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How to simplify sin x2 cos x2?
If you mean: sin2(x) cos2(x) then it can be simplified by noting that the square of the sine of x is equal to (1 - cos(2x)) ÷ 2 and the square of the cosine of x is equal to (1 + cos(2x)) ÷ 2. We can then simplify further: sin(x)2cos(x)2 = [(1 - cos(2x)) / 2][(1 + cos(2x)) / 2] = (1 - cos(2x))(1 + cos(2x)) / 2 = (1 - cos2(2x)) / 2 Also note that 1 - cos2(x) = sin2(x), so we can then say: = sin2(2x) / 2
What is the integral of cos raised to 2 times x?
∫cos2(x).dxUse the identity cos2(x) = (1/2)(1+cos(2x))∫(1/2)(1+cos(2x))dxPull out constant:(1/2)∫(1+cos(2x))dxIntegrate:(1/2)(x + sin(2x)/2) + CSimplify:x/2 + sin(2x)/4 + CThe identity sin(2x) = 2sin(x)cos(x) can be used to rewrite it as(x + sin(x)cos(x))/2 + C
How do you solve the trigonomic equation sin2x equals negative cosx?
Sin(2x) = -cos(x)But sin(2x) = 2 sin(x) cos(x)Substitute it:2 sin(x) cos(x) = -cos(x)Divide each side by cos(x):2 sin(x) = -1sin(x) = -1/2x = 210°x = 330°
How would you solve the integral of 1 plus tan2x plus tan squared 2x?
Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C
What is (1 cos x)(1- cos x)?
(1+cosx)(1-cosx)= 1 +cosx - cosx -cos^2x (where ^2 means squared) = 1-cos^2x = sin^2x (sin squared x)
Find the derivative of y x2 sin x 2xcos x - 2sin x?
y = (x^2)(sin x)(2x)(cos x) - 2sin xy' = [[(x^2)(sin x)][(2x)(cos x)]]' - (2sin x)'y' = [[(x^2)(sin x)]'[(2x)(cos x)] + [(2x)(cos x)]'[(x^2)(sin x)]]- (2sin x)'y' = [[(x^2)'(sin x) + (sin x)'(x^2)][(2x)(cos x)] + [(2x)'(cos x) + (cos x)'(2x)][(x^2)(sin x)] ] - 2(cos x)y' = [[(2x)(sin x )+ (cos x)(x^2)][(2x)(cos x)] + [2cos x - (sin x)(2x)][(x^2)(sin x)]] - 2(cos x)y' = (4x^2)(sin x cos x) + (2x^3)(cos x)^2 + (2x^2)(sin x cos x) - (2x^3)(sin x)^2 - 2cos xy' = (6x^2)(sin x cos x) + (2x^3)(cos x)^2 - (2x^3)(sin x)^2 - 2cos x (if you want, you can stop here, or you can continue)y' = (3x^2)(2sin x cos x) + (2x^3)[(cos x)^2 - (sin x)^2] - 2cos xy' = (3x^2)(sin 2x) + (2x^3)(cos 2x) - 2 cos xy' = (2x^3)(cos 2x) + (3x^2)(sin 2x) - 2 cos x
How do you prove tan x plus tan x sec 2x equals tan 2x?
tan x + (tan x)(sec 2x) = tan 2x work dependently on the left sidetan x + (tan x)(sec 2x); factor out tan x= tan x(1 + sec 2x); sec 2x = 1/cos 2x= tan x(1 + 1/cos 2x); LCD = cos 2x= tan x[cos 2x + 1)/cos 2x]; tan x = sin x/cos x and cos 2x = 1 - 2 sin2 x= (sin x/cos x)[(1 - 2sin2 x + 1)/cos 2x]= (sin x/cos x)[2(1 - sin2 x)/cos 2x]; 1 - sin2 x = cos2 x= (sin x/cos x)[2cos2 x)/cos 2x]; simplify cos x= (2sin x cos x)/cos 2x; 2 sinx cos x = sin 2x= sin 2x/cos 2x= tan 2x
What is Cos 2 x divided by cos x?
cos 2x = cos2 x - sin2 x = 2 cos2 x - 1; whence, cos 2x / cos x = 2 cos x - (1 / cos x) = 2 cos x - sec x.
