How to prove <span title="sin(x+y)=sinxcosy+cosxsiny?">sin(x+y)=sinxcosy+cosxsin...</span>

Question

thank you very much

Answer 1

If you could tell me what level of math this is for, I will think about it for you.  The method I know uses exponential and imaginary numbers, but you may be at a different math level.  I want to help you the correct way. Please let me know.  Thanks.

Answer 2

I&#039;ll try to explain it to you, though it&#039;s quite hard without drawing....

(the proof is somewhat complicated...)

Imagine the trigonometric circumference: as you know, any arc begins at A Ξ (1;0); then consider the arcs x and y (assuming x &gt; y), whose ending points (belonging to the trigonometric circumference), X and Y, respectively, have the following coordinates:

X Ξ (cosx; sinx)

Y Ξ (cosy; siny)

then imagine to draw the arc (x - y) beginning at A Ξ (1;0) and ending at B Ξ (cos(x - y); sin(x - y))

well, due to the construction, the chords XY and AB are equal to each other:

XY = AB

then, recalling the evaluation of the distance between two points as a

function of their coordinates, you get:

XY = √[(cosx - cosy)² + (sinx - siny)²]

AB = √{[cos(x - y) - 1]² + [sin(x - y) - 0]²} = √{[cos(x - y) - 1]² + sin²(x - y)}

thus equating them, you get:

XY = AB →

√[(cosx - cosy)² + (sinx - siny)²] = √{[cos(x - y) - 1]² + sin²(x - y)} →

squaring both sides:

(cosx - cosy)² + (sinx - siny)² = [cos(x - y) - 1]² + sin²(x - y) →

expanding the squares:

cos²x - 2cosx cosy + cos²y + sin²x - 2sinx siny + sin²y = cos²(x - y) - 2cos(x - y) +

1 + sin²(x - y) →

note that:

cos²(x - y) + sin²(x - y) = 1 (pythagorean identity)

cos²x + sin²x = 1

cos²y + sin²y = 1

therefore:

1 - 2cosx cosy + 1 - 2sinx siny  = 1 - 2cos(x - y) + 1  ÃƒÂ¢Ã‚†Â’

2 - 2cosx cosy - 2sinx siny  = 2 - 2cos(x - y)  ÃƒÂ¢Ã‚†Â’

- 2cosx cosy - 2sinx siny  = - 2cos(x - y)  ÃƒÂ¢Ã‚†Â’

dividing both sides by (- 2):

cosx cosy + sinx siny  = cos(x - y)

that is the cosine subtraction rule indeed;

in order to get sine addiction rule from this, simply replace x with u = (π/2) - x, yielding:

cos(u - y) = cosu cosy + sinu siny  ÃƒÂ¢Ã‚†Â’

cos[(π/2) - x - y] = cos[(π/2) - x] cosy + sin[(π/2) - x] siny  ÃƒÂ¢Ã‚†Â’

cos[(π/2) - (x + y)] = cos[(π/2) - x] cosy + sin[(π/2) - x] siny  ÃƒÂ¢Ã‚†Â’

recall the identities:

cos[(π/2) - α] = sinα

sin[(π/2) - α] = cosα

thus, in conclusion:

sin(x + y) = sinx cosy + cosx siny

your identity being proved

I hope it helps...

Bye!!

Answer 3

this actually came from the formula of addition for cosine .. . .

cos(x ± y) = cosx cosy ∓ sinx siny

then

sin(x + y) = cos(π/2 - (x+y)) = cos((π/2 - x) - y)

= cos(π/2 - x) cosy + sin(π/2 - x) siny

= sinx cosy + cosx siny

now, to prove that cos(x + y) = cosx cosy - sinx siny .. . .. .

this used the distance formula:

let P(1,0), A(cosx, sinx) , B(cosy, siny) , C(cos(x-y), sin(x-y))

the angle from the positive x-axis to A is x.

the angle from the positive y-axis to B is y.

thus the angle between them is x - y , say, x &gt; y.

thus the distance bet AB is the same distance as CP

then

d(AB) = √[(cosx - cosy)^2 + (sinx - siny)^2]

= √[cos^2x - 2cosx cosy + cos^2y + sin^2x - 2sinxsiny + sin^2y]

= √[2 - 2(cosx cosy + sinx siny)]

meanwhile

d(CP) = √[(1- cos(x-y)^2 + sin^2(x-y)]

= √[1 - 2cos(x-y) + cos^2(x-y) + sin^2(x-y)]

= √[2 - 2cos(x-y)]

thus

cos(x - y) = (cosx cosy + sinx siny)

and the rest follows.