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What is cos theta times cos theta?
Cos theta squared
Why sin integral is cos?
sin integral is -cos This is so because the derivative of cos x = -sin x
What is cos(33)?
If the angle is measured in degrees, then cos(33) = 0.8387, approx.
Does heat rise in a solid?
no cos i say so
Prove that sin A cos A2?
Not correct. sin2alpha + cos2alpha = 1
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 the range for y equals 4 cos parantheses2xparantheses?
The range for y = 4 cos (2x) is [-4, +4].Not asked, but answered for completeness sake, the domain is [-infinity, +infinity].
What is sec2x in relationship to cos?
Sec(2x) = 1/Cos(2x)
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
Integral of tan square x secant x?
convert tan^2x into sin^2x/cos^2x and secant x into 1/cos x combine terms for integral sin^2x/cos^3x dx then sub in u= cos^3x and du=-2sin^2x dx
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)
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
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
How do you solve double angle equations for trigonometry?
There are two ways to solve for the double angle formulas in trigonometry. The first is to use the angle addition formulas for sine and cosine. * sin(a + b) = sin(a)cos(b) + cos(a)sin(b) * cos(a + b) = cos(a)cos(b) - sin(a)sin(b) if a = b, then * sin(2a) = sin(a)cos(a) + cos(a)sin(a) = 2sin(a)cos(a) * cos(2a) = cos2(a) - sin2(b) The cooler way to solve for the double angle formulas is to use Euler's identity. eix = cos(x) + i*sin(x). Yes, that is "i" as in imaginary number. we we put 2x in for x, we get * e2ix = cos(2x) + i*sin(2x) This is the same as * (eix)2 = cos(2x) + i*sin(2x) We can substitute our original equation back in for eix. * (cos(x) + i*sin(x))2 = cos(2x) + i*sin(2x) We can distribute the squared term. * cos2(x) + i*sin(x)cos(x) + i*sin(x)cos(x) + (i*sin(x))2 = cos(2x) + i*sin(2x) And simplify. Because i is SQRT(-1), the i squared term becomes negative. * cos2(x) + 2i*sin(x)cos(x) - sin2(x) = cos(2x) + i*sin(2x) * cos2(x) - sin2(x) + 2i*sin(x)cos(x) = cos(2x) + i*sin(2x) Now you can plainly see both formulas in the equation arranged quite nicely. I don't yet know how to get rid of the i, but I'm working on it.
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
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.
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)
