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they do have calculators for these questions you know
sin 2x = (sin x)/2
2 sin x cos x - (1/2)sin x = 0
2 sin x(cos x - 1/4) = 0
2 sin x = 0 or cos x - 1/4 = 0
sin x = 0 or cos x = 1/4
in the interval [0, 360)
sin x = 0, when x = 0, 180
cos x = 1/4, when x = 75.52, 284.48
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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.
Is sin 2x equals 2 sin x cos x an identity?
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
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 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.
What does sine times cosine equal?
Sin(x) cos(x) = 1/2 of sin(2x)
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 do you solve solve for x 2x plus a equals p?
2x+a=p 2x=p-a x=.5p-.5a
Can you solve the differential equation dy over dx equals y squared over e to the 2x power?
-2y square exp power -2x-1
How do you solve 2x-5y9?
You can't solve a formula with no equals sign
Y plus 2x equals 0 solve for y?
y= -2x
Solve 2x-3 equals 9?
2x-3=9 so 2x=12 and x=6
What is the period of the equation y equals -6 sin 2x?
Pi
How can you solve -2sin2x plus 14sin2x equals 9?
-2sin(2x)+14 sin(2x)=9 12 sin(2x)=9 sin(2x)=9/12 2x= sin^-1(3/4) = sin^-1(0.75) 2x=0.84806 radian x = 0.42403 radian or 24.295 degrees The other solution is x=2.7176 radian or 155.705 degrees Since the sine function has period 2PI, the general solution may be written as x= 0.43403+ 2nPI, where n is any integer and x=2.7176 + 2nPI, where n is any integer
Solve 2x plus 3 equals 9?
3
Solve for 2x-8 equals -24?
x = -8
Solve ex equals 2e1-2x?
meant to be e^x = 2e^1-2x
What is the derivative of y equals sin times x to the power of 2?
D(y)= sin 2x
