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Prove that cos4x=8cos^4x-8cos^2x+1 so I'm
Prove that
cos4x=8cos^4x-8cos^2x+1 so I'm guessing you begin with the left side cos4x=cos2(2x) then im kinda lost
2 answers
asked by
Natash
1,292 views
Prove:
cos4x = 8cos^4x - 8cos^2x + 1 My Attempt: RS: = 4cos^2x (2cos^2x - 1) + 1 = 4 cos^2x (cos2x) + 1 LS: = cos2(2x) =
38 answers
asked by
Anonymous
2,751 views
Prove that it is an identity: cos (4x)=8cos^4(x)-8cos^2(x)+1
1 answer
asked by
Anonymus
2,035 views
Show that cos 4 alpha =8cos^4alpha - 8cos^2alpha + 1
1 answer
asked by
Joshua
135 views
Find all solutions on the interval [0.2pi)
A) -3sin(t)=15cos(t)sin(t) I have no clue... b) 8cos^2(t)=3-2cos(t) All i did was move
1 answer
asked by
Dan
844 views
If x=2cos (z) and y= sin (z), then d^2y/dx^2=?
What I got -sin (z) / -8cos (z)*sin (z) simplifies to 1/8cos (z) I don't think I'm
1 answer
asked by
Liam
453 views
Does anybody know how I can graph these on wolfram? x=cos(8cos(t))sin(t),y=sin(8cos(t))sin(t),z=cos(t).I type as it on
3 answers
asked by
Anonymous
450 views
Show that Cos 4¤=8cos^4¤ - 8cos^2 + 1
1 answer
asked by
Azeez
95 views
Why does 1/8cos(2y+6) become 1/8cos(arcsinx/4)?
1 answer
asked by
Aleks
443 views
Prove the identity.
6cos^2(x)-1 -12cos^4(x) +8cos^6(x)= cos^3(2x) Thanks for your help :)
3 answers
asked by
Tommy
609 views