Question

If cos(x)cos(pi/7)+sin(x)sin(pi/7)=-V2/2

Answer 1

1. ∫ cos³(πx) dx
2. ∫ cos²(πx) cos(πx) dx
3. ∫(1 - sin²(πx)) cos(πx) dx
4. ∫ cos(πx) dx - ∫ sin²(πx) cos(πx) dx

Set u=πx, du=π dx
Set v=sin(u), dv=cos(u) du

5. 1 ⁄ π ∫ cos(u) du - 1 ⁄ π ∫ v² dv
6. 1 ⁄ π [ ∫ cos(u) du - ∫ v² dv ]
7. 1 ⁄ π [ sin(u) - v³ ⁄ 3 ] + C
8. 1 ⁄ π [ sin(πx) - sin³(πx) ⁄ 3 ] + C
9. sin(πx) ⁄ π - sin³(πx) ⁄ (3π) + C

One day you will know Euler's as the most versatile tidbit of knowledge on the planet.

10. ∫ cos³(πx) dx
11. ∫ [(e^(iπx) + e^(-iπx)) ⁄ 2]³ dx
12. ⅛ ∫ [e^(iπx) + e^(-iπx)]³ dx
13. ⅛ ∫ [e^(3iπx) + e^(-3iπx) + 3e^(iπx) + 3e^(-iπx)] dx
14. ⅛ [∫ e^(3iπx) dx + ∫ e^(-3iπx) dx + 3∫ e^(iπx) dx + 3∫ e^(-iπx) dx]
15. ⅛ [e^(3iπx) ⁄ (3iπ) + e^(-3iπx) ⁄ (-3iπ) + 3e^(iπx) ⁄ (iπ) + 3e^(-iπx) ⁄ (-iπ)] + C
16. ⅛ [(1 ⁄ (3iπ))(e^(3iπx) - e^(-3iπx)) + (3 ⁄ (iπ))(e^(iπx) - e^(-iπx))] + C
17. ⅛ [(1 ⁄ (3iπ))(2i sin(3πx)) + (3 ⁄ (iπ))(2i sin(πx))] + C
18. ⅛ [(2 sin(3πx)) ⁄ (3π) + (6 sin(πx)) ⁄ π] + C
19. [sin(3πx) + 9 sin(πx)] ⁄ (12π) + C

Which is another way of saying the same thing.

Pick your poison. But note the link below. Wolfram's alpha sets it up with Euler's.

Answer 2

If \[\cos(x)\cos(\frac{ \pi }{ 7 })+\sin(x)\sin(\frac{ \pi }{ 7 })=-{\frac{ \sqrt{2} }{ 2 }}\]
then x can equal:(check all that apply)

Answer 3

Does that help?

Answer 4

LHS = cos (x - pi/7)

Answer 5

cos (x - pi/7) = -sqrt2/2

Answer 6

Woud you like me to put the choices?

Answer 7

ang;es whose cosine is -sqrt2/2 = 3pi/4 and 5pi/4 ( in the range 0 to 2pi)