Algebra and sqrs n sqrrt

  • Thread starter brandy
  • Start date
  • #1
brandy
161
0
my problem is this: √(4800)=√((80cos ̂)^2+(80sin ̂)^2)
pretty simple i no but i still keep making mistakes somewhere.
i can get it down to: 4800/6400=(cos ̂)^2 + (sin ̂)^2
but then i get a little stuck.
by the way ̂=thetre=unkown angle.
good luck
i need it.
 

Answers and Replies

  • #2
d_leet
1,077
1
It's false... sin2(x) + cos2(x) = 1 for all values of x and 4800/6400 is not equal to 1, thus the equation is simply false.
 
  • #3
brandy
161
0
It's false... sin2(x) + cos2(x) = 1 for all values of x and 4800/6400 is not equal to 1, thus the equation is simply false.

yea i figured that out just before. oops. mybad. but how could you rearange a similar equation so that it equalled thetre.
im not very intelligent if u haven't already noticed.
no wait scratch that. how could u rearrange the equation x=(cos thetre)^2 + (sin thetre)^2
even though my original numbers were incorrect.
 
Last edited:
  • #4
d_leet
1,077
1
yea i figured that out just before. oops. mybad. but how could you rearange a similar equation so that it equalled thetre.
im not very intelligent if u haven't already noticed.
no wait scratch that. how could u rearrange the equation x=(cos thetre)^2 + (sin thetre)^2
even though my original numbers were incorrect.

in that situation x must be 1, it cannot have any other value if that is to be a true statement.
 

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