Calc.
Differentiate. y= (cos x)^x u= cos x du= sin x dx ln y = ln(cos x)^x ln y = x ln(cos x) (dy/dx)/(y)= ln(cos x) (dy/dx)= y ln(cos x) = (cos x)^x * (ln cos x) (dx/du)= x(cos x)^(x1) * (sin x) =  x sin(x)cos^(x1)(x)
asked by Chelsea on March 9, 2011 
calculus
Differentiate. y= (cos x)^x u= cos x du= sin x dx ln y = ln(cos x)^x ln y = x ln(cos x) (dy/dx)/(y)= ln(cos x) (dy/dx)= y ln(cos x) = (cos x)^x * (ln cos x) (dx/du)= x(cos x)^(x1) * (sin x) =  x sin(x)cos^(x1)(x)
asked by Chelsea on March 10, 2011 
Math  Solving Trig Equations
What am I doing wrong? Equation: sin2x = 2cos2x Answers: 90 and 270 .... My Work: 2sin(x)cos(x) = 2cos(2x) sin(x) cos(x) = cos(2x) sin(x) cos(x) = 2cos^2(x)  1 cos(x) (+/)\sqrt{1  cos^2(x)} = 2cos^2(x)  1 cos^2(x)(1 
asked by Anonymous on November 24, 2007