Another trigonometry problem help

[dolina dahani]

I need help again hehe 12. Question
Solve equation:

 

[Deleted member 4993]

I need help again hehe 12. Question
Solve equation:

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING

Please share your work/thoughts about this assignment.

 

[dolina dahani]

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING

Please share your work/thoughts about this assignment.
View attachment 17398

I'm not sure what did I do wrong? Sorry I'm new but I read the before posting. I need an answer for the 12 question not all of them

 

[dolina dahani]

sin^4(x)+cos^4(x)=cos4x This was the question

 

[Deleted member 4993]

I'm not sure what did I do wrong? Sorry I'm new but I read the before posting. I need an answer for the 12 question not all of them

We understand that you need an answer.

However, to help you effectively, you need to share your work regarding this problem and tell us exactly where you are lost.

 

[dolina dahani]

We understand that you need an answer.

However, to help you effectively, you need to share your work regarding this problem and tell us exactly where you are lost.

Ooh my bad sorry sir, the problem is that I did not do much of the stuff that would be helpful. I could do some stuff with sin^4(x)+cos^4(x) but im completely lost in cos4x . Sorry again, thank you for your patience.

 

[Steven G]

Please show us what you did.

Maybe use the double angle formula twice.

sin(4x) = sin (2*(2x))

 

[Deleted member 4993]

sin^4(x)+cos^4(x)=cos4x This was the question

sin⁴(x) + cos⁴(x) = cos(4x)

sin⁴(x) + cos⁴(x) + 2 * sin²(x) * cos²(x) = cos(4x) + 2 * sin²(x) * cos²(x)

continue....

 

[pka]

sin^4(x)+cos^4(x)=cos(4x) This was the question

\(\begin{gathered}
\cos (4x) = \\ \cos [2(2x)] = \\ {\cos ^2}(2x) - {\sin ^2}(2x) = \\ {\left[ {{{\cos }^2}(x) - {{\sin }^2}(x)} \right]^2} - {\left[ {2\sin (x)\cos (x)} \right]^2} = \\ {\cos ^4}(x) - 2{\cos ^2}(x){\sin ^2}(x) + {\sin ^4}(x) - 4{\cos ^2}(x){\sin ^2}(x) = \\ {\cos ^4}(x) + {\sin ^4}(x) - 6{\cos ^2}(x){\sin ^2}(x) \\
\end{gathered} \)
This might help you?

 

[dolina dahani]

Please show us what you did.

Maybe use the double angle formula twice.

sin(4x) = sin (2*(2x))

sin⁴(x) + cos⁴(x) = cos(4x)

sin⁴(x) + cos⁴(x) + 2 * sin²(x) * cos²(x) = cos(4x) + 2 * sin²(x) * cos²(x)

continue....

\(\begin{gathered}
\cos (4x) = \\ \cos [2(2x)] = \\ {\cos ^2}(2x) - {\sin ^2}(2x) = \\ {\left[ {{{\cos }^2}(x) - {{\sin }^2}(x)} \right]^2} - {\left[ {2\sin (x)\cos (x)} \right]^2} = \\ {\cos ^4}(x) - 2{\cos ^2}(x){\sin ^2}(x) + {\sin ^4}(x) - 4{\cos ^2}(x){\sin ^2}(x) = \\ {\cos ^4}(x) + {\sin ^4}(x) - 6{\cos ^2}(x){\sin ^2}(x) \\
\end{gathered} \)
This might help you?

Thank you all for help yes i solved it,pka thanks for explanation!