Asked by Phoebe
Please help!!!!!! cosx-sin^2x-1
Answers
Answered by
Damon
So what is the question and I suspect you have a typo
Answered by
Phoebe
I always get confused working out this problem because I keep getting the wrong answer. cosine of x- sin squared x-1
Answered by
Phoebe
and it needs me to write the expression in factored form as an algebraic expression of a single trigonometric function.
Answered by
Damon
sin^2 x = 1 - cos^2 x
cos x - (1-cos^2 x) - 1
cos x - 2 + cos^2 x
cos^2 x + cos x - 2
as you wrote it
perhaps you mean
cos x - (sin^2 x -1)
cos x - (1 -cos^2 x - 1)
cos x + cos^2 x
cos x (cos x + 1)
cos x - (1-cos^2 x) - 1
cos x - 2 + cos^2 x
cos^2 x + cos x - 2
as you wrote it
perhaps you mean
cos x - (sin^2 x -1)
cos x - (1 -cos^2 x - 1)
cos x + cos^2 x
cos x (cos x + 1)
Answered by
Phoebe
Where did you get the 1-cosine squared from?
Answered by
Phoebe
Find an algebraic expression equivalent to the given expression. sine (tan inverse of u/2)
Answered by
Phoebe
For the problem cosine of x-sine squared x-1 the choices are A)sine squared x, B) (cot x+1)(cot x-1), C) (cos x+2)(sin x-1), D)(sin x+2)(sin x-1)
Answered by
Damon
sin^2 x + cos^2 x = 1 identity
so
sin^2 x = 1 - cos^2 x
so
sin^2 x = 1 - cos^2 x
Answered by
Damon
(cos/sin +1)(cos/sin-1) = cos^2/sin^2 - 1
====================================
sin[ tan^-1(u/2) ]
right triangle
opposite = u
adjacent = 2
so hypotenuse = sqrt(u^2 +4)
so
sin = u/sqrt(u^2+4)
or [u sqrt (u^2+4)] / (u^2+4)
====================================
sin[ tan^-1(u/2) ]
right triangle
opposite = u
adjacent = 2
so hypotenuse = sqrt(u^2 +4)
so
sin = u/sqrt(u^2+4)
or [u sqrt (u^2+4)] / (u^2+4)
Answered by
Anonymous
Thank you for your help. Can you please delete all the questions I have typed in today?
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