0
YES!!!!
Sin(2x) = Sin(x+x')
Sin(x+x') = SinxCosx' + CosxSinx'
I have put a 'dash' on an 'x' only to show its position in the identity. Both x & x' carry the same value.
Hence
SinxCosx' + CosxSinx' =
Sinx Cos x + Sinx'Cosx =>
2SinxCosx
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How do you derive sin2x?
If you are refering to the double-angle formula for sin(x), the best way is to use what is known as Euler's identity. Euler's identity is eix = cos(x) + i*sin(x) where x is any real angle in radians, e is Euler's constant 2.71828182845... and i is the imaginary number: SQRT(-1). Assuming that is true, then ei(2x) = cos(2x) + i*sin(2x) and that is the same as saying (eix)2= cos(2x) + i*sin(2x) and substituting from the original equation: (cos(x) + i*sin(x))2 = cos(2x) + i*sin(2x). By distribution, remembering that i2 = -1, we get cos2(x) + i*2*sin(x)*cos(x) - sin2(x) = cos(2x) + i*sin(2x). Now we can separate the equation into its real and imaginary parts. cos2(x) - sin2(x) = cos(2x) and i*2*sin(x)*cos(x) = i*sin(2x), and after i cancels, there's our good old double angle formula. If derive refers to derivative, then use the chain rule. d(sin(2x))/dx=2cos(2x)
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.
How do you confirm this trig identity?
sin^5 2x = 1/8 sin2x (cos(8x) - 4 cos(4x)+3)
How do you prove that 2 sin 3x divided by sin x plus 2 cos 3x divided by cos x equals 8 cos 2x?
You need to know the trigonometric formulae for sin and cos of compound angles. sin(x+y) = sin(x)*cos(y)+cos(x)*sin(y) and cos(x+y) = cos(x)*cos(y) - sin(x)*sin(y) Using these, y = x implies that sin(2x) = sin(x+x) = 2*sin(x)cos(x) and cos(2x) = cos(x+x) = cos^2(x) - sin^2(x) Next, the triple angle formulae are: sin(3x) = sin(2x + x) = 3*sin(x) - 4*sin^3(x) and cos(3x) = 4*cos^3(x) - 3*cos(x) Then the left hand side = 2*[3*sin(x) - 4*sin^3(x)]/sin(x) + 2*[4*cos^3(x) - 3*cos(x)]/cos(x) = 6 - 8*sin^2(x) + 8cos^2(x) - 6 = 8*[cos^2(x) - sin^2(x)] = 8*cos(2x) = right hand side.
How do you solve sin 2x equals sin x over 2?
they do have calculators for these questions you knowsin 2x = (sin x)/22 sin x cos x - (1/2)sin x = 02 sin x(cos x - 1/4) = 02 sin x = 0 or cos x - 1/4 = 0sin x = 0 or cos x = 1/4in the interval [0, 360)sin x = 0, when x = 0, 180cos x = 1/4, when x = 75.52, 284.48Check:
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
If fx equals cossinx2 then f prime equals?
f(x)=cos(sin(x2)) [u(v)]' = u'(v) * v' so f'(x) = cos'(sinx(x2)) * sin'(x2) * (x2)' f'(x) = -sin(sin(x2)) * cos(x2) * 2x = -2x sin(sin(x2)) cos(x2)
How do you derive sin2x?
If you are refering to the double-angle formula for sin(x), the best way is to use what is known as Euler's identity. Euler's identity is eix = cos(x) + i*sin(x) where x is any real angle in radians, e is Euler's constant 2.71828182845... and i is the imaginary number: SQRT(-1). Assuming that is true, then ei(2x) = cos(2x) + i*sin(2x) and that is the same as saying (eix)2= cos(2x) + i*sin(2x) and substituting from the original equation: (cos(x) + i*sin(x))2 = cos(2x) + i*sin(2x). By distribution, remembering that i2 = -1, we get cos2(x) + i*2*sin(x)*cos(x) - sin2(x) = cos(2x) + i*sin(2x). Now we can separate the equation into its real and imaginary parts. cos2(x) - sin2(x) = cos(2x) and i*2*sin(x)*cos(x) = i*sin(2x), and after i cancels, there's our good old double angle formula. If derive refers to derivative, then use the chain rule. d(sin(2x))/dx=2cos(2x)
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.
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°
Sin2x equals cos3x find x?
0. sin 2x = cos 3x 1. sin 2x = sin (pi/2 - 3x) [because cos u = sin (pi/2 - u)] 2. [...]
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
Integral of sin squared x?
First we look at the double-angle identity of cos2x. We know that: cos2x = cos^2x - sin^2x cos2x = [1-sin^2x] - sin^2x.............. (From sin^2x + cos^2x = 1, cos^2x = 1 - sin^2x) Therefore: cos2x = 1 - 2sin^2x 2sin^2x = 1 - cos2x sin^2x = 1/2(1-cos2x) sin^2x = 1/2 - cos2x/2 And intergrating, we get: x/2 - sin2x/4 + c...................(Integral of cos2x = 1/2sin2x; and c is a constant)
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 confirm this trig identity?
sin^5 2x = 1/8 sin2x (cos(8x) - 4 cos(4x)+3)
What is the equation of the line tangent to the graph y equals sin 2x?
That line is [ y = 2 cos(2x) ].
How do you prove that 2 sin 3x divided by sin x plus 2 cos 3x divided by cos x equals 8 cos 2x?
You need to know the trigonometric formulae for sin and cos of compound angles. sin(x+y) = sin(x)*cos(y)+cos(x)*sin(y) and cos(x+y) = cos(x)*cos(y) - sin(x)*sin(y) Using these, y = x implies that sin(2x) = sin(x+x) = 2*sin(x)cos(x) and cos(2x) = cos(x+x) = cos^2(x) - sin^2(x) Next, the triple angle formulae are: sin(3x) = sin(2x + x) = 3*sin(x) - 4*sin^3(x) and cos(3x) = 4*cos^3(x) - 3*cos(x) Then the left hand side = 2*[3*sin(x) - 4*sin^3(x)]/sin(x) + 2*[4*cos^3(x) - 3*cos(x)]/cos(x) = 6 - 8*sin^2(x) + 8cos^2(x) - 6 = 8*[cos^2(x) - sin^2(x)] = 8*cos(2x) = right hand side.
