There exist constants a, b, c, and d such that(sin x)^7 = asin 7x + bsin 5x + csin 3x + dsin x for all angles x. Find d.
I tried rewriting them in the exponential form, but it becomes ugly as there are many negative cosines and divisions of i. Help?
http://www.wolframalpha.com/input/?i=%28sinx%29%5E7
will show the answer you need.
sin^7 = sin(1-cos^2)^3
Now you can start expanding and using the sum formulas.
I'm sure a google search will show up all the gory details.
Or, you can go the other way
sin(7x) = sin(6x)cos(x) + cos(6x)sin(x)
and start expanding those as well, till all you have left is a bunch of sin(kx).
To find the constant d, we can compare the coefficients of sin x on both sides of the equation.
On the right-hand side of the equation, the term dsin x only appears once. Let's focus on finding the coefficient of sin x on the left-hand side of the equation, which corresponds to (sin x)^7.
Using the binomial theorem, we can expand (sin x)^7 as follows:
(sin x)^7 = (sin x)^6 * sin x
= ((sin x)^2)^3 * sin x
= (1 - cos^2(x))^3 * sin x
Expanding further using the binomial theorem:
= (1 - 3cos^2(x) + 3cos^4(x) - cos^6(x)) * sin x
Notice that the coefficient of sin x in this expansion is 1 - 3cos^2(x) + 3cos^4(x) - cos^6(x).
Now let's compare this to the right-hand side of the equation:
asin 7x + bsin 5x + csin 3x + dsin x
Since we are only interested in the coefficient of sin x, let's write the terms in terms of sin x using the identity sin^2(x) + cos^2(x) = 1:
asin 7x + bsin 5x + csin 3x + dsin x
= asin(7x) + bsin(5x) + csin(3x) + dsin(x)
Comparing the coefficient of sin x, we get:
1 - 3cos^2(x) + 3cos^4(x) - cos^6(x) = d
Therefore, the constant d is equal to 1 - 3cos^2(x) + 3cos^4(x) - cos^6(x).
To find the constant d in the equation (sin x)^7 = asin 7x + bsin 5x + csin 3x + dsin x, we can use the fact that for any angle x, sin(7x), sin(5x), sin(3x), and sin(x) can be expressed as follows:
sin(7x) = Im(e^(7ix))
sin(5x) = Im(e^(5ix))
sin(3x) = Im(e^(3ix))
sin(x) = Im(e^(ix))
Let's rewrite the given equation using these expressions:
(sin x)^7 = asin(7x) + bsin(5x) + csin(3x) + dsin(x)
(sin x)^7 = aIm(e^(7ix)) + bIm(e^(5ix)) + cIm(e^(3ix)) + dIm(e^(ix))
To simplify further, let's use Euler's formula, which states that e^(ix) = cos(x) + isin(x):
(sin x)^7 = aIm((cos(7x) + isin(7x))) + bIm((cos(5x) + isin(5x))) + cIm((cos(3x) + isin(3x))) + dIm((cos(x) + isin(x)))
Now, we will distribute the imaginary part (Im) to each term:
(sin x)^7 = a(cos(7x)sin(x) + i(sin(7x)cos(x))) + b(cos(5x)sin(x) + i(sin(5x)cos(x))) + c(cos(3x)sin(x) + i(sin(3x)cos(x))) + d(cos(x)sin(x) + i(sin(x)cos(x)))
Next, we will combine the real and imaginary parts separately:
(sin x)^7 = (a(cos(7x)sin(x) + b(cos(5x)sin(x) + c(cos(3x)sin(x) + d(cos(x)sin(x))) + i(a(sin(7x)cos(x) + b(sin(5x)cos(x) + c(sin(3x)cos(x) + d(sin(x)cos(x)))
Now, let's equate the real and imaginary parts of the equation to find the values of a, b, c, and d:
Real part:
(sin x)^7 = acos(7x)sin(x) + bcos(5x)sin(x) + ccos(3x)sin(x) + dcos(x)sin(x)
Imaginary part:
0 = asin(7x)cos(x) + bsin(5x)cos(x) + csin(3x)cos(x) + dsin(x)cos(x)
Since the equation holds for all values of x, both the real and imaginary parts must be equal to zero. Let's focus on the Imaginary part and find the constant d.
0 = asin(7x)cos(x) + bsin(5x)cos(x) + csin(3x)cos(x) + dsin(x)cos(x)
Now, we will separate the terms involving d and rearrange the equation:
0 = d(sin(x)cos(x)) + (asin(7x)cos(x) + bsin(5x)cos(x) + csin(3x)cos(x))
Since sin(2x) = 2sin(x)cos(x), we can rewrite the first term as:
0 = d(1/2)sin(2x) + (asin(7x)cos(x) + bsin(5x)cos(x) + csin(3x)cos(x))
Now, we can rewrite each trigonometric product as the sum:
0 = (d/2)sin(2x) + (1/2)(asin(7x) + bsin(5x) + csin(3x))cos(x)
Since this equation holds for all values of x, the coefficients of sin(2x) and cos(x) must be zero. Therefore:
Coefficient of sin(2x) = (d/2) = 0
Coefficient of cos(x) = (1/2)(asin(7x) + bsin(5x) + csin(3x)) = 0
From the first equation, we have d/2 = 0, which means d = 0.
Therefore, the constant d in the given equation is 0.