Factor and simplify: sin^4 x - cos^4 x?
sin^4 x - cos^4 x = (sin^2 x - cos^2 x)(sin^2 x + cos^2 x) = (sin x - cos x)(sin x + cos x) * 1 = (sin x - cos x)(sin x + cos x) The given answer, obviously is not for this question ... as you can see it has "y" !!
Factor and simplify: 1-2sin^2x+sin^4x?
= sin^4(x) - 2sin²(x) + 1 = (sin²(x) - 1)(sin²(x) - 1) = (sin(x) + 1)(sin(x) - 1)(sin(x) + 1)(sin(x) - 1) = (sin(x) + 1)²(sin(x) - 1)²
Factor and simplify 1-2sin^2x+sin^4x?
1. Let y=sin x 1-2y^2+y^4 = (1-y^2)^2 replace y by sin x = (1-sin^2 x)^2 2. sin(46.4) /20.6 =sin(100.1) / c cross multiply c sin(46.1) = (20.6) sin(100.1) divide both sides by sin(46.1) c = (20.6) sin(100.1) /sin(46.1) c = 28.15 sin(33.5) / a = sin(46.4) / 20.6 cross multiply sin(33.5) (20.6) = a sin(46.4) divide both sides by sin(46.4) sin(33.5) (20.6) /sin(46.4) = a a = 15.7 3. cos θ = 0.0872 θ = cos^-1(0.0872) θ = 1.4835 radian = (1.4835*180/pi) = 84.997 = 85 degrees But we want the angle in the fourth quadrant 270 degrees < θ < 360 degrees . θhe required angle is 360-85 = 275 degrees
Factor 1-2sin^2(x) + sin^4(x) & simplify?
let y = sin2^(x) 1-2sin^2(x) + sin^4(x) => 1 - 2y + y^2 => (y - 1)^2 => [(sin^2(x) -1]^2 => (sin(x) - 1)^2 (sin(x) + 1)^2 ; <== factorized ============================= or => [(sin^2(x) -1]^2 => [(sin^2(x) - (sin^2(x) + cos^2(x))]^2 => [(sin^2(x) - sin^2(x) - cos^2(x))]^2 => cos^4(x) ; <== single term
How to factor and simplify sec^4x-tan^4x?
Remember that secx = 1/cosx and tanx=sinx/cosx. So we can change the expression using this: sec^4x-tan^4x = 1/cos^4x - sin^4x/cos^4x 1/cos^4x - sin^4x/cos^4x = (1 - sin^4x)/(cos^4x) (1 - sin^4x) is the same as (1-sin^2x)(1+sin^2x), so: (1 - sin^4x)/(cos^4x) = (1-sin^2x)(1+sin^2x)/(cos^4x) (1-sin^2x) is the same as cos^2x, so: (1-sin^2x)(1+sin^2x)/(cos^4x) = cos^2x(1+sin^2x)/cos^4x cos^2x(1+sin^2x)/cos^4x = (1+sin^2x)/cos^2x
Factor and Simplify: cot^4 x+2 cot^2 x+1?
the answer is D rewrite terms using sin and cos (cos^4 x)/(sin^4 x) + (2 cos^2 x)/(sin^2 x) + 1 find common denominator (cos^4 x)/(sin^4 x) + (2 cos^2 x sin^2)/(sin^4) + sin^4 x/sin^4 x combine cos^4 x + 2 cos^2 x sin^2 x + sin^4 x --------------------------------------... sin^4 x factor the numerator (cos^2 x + sin^2 x)(cos^2 x + sin^2 x) -------------------------------------- sin^4 x remember sin^2 x +cos^2 x = 1 1/sin^4 x = csc^4 x
Simplify: sin^4 x - cos^4 x?
