Prove that
Sin^2 A +Cos^2 A = 1
Answers
Step-by-step explanation:
Use the formula for a circle
(
x
2
+
y
2
=
r
2
)
, and substitute
x
=
r
cos
θ
and
y
=
r
sin
θ
.
Explanation:
The formula for a circle centred at the origin is
x
2
+
y
2
=
r
2
That is, the distance from the origin to any point
(
x
,
y
)
on the circle is the radius
r
of the circle.
Picture a circle of radius
r
centred at the origin, and pick a point
(
x
,
y
)
on the circle:
graph{(x^2+y^2-1)((x-sqrt(3)/2)^2+(y-0.5)^2-0.003)=0 [-2.5, 2.5, -1.25, 1.25]}
If we draw a line from that point to the origin, its length is
r
. We can also draw a triangle for that point as follows:
graph{(x^2+y^2-1)(y-sqrt(3)x/3)((y-0.25)^4/0.18+(x-sqrt(3)/2)^4/0.000001-0.02)(y^4/0.00001+(x-sqrt(3)/4)^4/2.7-0.01)=0 [-2.5, 2.5, -1.25, 1.25]}
Let the angle at the origin be theta (
θ
).
Now for the trigonometry.
For an angle
θ
in a right triangle, the trig function
sin
θ
is the ratio
opposite side
hypotenuse
. In our case, the length of the side opposite of
θ
is the
y
-coordinate of our point
(
x
,
y
)
, and the hypotenuse is our radius
r
. So:
sin
θ
=
opp
hyp
=
y
r
⇔
y
=
r
sin
θ
Similarly,
cos
θ
is the ratio of the
x
-coordinate in
(
x
,
y
)
to the radius
r
:
cos
θ
=
adj
hyp
=
x
r
⇔
x
=
r
cos
θ
So we have
x
=
r
cos
θ
and
y
=
r
sin
θ
. Substituting these into the circle formula gives
x
2
+
y
2
=
r
2
(
r
cos
θ
)
2
+
(
r
sin
θ
)
2
=
r
2
r
2
cos
2
θ
+
r
2
sin
2
θ
=
r
2
The
r
2
's all cancel, leaving us with
cos
2
θ
+
sin
2
θ
=
1
This is often rewritten with the
sin
2
term in front, like this:
sin
2
θ
+
cos
2
θ
=
1
And that's it. That's really all there is to it. Just as the distance between the origin and any point
(
x
,
y
)
on a circle must be the circle's radius, the sum of the squared values for
sin
θ
and
cos
θ
must be 1 for any angle
θ
.