what is the sum of sin^2 x + cos^2 x? can you proove it also?

Question

if seen a lot of proof using a right angled triangle with pythagoras theorem, and that's true...

but, it seems that this statement above is only valid on a right angled triangle, so, try other proofs... :)

i've seen a lot of proof using a right angled triangle with pythagoras theorem, and that's true...

but, it seems that this statement above is only valid in a right angled triangle, so, it's better if you try other prooving method...

Answer 1

e^ix = cos x + i sin x
1 / e^ix = e^-ix = cos x - i sin x

So you can derive

cos x = (e^(ix) + e^(-ix)) / 2
sin x = (e^(ix) - e^(-ix)) / (2i)

Let u = e^ix so that e^(-ix) = 1 / e^(ix) = 1/ u

Then cos x = (u + 1/u) / 2 and sin x = (u - 1/u) / (2i) and so

sin^2 x + cos^2 x = ((u - 1/u) / (2i))^2 + ((u + 1/u) / 2)^2
= ((u^2 - 2 + 1/u^2) / (-4)) + ((u^2 + 2 + 1/u^2) / 4)
= (1/4) ( -u^2 + 2 - 1/u^2 + u^2 + 2 + 1/u^2 )
= (1/4) (2 + 2)
= (1/4) 4
= 4/4
= 1

:-)

Answer 2

1

Answer 3

sin^2 x + cos^2 x = 1

The proof is easy: remember that in a right-angled triangle ABC (right angle at corner B) then AB^2 + BC^2 = AC^2 (the Pythagoras equation). Now, let's look at the definitions of SIN and COS:
SIN A = BC/AC and COS A = AB/AC.
Now we can substitute these definitions:

SIN^2 A = BC^2/AC^2 and COS^2 A = AB^2/AC^2
so SIN^2 A + COS^2 A = (BC^2 + AB^2)/AC^2
but BC^2 + AB^2 = AC^2 (from Pythagoras) so
SIN^2 A + COS^2 A = AC^2/AC^2 = 1

Answer 4

sin^2 x+ cos^2 x= 1; is the algebraic version of Pythagoras' Theorem.
Proof:
For any right angled traingle xyz, with angle z= 90 deg, sides X, Y, Z such that side X is opposite angle x and Y is opposite y, etc:
X^2+ Y^2 = Z^2 ---(Pythagoras' Theorem)
Didividing through both sides of the above equation by Z^2 gives
(X/Z)^2+ (Y/Z)^2= 1.
Recall that sin x= X/Z and cos x=Y/Z, implying sin^2 x= (^2X/Z) and cos^2 x =(Y/Z)^2, then your identity follows quite easily.

Answer 5

Consider a rt angled triangel ABC with rt angle B, let /_A =x
sin x = BC/AC
cos x = AB/AC
square and add
sin^2 x+cos^2 x =( BC^2 + AB^2) / AC^2
In rt angled triangle, sum of squares of other sides is equal to square of hypotenuese
Hence BC^2+AB^2=AC^2
hence sin^2 x + cos^x = AC^2/ AC^2
hence the result is 1

Answer 6

sin squared plus cosine squared is equal to one. To prove it, draw a right angled triangle with hypotenuse 1 and angle x, resolve to find the lengthe of the opposite and adjacent sides, then apply Pythagors.

Answer 7

sin² x + cos² x = 1

you can prove it by Pitagoras' Theorem:

just consider a right triangle with
angle x,
opossite cathetus a,
adjacent cathetus b,
and hypotenuse equal c.

by definition: sin x = a/c; cos x = b/c
power square and sum:

sin² x + cos² x = a²/c² + b²/c² = (a² + b²)/c²

but by Pitagoras' Theorem: a² + b² = c²

sin² x + cos² x = c²/c² = 1

Answer 8

1

It can be proved for right angled triangles because all trignometric functions / ratios are defined for right angled triangles.
sin = opposite side / hypotenuse

Answer 9

i dont remember the proof, but its 1.