sin^4x - cos^4x = (sin^2x)(sin^2x) - (cos^2x)(cos^2x) = (sin^2x)^2 - (cos^2x)^2 = (sin^2x - cos^2x)(sin^2x + cos^2x) = (sin^2x - cos^2x)(1) --->From the identity cos^2x + sin^2x = 1 = sin^2x - cos^2x Now, there are several ways of simplifying this trigonometric expression depending on whether you want it to be in terms of sin x, cos x, tan x or multiple angles. 1st option: Using the identity cos^2x + sin^2x = 1 again, we get sin^2x = 1 - cos^2x, or cos^2x = 1 - sin^2x Hence, sin^4x - cos^4x = sin^2x - cos^2x = 1 - cos^2x - cos^2x = 1 - 2cos^2x Similarly, sin^4x - cos^4x = sin^2x - cos^2x = sin^2x - (1 - sin^2x) = 2sin^2x - 1 2nd option: Using the double angle formula: cos (2x) = cos^2x - sin^2x sin^4x - cos^4x = sin^2x - cos^2x = - (cos^2x - sin^2x) = - cos (2x) Hope this helps.
How do I simplify [math]\sin^{2}{x} +\cos^{4}{x}-\sin^{4}{x}[/math]?
my favourite way:I would just repeatedly use the “raised cosine” identities:cos^2(x) = 0.5 + 0.5 cos(2x), (formula A)sin^2(x) = 0.5 - 0.5 cos(2x), (formula B)(can be derived from the first one, setting x->pi/2 - x)Squaring these identities, and using them again to simplify their own squares yields formulas for cos^4(x) and sin^4(x) :cos^4(x) = [cos^2(x)]^2 = [0.5 + 0.5 cos(2x)]^2 = ….When you are done, substitute these fourth powers and the squares (formulas A,B ) into your expression and also into your alleged answer and compare.Well, you carry on, else if you don't do it actively but rather read it passively you might won't really assimilate it. …Formulas A,B is what you are ever gonna need (plus a little algebra, expanding squares and collecting coefficients in polynomial
How can you simplify [math]\cos^6x + \sin^6x [/math] to [math]1- 3\sin^2x \cos^2x ?[/math]
cos^6 (x) + sin^6 (x) can be written as [cos^2 (x) + sin^2 (x)] * [cos^4 (x) - cos^2 (x) sin^2 (x) + sin^4 (x)]= [cos^4 (x) - cos^2 (x) sin^2 (x) + sin^4 (x)] as cos^2 (x) + sin^2 (x) = 1 Now we can write cos^4 (x) - cos^2 (x) sin^2 (x)+ sin^4 (x) as: cos^4 (x) - cos^2 (x) sin^2 (x) + sin^4 (x) +3cos^2 (x) sin^2 (x) - 3cos^2 (x) sin^2 (x)= cos^4 (x) + 2cos^2 (x) sin^2 (x) + sin^4 (x) - 3cos^2 (x) sin^2 (x)= [cos^2 (x) + sin^2 (x)]^2 - 3cos^2 (x) sin^2 (x)= 1 - 3cos^2 (x) sin^2 (x) as cos^2 (x) + sin^2 (x) = 1
How do you factor sin(3x)?
Simply Googleing sin(3x) yields a bazillion answers; you should learn how to use Google.Here's one that I could copy&paste (ie: w/o rewriting):Given sin (s + t) = sin(s)cos(t) + cos(s)sin(t) => sin(3x) = sin(x + 2x) => sin(3x) = sin(x)cos(2x) + cos(x)sin(2x) given cos(2t) = 1 - 2sin²(t) => sin(3x) = sin(x)( 1 - 2sin²(x) ) + cos(x)sin(2x) => sin(3x) = sin(x) - 2sin³(x) + cos(x)sin(2x) given sin(2t) = 2sin(t)cos(t) => sin(3x) = sin(x) - 2sin³(x) + cos(x)2sin(x)cos(x) => sin(3x) = sin(x) - 2sin³(x) + 2sin(x)cos²(x) given sin²(t) + cos²(t) = 1 => cos²(t) = 1 - sin²(t) => sin(3x) = sin(x) - 2sin³(x) + 2sin(x)(1 - sin²(x)) => sin(3x) = sin(x) - 2sin³(x) + 2sin(x) - 2sin³(x) => sin(3x) = 3sin(x) - 4sin³(x)Source(s):http://www.clarku.edu/~djoyce/trig/ident...(from How do you prove sin3x=3sinx-4sin³x?)