trigonometry?

Question

find the exact value between 0 and 2pi:

(sin2x + cos 2x)^2 = 1

Answer 1

0, 1/4pi, 1/2pi, 3/4pi, pi, 5/4pi, 3/2pi, 7/4pi

First, forget about the "2x" part for a moment and don't square the left side! Take, instead, square roots and you have:

sin y + cos y = +/- 1

This can only happen at 0, 1/2pi, pi, and 3/4 pi. That is because at those points, one will equal +/- 1 and the other 0. At all other points, the sum will be greater than 1, with a maximum halfway between the listed angles of the square root of 2.

Cool, you know what "y" can equal. Continue the listing for another rotation around the origin: 2pi, 5/2pi, 3pi, and 7/2pi. That's because y = 2x so x = y/2. So we need that second rotation to get the 3rd and 4th quadrant values!

Then, since the angle desired, "x" equals "y" divided by 2, divide each of the angles in the two rotations by 2 to get:

0, 1/4pi, 1/2pi, 3/4pi, pi, 5/4pi, 3/2pi, 7/4pi

And you're done without knowing anything at all about double angle relationships!

(In fact, you could use this example to show those relationships, rather than use them to prove it, if you liked.)


Added:

One could also include 2pi in the list, but usually one wouldn't as 0 ("0 pi") is the same angle.

Answer 2

(sin2x + cos2x)^2 = sin^2(2x) + 2*cos2x*sin2x + cos^2(2x)
= (sin^2(2x) + cos^2(2x)) + 2*cos2x*sin2x

= 1 + 2*cos2x*sin2x

So we have 1 + 2*cos2x*sin2x = 1 or
2*cos2x*sin2x = 0

But by the double angle formula for sin

2*cos2x*sin2x = sin(4x)

So we must have

sin(4x) = 0

since we want x between 0 and 2pi, we have 4x
between 0 and 8pi. The solutions aer

4x = 0, pi, 2pi, 3pi, 4pi, 5pi, 6pi, 7pi

therefore

x = 0, pi/4, 2pi/4, 3pi/4, 4pi/4, 5pi/4, 6pi/4, 7pi/4

Answer 3

First, multiply out the terms on the left:

sin² (2x) + 2 sin (2x) cos (2x) + cos² (2x) = 1

Simplify using the Pythagorean theorem and double-angle identities:

1 + sin (4x) = 1
sin (4x) = 0

So 4x = πk for some integer k, thus x = π/4 k. The values of x between 0 and 2π are thus:

0, π/4, π/2, 3π/4, π, 5π/4, 3π/2, 7π/4, and 2π.

Answer 4

(sin2x + cos2x)^2=1
2cos(2x-45)^2=1
cos(2x-45)=+or- 1/squt2
2x-45=2npi +_ 45
(+)
x=n(pi) + 45
x= 45, 225(value between 0&2pi)
(-)
x=m(pi)
x=0,180,360 (value between 0&2pi, I consider 0,360 also. I mean 0<= x<=360)
so answers are 0,45,180, 225,360
but if you take
cos(2x-45)=-1/sqrt2
2x-45=2n(pi) +_ 135
(+)
x=n(pi)+90
x=90,270
(-)
x=m(pi)-45
x=135,315
so
final answers are 0,45,90,135,180,225,270,315,360

Answer 5

if you mean solving for x
(sin2x + cos 2x)² = 1
implies
sin²(2x) +cos²(2x) + 2sin2x cos2x = 1
1 + sin4x = 1
sin4x = 0
implies
4x = 0,±π, ±2π, ±3π, ±4π, ±5π, ±6π, ±7π, ±8π, ...
in your interval
x=0,(π/4),(π/2),(3π/4), π, (5π/4), (3π/2), (7π/4), 2π

exclude 0 and 2π if you mean the open interval