Prove that sin^2A+cos^2A = 1...?

Question

Please help me... Please oh please oh please... I really need help... Thank you in advance and God bless!

Answer 1

First, draw a triangle
..../|
c / | b
../_|
a
A= The angle between a and c
sin(A)= b/c
cos(A)= a/c

Therefore
sin^2(A) = b^2/c^2
cos^2(A) = a^2/c^2

So...
sin^2(A) + cos^2(A)
...is basically
b^2/c^2 + a^2/c^2

It can be rewritten to
a^2 + b^2
--------------
c^2

and because of phytagorous's theorem
where a^2 + b^2 = c^2
substitute (a^2 + b^2) with c^2
so then you get
c^2/c^2

which is basically

1

Answer 2

Im letting the 3 sides of the triangle be:
1) opp : the side opposite A
2) adj : the side adjacent to A
3) hypo : the hypotenuse of the triangle

Proof:
sin^2A + cos^2A
= opp^2/hypo^2 + adj^2/hypo^2
= (opp^2 + adj^2)/ hypo^2 {taking hypo^2 as the LCM)
= hypo^2/hypo^2 { as per Pythagoras' theorem, hypo^2 = opp^2 + adj^2}
= 1 {as both cancel out}

TADA!!

P.S. Hope this helps. I would appreciate if u choose this as the best asnwer. If any problem contact me : [e-mail]

Answer 3

sin A = y / r
cos A = x / r

sin ² A + cos ² A = (x ² + y ²) / r ²
sin ² A + cos ² A = r ² / r ²
sin ² A + cos ² A = 1

Answer 4

Try it with this:

sin^2(A)=1/2[1-cos(2A)]
cos^2(A)=1/2[1+cos(2A)]
sin^2(A)+cos^2(A)=1/2 - (1/2)cos(2A) +1/2 + (1/2)cos(2A)
=1/2 + 1/2 =1

Answer 5

From triangle ABC with sides a,b,c as usually drawn,
find sin(A) = a/c and cos(A) = b/c.
Now work out sin^2(A) and cos^2(A), then add them
and you'll see that you now have to invoke Pythagoras.

Answer 6

dont u need like a picture of a triangle to solve that?