how do you prove that sin^2 (x) + cos^2(x) = 1 ? (no matter what x is)?

Question

how do you prove that sin^2 (x) + cos^2(x) = 1 ? no matter what the value of "x" is???
please give an example

Answer 1

Pythagoras' theorem:

O^2 + A^2 = H^2
divide by H^2
(O/H)^2 + (A/H)^2 = 1

sin^2 + cos^2 = 1

Answer 2

Dr D has presented a patrial answer (it extremely is incomplete if we can not locate coefficient of x^3 and make a wager) hence i'm offering the excellent answer enable the different 2 roots be c and d then we get comparing coefficients (it extremely is given via viete;s relation additionally) a+b+c+d = -a million ab+ac+advert+bc+bd+cd = 0 abc+abd + acd + bdc = 0 abcd = -a million putting a+ b= s c+d = t, ab = p and cd = q we get s+ t = -a million ...a million p + q + st = 0....2 pt + qs = 0...3 and pq = -a million...4 q = - a million/p (from 4) and t = -a million -s(from a million) we get p - a million/p - s^2 -s = 0 ,.. 5 and p(-a million-s) - s/p = 0 => p^(a million-s) - s = 0 => s = =p^2/(a million+p^2) .. 6 from 6 positioned s in 5 to get p - a million/p -p^4/(p^2+a million)^2 +p^2/(a million+p^2) = 0 or p^2(p^2+a million)^2 -(p^2+a million)^2 -p^5 +p^3(p^2+a million) = 0 or p^2(p^4+2p^2+a million) - (p^4 + 2p^2 + a million) -p^5 + p^5 + p^3 = 0 or p^6 +p^4 +p^3 -p^2 -a million = 0 so p is a root of x^6 +x^4 + x^3 -x^2 -a million so ab (it extremely is p) a root of x^6 +x^4 + x^3 -x^2 -a million

Answer 3

Pythagoras said he square of the length of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides.

If you construct your right triangle by picking a point on a ray beginning and dropping a perpendicular to the x axis, the length of the vertical side will be "y" and the length of the horizontal side will be "x." Call the distance from the origin to the point on the ray "r," and the angle formed by the ray and the positive x axis "a"/

Ok, now from Pythagoras, r² = x² + y²
If we multiply both sides by 1/r²
1 = x²/r² + y²/r²
1 = (x/r)² + (y/r)²
From trig,
sin(a) = y/r, and cos(a) = x/r

By substituting the identities from trig, you get
1 = cos² (a) + sin² (a)
And, it doesn't make any difference what a is... it can even be x, so,
1 = cos² (x) + sin² (x)

Answer 4

By Pythagorean's Theorem:
When x and y are two sides of the right triangle and h is the hypotenuse, a^2 + b^2 = h^2.
Divide everything by h^2.
So (a/h)^2 + (b/h)^2 = 1
By definition, sin x = a/h and cos x = b/h
So sin^2 x + cos^2 x = 1.

Answer 5

by Pythagorean theorem
draw a triangle. . . say the side b is the x axis , side a is the vertical line . . and the hypothenuse
let the length of hypotenuse is 1
the computing the sides : a = 1 sin x . . . . and . . b = 1 cos x

by Pythagorean theorem
a² + b² = 1²
sin ² x + cos ² x = 1

Answer 6

Let the three sides of a right angled triangle be b(base),p(perpendicular) and h(hypotenuse)
sin =p/h
cos=b/h
sin^2+c0s^2
(p/h)^2+(b/h)^2
=p^2/h^2 +b^2/h^2
=(p^2+b^2)/h^2
=h^2/h^2 [as p^2+b^2=h^2 as per pythgores theorem]
=1

Answer 7

Consider an angle θ in XY plane.
OR is a rotating vector , of length r, in this plane.
OR is vector (x , y)
sin θ = y / r
cos θ = x / r
sin ² θ + cos ² θ
= y ² / r ² + x ² / r ²
= ( y ² + x ² ) / r ²
= (r ²) / r ²
= 1
Thus sin ² x + cos ² x = 1 for all x.

Answer 8

that's one nasty proof, you have to get points from the unit circle (circle with radius 1, equation x² + y² = 1) and manipulate things there. you should end up with x = cos theta and y = sin theta...

Answer 9

My teacher showed me a geometric proof using a unit circle when I was in high school. And I still remember it. Here it is:

http://s5.photobucket.com/albums/y164/azianshrimp/?action=view¤t=sinxsquaredcosxsquaredproof.jpg