given the identity sin(x+y)=sinx cosy + siny cosx
sin2x = 2 sinx cosx and
sin(2(x)+x) = sin 2x cos x + sinx cos 2x
using the last two identities gives
sin3x= 2 sinx cosx cosx + sinx cos2x
factoring the sinx we have
sin3x = sinx(2cosx cosx+cos2x)
which satisfies the requirement.
However, we can simplify further since cos 2x = cosx cosx - sinx sinx (a well known identity)
sin3x = sinx (2cosx cosx +cosx cosx - sinx sinx)
so sin3x= sinx(3cosx cosx - sinx sinx)
or sin 3x = 3.cos²x.sinx - sin³x
* * * * *
Good, but not good enough. The answer was required in terms of sin, not a mixture of sinx and cosx. Easily recitified, though, since cos²x = 1 - sin²x
Therefore sin3x = 3*(1-sin²x)*sinx - sin³x
= 3sinx - 4sin³x
The derivative of y = sin(3x + 5) is 3cos(3x + 5) but only if x is measured in radians.
Yes, 3x and x are like terms because they have the same variable raised to the same power.
The expression "x minus 4x" can be simplified by combining like terms. When subtracting 4x from x, we are essentially subtracting 4 of the x terms from 1 of the x terms, resulting in a final answer of -3x. So, x minus 4x simplifies to -3x.
Multiply each term of the second expression by each of the terms in the first expression and add them together, collecting like terms together: (x - 6)(3x² + 10x - 1) = x(3x² + 10x - 1) + (-6)(3x² + 10x - 1) = (3x³ + 10x² - x) + (-18x² - 60x + 6) = 3x³ + 10x² - 18x² - x - 60x + 6 = 3x² + (10 - 18)x² + (-1 - 60)x + 6 = 3x² - 8x² - 61x + 6
3x**2+x-10 Multiply first and last terms 3*-10=-30 Find factors of -30 that add up to second term 6+-5=1 rewrite 3x**2-5x+6x-10 Factor pairs of terms 3x**2-5x+6x-10 = x(3x-5)+2(3x-5) Factor x(3x-5)+2(x-5) = (x+2)(3x-5)
Using Euler's Formula, you use (cos(x) + i sin(x))^n = cos (nx) + i sin(nx) Now you let n=3 (cos(x) + i sin (x))3 = cos(3x) + i sin (3x) (cos(x))3 + 3(cos(x))2 * i sin(x) + 3cos(x) * i2 (sin(x))3 = cos(3x)+ i sin(3x) (cos(x))3 + i(3sin(x)(cos (x))2) - 3cos(x)(sin(x)2) - i(sin(x))3 = cos (3x) + i sin(3x) Now only use the terms with i in them to figure out what sin(3x) is... 3sin(x)(cos(x))2 - (sin(x))3 = sin(3x) Hope this helps! :D
I think you mean to solve:(-2 cos23x) - (3 sin 3x) = 0cos2 x + sin2 x = 1⇒ cos2x = 1 - sin2 x⇒ -2 cos2 3x - 3 sin 3x = -2(1 - sin2 3x) - 3 sin 3x = 0⇒ 2 sin2 3x - 3 sin 3x - 2 = 0⇒ (2 sin 3x + 1)(sin 3x - 2) = 0⇒ sin 3x = -1/2 or 2sin 3x = 2 is impossible as the range of sine is -1 ≤ sine ≤ 1Thus:sin 3x = -1/2⇒ 3x = 2nπ - π/6 or 2nπ - 5π/6⇒ x = 2/3nπ -π/18 or 2/3nπ -5π/18
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
y= sin 3x
The derivative of y = sin(3x + 5) is 3cos(3x + 5) but only if x is measured in radians.
3x + x = 4x
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.
Yes, 3x and x are like terms because they have the same variable raised to the same power.
sin(3x) = 3sin(x) - 4sin^3(x)
The expression "x minus 4x" can be simplified by combining like terms. When subtracting 4x from x, we are essentially subtracting 4 of the x terms from 1 of the x terms, resulting in a final answer of -3x. So, x minus 4x simplifies to -3x.
sin7x-sin6x+sin5x
0. sin 2x = cos 3x 1. sin 2x = sin (pi/2 - 3x) [because cos u = sin (pi/2 - u)] 2. [...]