Sin³ x-sin x=cos ² x
we know that:
sin²x + cos²x=1 ⇒cos²x=1-sin²x
Therefore:
sin³x-sin x=1-sin²x
sin³x+sin²x-sin x-1=0
sin³x=z
z³+z²-z-1=0
we divide by Ruffini method:
1 1 -1 -1
1 1 2 1 z=1
-------------------------------------
1 2 1 0
-1 -1 -1 z=-1
--------------------------------------
1 1 0 z=-1
Therefore; the solutions are z=-1 and z=1
The solutions are:
if z=-1, then
sin x=-1 ⇒x= arcsin -1=π+2kπ (180º+360ºK) K∈Z
if z=1, then
sin x=1 ⇒ x=arcsin 1=π/2 + 2kπ (90º+360ºK) k∈Z
π/2 + 2kπ U π+2Kπ=π/2+kπ k∈Z ≈(90º+180ºK)
Answer: π/2 + Kπ or 90º+180ºK K∈Z
Z=...-3,-2,-1,0,1,2,3,4....
Answer:
No
Step-by-step explanation:
It is not a function because the x should never repeat the y can but not the x.
Answer:
Boden's conjecture is correct
Step-by-step explanation:
we know that
If two angles are complementary, the cofunction identities state that the sine of one equals the cosine of the other and vice versa.
A right triangle, has two acute angles.
The two acute angles are complementary
so
In any right triangle, the sine of one acute angle is equal to the cosine of the other acute angle
therefore
Boden's conjecture is correct
Answer:
Step-by-step explanation:
The first is 
<em>The second is </em>
So 
{Simple Subtraction}
<em>I hope this helps you</em>
<em>:)</em>
