sin cos

Trigonometry equalities, inequalities and expressions - sin, cos, tan, cot

sin cos

Postby yppetrov » Sat Dec 16, 2006 3:02 am

Why sin^2 + cos^2 = 1 ?
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Postby babyj » Mon Dec 03, 2007 6:36 pm

I dont know...i have never seen it like that in my life

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Re: sin cos

Postby dduclam » Sun May 04, 2008 5:10 am

yppetrov wrote:Why sin^2 + cos^2 = 1 ?


Because Pitago theorem show that [tex]a^2+b^2=c^2[/tex] :D ([tex]0\le \alpha \le90^\circ[/tex])

Nother value of [tex]\alpha[/tex] ,using trigonometry circle !

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Postby anuj94 » Wed Jul 16, 2008 11:46 pm

Make a Triangle ABC
In▲ABC take A as a referer and B as 90°
So, AB = Base , BC= Perpendicular and AC = Hypotenous
Sin = P/H = BC/AC
cos = B/H = AB/AC
if AB = ABx , BC = BCx , AC=ACx taking x as a constant
sin^2 + cos^2 = 1
(BCx/ACx)^2 + (ABx/ACx)^2 =1
BC^2 x^2 +AC^2 x^2 + AB^2 x^2 /AC^2 x^2
BC^2 x^2 + AB^2 x^2/AC^2 x^2 =1
cross multiply
BC^2 x^2 + AB^2 x^2=AC^2 x^2
x^2( BC^2 + AB^2 ) = AC^2 x^2
BC^2 + AB^2 = AC^2 x^2/x^2
BC^2 + AB^2 = AC^2
a2+b2=c2
hence proved

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Re: sin cos

Postby Guest » Thu Mar 05, 2015 7:30 am

The question is wrong...............
Why sin^2 + cos^2 = 1 ?

sin^2 of what.......cos^2 of what...........

It should read sin^2(theta) + cos^2(theta) = 1

sin^2(theta) is same as writing sin(theta) multiplied by sin(theta)

for ease of typing here I wiil call it sin^2(B) is same as sin(B) x sin(B)......
The proof.........take a line of unit length and at an angle (B) to horizontal
the coordinates of the ends of the line are x1,y1 and x2,y2
resolve this into horizontal distance plus vertical distance between ends, like a right angles triangle
x distance is x2 - x1 and vertical distance is y2 - y1.
By Pathagoras theorem the length of the line is sqroot(x^2 + y^2) and this is equal to 1 .....it was unit length.
also by trig. x = 1 x cos(B) and y = 1 x sin(B) this is same as x = cos(B) and y = sin(B)
So by Pathagoras the length of the line is sqroot(cos^2(B) + sin^2(B)) and this equals 1

OR cos^2(B) + sin^2(B) = 1
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