Respuesta :
[tex]\bf cos(6\theta)\implies cos(3\theta+3\theta)\implies cos[2(3\theta)]
\\\\\\
now\implies
\begin{cases}
cos(3x)=4cos^3(x)-3cos(x)\\\\
cos(2x)=2cos^2(x)-1
\end{cases}
\\\\\\
cos[2(3\theta)]\implies 2\left[ 4cos^3(\theta)-3cos(\theta) \right]^2-1[/tex]
so.. just expand the binomial, and simplify if any
[tex]\bf sin(6\theta)\implies sin(3\theta+3\theta)\implies sin[2(3\theta)] \\\\\\ now\implies \begin{cases} sin(2x)=2cos(x)sin(x)\\\\ sin(3x)=3sin(x)-4sin^3(x) \end{cases} \\\\\\ sin[2(3\theta)]\implies 2cos(3\theta)sin(3\theta) \\\\\\ 2\ [4cos^3(\theta)-3cos(\theta)]\ [3sin(\theta)-4sin^3(\theta)][/tex]
multiply and distribute and simplify if any
so.. just expand the binomial, and simplify if any
[tex]\bf sin(6\theta)\implies sin(3\theta+3\theta)\implies sin[2(3\theta)] \\\\\\ now\implies \begin{cases} sin(2x)=2cos(x)sin(x)\\\\ sin(3x)=3sin(x)-4sin^3(x) \end{cases} \\\\\\ sin[2(3\theta)]\implies 2cos(3\theta)sin(3\theta) \\\\\\ 2\ [4cos^3(\theta)-3cos(\theta)]\ [3sin(\theta)-4sin^3(\theta)][/tex]
multiply and distribute and simplify if any
The expansion for [tex]\rm cos\; 6\theta[/tex] and [tex]\rm sin\; 6\theta[/tex] is given below and this can be evaluated by using the trigonometric properties.
Given :
Trigonometric Functions -- [tex]\rm cos\; 6\theta[/tex] and [tex]\rm sin\; 6\theta[/tex]
The following steps can be used in order to determine the expansion for [tex]\rm cos\; 6\theta[/tex] and [tex]\rm sin\; 6\theta[/tex]:
Step 1 - The trigonometric properties can be used in order to determine the expansion for [tex]\rm cos\; 6\theta[/tex] and [tex]\rm sin\; 6\theta[/tex].
Step 2 - The properties are used is mentioned below:
[tex]\rm cos\;2\theta = 2cos^2\theta - 1[/tex]
[tex]\rm cos\; 3\theta=4cos^3\theta - 3cos\theta[/tex]
[tex]\rm sin\;2\theta = 2\;sin\;\theta \;cos\;\theta[/tex]
[tex]\rm sin\; 3\theta=3sin\theta-4sin^3\theta[/tex]
Step 3 - The expansion of [tex]\rm cos\; 6\theta[/tex] is evaluated below:
[tex]\rm cos\; 6\theta = cos 2(3\theta)[/tex]
[tex]\rm cos\; 6\theta = 2(4cos^3\theta-3cos\theta)^2-1[/tex]
[tex]\rm cos\; 6\theta = 2(16cos^6\theta+9cos^2\theta-24cos^4\theta)-1[/tex]
[tex]\rm cos\; 6\theta = 32cos^6\theta+18cos^2\theta-48cos^4\theta-1[/tex]
Step 4 - The expansion of [tex]\rm sin\; 6\theta[/tex] is evaluated below:
[tex]\rm sin\; 6\theta = sin2(3\theta)[/tex]
[tex]\rm sin\; 6\theta = 2\;sin3\theta \; cos3\theta[/tex]
[tex]\rm sin\; 6\theta = 2(4cos^3\theta-3cos\theta)(3sin\theta -4sin^3\theta)[/tex]
[tex]\rm sin\; 6\theta = 2(12cos^3\theta sin\theta-16cos^3\theta sin^3\theta-9cos\theta sin\theta+12cos\theta sin^3\theta)[/tex]
For more information, refer to the link given below:
https://brainly.com/question/13710437