How to simplify sin x2 cos x2?
If you mean: sin2(x) cos2(x) then it can be simplified by noting that the square of the sine of x is equal to (1 - cos(2x)) ÷ 2 and the square of the cosine of x is equal to (1 + cos(2x)) ÷ 2. We can then simplify further: sin(x)2cos(x)2 = [(1 - cos(2x)) / 2][(1 + cos(2x)) / 2] = (1 - cos(2x))(1 + cos(2x)) / 2 = (1 - cos2(2x)) / 2 Also note that 1 - cos2(x) = sin2(x), so we can then say: = sin2(2x) / 2
What is the integral of cos raised to 2 times x?
∫cos2(x).dxUse the identity cos2(x) = (1/2)(1+cos(2x))∫(1/2)(1+cos(2x))dxPull out constant:(1/2)∫(1+cos(2x))dxIntegrate:(1/2)(x + sin(2x)/2) + CSimplify:x/2 + sin(2x)/4 + CThe identity sin(2x) = 2sin(x)cos(x) can be used to rewrite it as(x + sin(x)cos(x))/2 + C
Integrate sin x cos 2x?
By Angle-Addition, cos(2x) = 2cos(x)^2-1 So, sin(x)cos(2x) = [2cos(x)^2-1]sin(x) = 2sin(x)cos(x)^2 - sin(x) Int[2sin(x)cos(x)^2 - sin(x)] = (-2/3)cos(x)^3 + cos(x) +K
How do you differentiate cos squared x?
use the double angle formula for cos(2x) which is: cos(2x)=2cos^2(x)-1 by this relation cos^2(x)=(cos(2x)+1)/2 now we'd integrate this instead this will give sin(2x)/4+x/2 =) hope this helps
What is (1 plus cos x)(1- cos x)?
(1+cosx)(1-cosx)= 1 +cosx - cosx -cos^2x (where ^2 means squared) = 1-cos^2x = sin^2x (sin squared x)
How do you solve the trigonomic equation sin2x equals negative cosx?
Sin(2x) = -cos(x)But sin(2x) = 2 sin(x) cos(x)Substitute it:2 sin(x) cos(x) = -cos(x)Divide each side by cos(x):2 sin(x) = -1sin(x) = -1/2x = 210°x = 330°
Integral of cosine squared x?
Integral of cos^2x=(1/2)(cosxsinx+x)+CHere is why:Here is one method: use integration by parts and let u=cosx and dv=cosxdxdu=-sinx v=sinxInt(udv)=uv-Int(vdu) so uv=cosx(sinx) and vdu=sinx(-sinx)so we have:Int(cos^2(x)=(cosx)(sinx)+Int(sin^2x)(the (-) became + because of the -sinx, so we add Int(vdu))Now it looks not better because we have sin^2x instead of cos^2x,but sin^2x=1-cos^2x since sin^2x+cos^2x=1So we haveInt(cos^2x)=cosxsinx+Int(1-cos^2x)=cosxsinx+Int(1)-Int(cos^2x)So now add the -Int(cos^2x) on the RHS to the one on the LHS2Int(cos^2x)=cosxsinx+xso Int(cos^2x)=1/2[cosxsinsx+x] and now add the constant!final answerIntegral of cos^2x=(1/2)(cosx sinx + x)+C = x/2 + (1/4)sin 2x + C(because sin x cos x = (1/2)sin 2x)Another method is:Use the half-angle identity, (cos x)^2 = (1/2)(1 + cos 2x). So we have:Int[(cos x)^2 dx] = Int[(1/2)(1 + cos 2x)] dx = (1/2)[[Int(1 dx)] + [Int(cos 2x dx)]]= (1/2)[x + (1/2)sin 2x] + C= x/2 +(1/4)sin 2x + C
How would you solve the integral of 1 plus tan2x plus tan squared 2x?
Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C
What is (1 cos x)(1- cos x)?
(1+cosx)(1-cosx)= 1 +cosx - cosx -cos^2x (where ^2 means squared) = 1-cos^2x = sin^2x (sin squared x)
