sin2 + cos2 = 1, how is it solved, i saw it solved once long time ago.?

Answer 1

sinΘ = opposite / hypotenuse (a/c)
cosΘ = adjacent / hypotenuse (b/c)

a² + b² = c² (from pytagorean theorem)
divide the equation by c²
a²/c² + b²/c² = c²/c²
(a/c)² + (b/c)² = 1
sin²Θ + cos²Θ = 1

Answer 2

Sin 2 Cos 2 1

Answer 3

sin2x +cos2x =1
This derivation can be solved using the pythagorean theorem:
c2 = a2 +b2
where c = the square root of a2 +b2
Using the angle x with sides of a right triangle as a, b, and c.
sinx=the opposite side a/ hypotenuse c
sin x =a/c
cos x = b/c, the side of angle x/the hypotenuse c.
c2 = a2 + b2
from above equations a =c sin x and b = c cos x, substitute the value of a and b into the aboive equation:
c2 =(c sinx)2 +(ccosx)2
c2 =c2 sin2 x + c2 cos2 x
Dividing the equation by c2:
1 = sin2 x +cos2X, or sin2 x +cos2x =1

Answer 4

u mean sin^2 + cos ^ 2 = 1

its a trig. identity. what do you mean solved?

do you mean u want it proved?

if you make a unit circle, the radius of that circle is one.

then make a triangle with an angle.

sin(angle) squared + cos(angle) squared = 1

since x^2 + y^2 = r^2 and r = 1 r^ = 1

the proof is something like that. i did it like 20 years ago so i might have forgotten most of it.

Answer 5

from right triangle with "a" and "b" as perpendicular sides and "c" the hypotenuse.

Let "x" be the acute angle opposite to "a"

sin x = a/c
cos x = b/c

by Pythagorean theorem
a^2 + b^2 = c^2

divide both sides of the equation by c^2
(a^2 / c^2) + (b^2 / c^2) = (c^2 / c^2)
(a / c)^2 + (b / c)^2 = 1

Recall that sin x = a/c and cos x = b/c

Therefore
(sin x)^2 + (cos x)^2 = 1

Answer 6

Sin2 Cos2

Answer 7

Do you mean to have an x in there?

ie solve for x
sin2x +cos2x =1

sin2x = 2 sinxcosx
cos2x = 1-2sin^2 x

sin2x + cos2x =1
2sinxcosx + 1-2sin^2 x = 1
2sinx( cosx -sinx)=0
sinx =0 or tanx =0
ie x=0 principal solution (or x =n x pi)

Answer 8

if i were u. i'd take a right angled triangle and substitute the values for sin and cos to prove the relationship!

Answer 9

To prove sin ² θ + cos² θ = 1:-
Consider circle of radius , r
Centre the origin.
r is a rotating vector components (x , y)
r makes angle θ with x axis
sin θ = y/r
cos θ = x/r
sin ² θ + cos ² θ = x ² / r ² + y ² / r ²
sin ² θ + cos ² θ = (x ² + y ²) / r ²
sin ² θ + cos ² θ = (r ²) / r ²
sin ² θ + cos ² θ = 1

Answer 10

sinx + cosx = 1
=> sqrt(2)[(sinx)1/sqrt(2) + (cosx)1/sqrt(2)] = 1
=> sin(x + pi/4) = 1/sqrt(2) = sin(pi/4)
=> x + pi/4 = (npi + (--1)^n (pi/4)
=> x = (n -- 1)pi + (--1)^n (pi/4)