Pre-CALC Help!! sin & cos arithmetic sequence

[questions]

X is an acute angle. find cos^3x-cosx 9f you know sinx sin2x sin4x is increasing arithmetic sequence?

 

[mmm4444bot]

?

Since you have not asked any questions of your own or made any statements about what you're thinking, there is no way for me to determine where you're stuck. That makes it difficult to know where to begin helping you.

Therefore, I'll just make some basic statements.

We're looking for one specific acute angle, yes? For this constant, I prefer to use the symbol C, where 0 < C < Pi/2.

Now, we have a sequence with three elements:

{sin(C), sin(2C), sin(4C)}

Somebody claims that this sequence of three numbers is arithmetic.

If so, then the differences between adjacent elements must be the same, yes?

In other words, we're looking for C such that the following is true.

sin(4C) - sin(2C) = sin(2C) - sin(C)

Solve this for C, and then evaluate cos(C)^3 - cos(C).

If you still need help, you need to find a way to be more specific.

 

[questions]

I'm not exactly sure on how to solve for c.

 

[soroban]

Hello, questions!

$\displaystyle x\text{ is an acute angle.}$

$\displaystyle \sin x,\:\sin2x,\:\sin4x\,\text{ is an increasing arithmetic sequence.}$

$\displaystyle \text{Find: }\; \cos^3x-\cos x$

$\displaystyle \text{Note: if }\sin x = 0\text{, we have the trivial sequence: }\:0,\:0,\:0$
. . . . . $\displaystyle \text{which is }not\text{ an increasing sequence.}$


$\displaystyle \text{So, we have: }\;\begin{Bmatrix} \sin x + d &=& \sin 2x & [1] \\ \sin2x + d &=& \sin4x & [2] \end{Bmatrix}$


$\displaystyle \text{From [1]: }\;d \;=\;\sin2x - \sin x \;=\;2\sin x\cos x - \sin x \;=\;\sin x(2\cos x - 1) \quad [3]$


$\displaystyle \text{From [2]: }\;d \;=\;\sin4x-\sin2x \;=\; 2\sin2x\cos2x - \sin2x \;=\;\sin2x(2\cos2x-1)$

. . . . . . . . . $\displaystyle = \;2\sin x\cos x\bigg[2(2\cos^2x-1) - 1\bigg] \;=\;2\sin x\cos x (4\cos^2x - 3)\quad [4]$


$\displaystyle \text{Equate [3] and [4]: }\;\sin x(2\cos x-1) \;=\;2\sin x\cos x(4\cos^2x - 3)$


$\displaystyle \text{Since }\sin x \neq 0,\,\text{ divide by }\sin x\!:\;\;2\cos x - 1 \;=\;2\cos x(4\cos^2x-3)$

. . $\displaystyle \text{and we have: }\;2\cos x - 1 \;=\;8\cos^3x - 6\cos x \quad\Rightarrow\quad 8\cos^3x - 8\cos x \:=\:-1$


$\displaystyle \text{Therefore: }\;\cos^3x - \cos x \;=\;-\frac{1}{8}$