Find the coordinates of the expression

[Logan Land]

Find the coordinates of the expression (cos x + sin x)^3 in the basis {1, sin x, cos x, sin 2x, cos 2x, sin3x, cos3x}.

I don't understand where to start since I am dealing with cos and sin now.

 

[topsquark]

Find the coordinates of the expression (cos x + sin x)^3 in the basis {1, sin x, cos x, sin 2x, cos 2x, sin3x, cos3x}.

I don't understand where to start since I am dealing with cos and sin now.

Okay, I'm going to write this out the easy way because I know darned well that anyone doing this problem is not going to ignore your calculator's ability to do integrations.

Given a (discrete) basis B, and calling the basis elements [math]b_n(x)[/math] we can say the following:
[math]( cos(x) + sin(x) )^3 = \sum_{n = 1}^{dim(B)} \frac{1}{\pi} \cdot b_n(x) \int_{0}^{2 \pi} (cos(x) + sin(x))^3 \cdot b_n(x)~dx[/math]

Find the integrals for each basis element. This is why you want the calculator...it's a messy time consuming task. You can guess at some of the coefficients, but it really does take a while.

Can you finish from here?
(Ans. [math](cos(x) + sin(x))^3 = \frac{3}{2}sin(x) + \frac{1}{2} sin(3x) + \frac{3}{2} cos(x) - \frac{1}{2} cos(3x)[/math])

-Dan

 

[Euge]

Find the coordinates of the expression (cos x + sin x)^3 in the basis {1, sin x, cos x, sin 2x, cos 2x, sin3x, cos3x}.

I don't understand where to start since I am dealing with cos and sin now.

Here's what I suggest. Using the sine addition formula, we can write

$$\cos x + \sin x = \sqrt{2}\left(\cos x \cdot \frac{1}{\sqrt{2}} + \sin x \cdot \frac{1}{\sqrt{2}}\right) = \sqrt{2}\sin\Bigl(x + \frac{\pi}{4}\Bigr).$$

Thus

$$(\cos x + \sin x)^3 = 2^{3/2}\sin^3\Bigl(x + \frac{\pi}{4}\Bigr).$$

Using the power-reduction formula

$$\sin^3 u = \frac{3}{4}\sin u - \frac{1}{4}\sin 3u,$$

we find

$$(\cos x + \sin x)^3 $$

$$= 2^{3/2}\left\{\frac{3}{4}\sin\Bigl(x + \frac{\pi}{4}\Bigr) - \frac{1}{4}\sin\Bigl(3x + \frac{3\pi}{4}\Bigr)\right\}$$

$$= \frac{1}{2}\left\{3\sqrt{2}\sin\Bigl(x + \frac{\pi}{4}\Bigr) - \sqrt{2}\sin\Bigl(3x + \frac{3\pi}{4}\Bigr)\right\}$$

$$ = \frac{1}{2}[3(\sin x + \cos x) - (-\sin 3x + \cos 3x)]$$

$$ = \frac{1}{2}(3\sin x + 3\cos x + \sin 3x - \cos 3x)$$

$$ = \frac{3}{2}\sin x + \frac{3}{2}\cos x + \frac{1}{2}\sin 3x - \frac{1}{2}\cos 3x.$$

 

[Logan Land]

Here's what I suggest. Using the sine addition formula, we can write

$$\cos x + \sin x = \sqrt{2}\left(\cos x \cdot \frac{1}{\sqrt{2}} + \sin x \cdot \frac{1}{2}\right) = \sqrt{2}\sin\Bigl(x + \frac{\pi}{4}\Bigr).$$

Thus

$$(\cos x + \sin x)^3 = 2^{3/2}\sin^3\Bigl(x + \frac{\pi}{4}\Bigr).$$

Using the power-reduction formula

$$\sin^3 u = \frac{3}{4}\sin u - \frac{1}{4}\sin 3u,$$

we find

$$(\cos x + \sin x)^3 $$

$$= 2^{3/2}\left\{\frac{3}{4}\sin\Bigl(x + \frac{\pi}{4}\Bigr) - \frac{1}{4}\sin\Bigl(3x + \frac{3\pi}{4}\Bigr)\right\}$$

$$= \frac{1}{2}\left\{3\sqrt{2}\sin\Bigl(x + \frac{\pi}{4}\Bigr) - \sqrt{2}\sin\Bigl(3x + \frac{3\pi}{4}\Bigr)\right\}$$

$$ = \frac{1}{2}[3(\sin x + \cos x) - (-\sin 3x + \cos 3x)]$$

$$ = \frac{1}{2}(3\sin x + 3\cos x + \sin 3x - \cos 3x)$$

$$ = \frac{3}{2}\sin x + \frac{3}{2}\cos x + \frac{1}{2}\sin 3x - \frac{1}{2}\cos 3x.$$

so this problem like the other one you helped me with we wouldn't want to use some sort of matrix form to solve? I see how you came to your conclusion but every time in class our teacher gives some sort of way to solve using matrices. She just never gave us an example with cos and sin.

Thanks so much for your help!

 

[Euge]

so this problem like the other one you helped me with we wouldn't want to use some sort of matrix form to solve? I see how you came to your conclusion but every time in class our teacher gives some sort of way to solve using matrices. She just never gave us an example with cos and sin.

Thanks so much for your help!

Even if you try to use a matrix form, you will still need power-reductions.