two prolems I really need to get done fast;
1. sin2(y)= cos4(y) when 0 greater or equal y less than or equal 360 degrees
2. solve algerbraically
sin^2(x)/ 1- cos(x) = sec (x) +1/ sec (x)
help fast please, thanks.
Hello alpina
Now or , where is any integer
So you now need to investigate , and put and .
For instance: gives or
or
(or , which is out-of-range).
Then try , then , etc until you have all the values of in the range.
2 I assume that you need to prove the identity:
So:
Grandad
Do you mean sin(2y)= cos(4y) or sin^2(y)= cos^4(y)? Since you use "^" in problem 2 I assume it is the first. cos(4y)= cos(2(2y))= cos^2(2y)- sin^2(2y)= (1- sin^2(2y))- sin^2(2y)= 1- 2sin^2(2y) so you can write the question as sin(2y)= 1- 2sin^2(2y). Let u= sin(2y) and the equation becomes u= 1- 2u^2 or 2u^2+ u- 1= 0. Solve that equation and, once you know u, solve sin(2y)= u for y.
I assume you mean sin^2(x)/(1- cos(x))= (sec(x)+ 1)/sec(x). Multiplying on both sides by 1- cos(x) and sec(x), sin^2(x)sec(x)= (sec(x)+ 1)(1- cos(x)).2. solve algerbraically
sin^2(x)/ 1- cos(x) = sec (x) +1/ sec (x)
help fast please, thanks.
Since sec(x)= 1/cos(x), that is sin^2(x)/cos(x)= (1/cos(x)+ 1)(1- cos(x)).
Multiplying out the right side gives 1/cos(x)- cos(x)+ 1- 1= 1/cos(x)- cos(x).
And on the left, sin^2(x)= 1- cos^2(x) so sin^2(x)/cos(x)= (1- cos^2(x)/cos(x)= 1/cos(x)- cos(x).
Were you asked to "solve" this equation or prove that this is an identity?