Prove the sum of angles identity for sine. That is,prove: sin(x+h)=sin(x)cos(h)+
ID: 3095224 • Letter: P
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Prove the sum of angles identity for sine. That is,prove: sin(x+h)=sin(x)cos(h)+cos(x)sin(h) Prove the sum of angles identity for sine. That is,prove: sin(x+h)=sin(x)cos(h)+cos(x)sin(h)Explanation / Answer
I'll try to explain it to you, though it's quite hard withoutdrawing.... (the proof is somewhat complicated...) Imagine the trigonometric circumference: as you know, any arcbegins at A (1;0); then consider the arcs x and h (assumingx > h), whose ending points (belonging to the trigonometriccircumference), X and H, respectively, have the followingcoordinates: X (cosx; sinx) H (cosh; sinh) then imagine to draw the arc (x - h) beginning at A (1;0)and ending at B (cos(x - h); sin(x - h)) well, due to the construction, the chords XH and AB are equal toeach other: XH = AB then, recalling the evaluation of the distance between two pointsas a function of their coordinates, you get: XH = [(cosx - cosh)² + (sinx - sinh)²] AB = {[cos(x - h) - 1]² + [sin(x - h) - 0]²} ={[cos(x - h) - 1]² + sin²(x - h)} thus equating them, you get: XH = AB [(cosx - cosh)² + (sinx - sinh)²] = {[cos(x- h) - 1]² + sin²(x - h)} squaring both sides: (cosx - cosh)² + (sinx - sinh)² = [cos(x - h) - 1]²+ sin²(x - h) expanding the squares: cos²x - 2cosx cosh + cos²h + sin²x - 2sinx sinh +sin²h = cos²(x - h) - 2cos(x - h) + 1 + sin²(x - h) note that: cos²(x - h) + sin²(x - h) = 1 (pythagorean identity) cos²x + sin²x = 1 cos²h + sin²h = 1 therefore: 1 - 2cosx cosh + 1 - 2sinx sinh = 1 - 2cos(x - h) + 1 2 - 2cosx cosh - 2sinx sinh = 2 - 2cos(x - h) - 2cosx cosh - 2sinx sinh = - 2cos(x - h) dividing both sides by (- 2): cosx cosh + sinx sinh = cos(x - h) that is the cosine subtraction rule indeed; in order to get sine addiction rule from this, simply replace xwith u = (/2) - x, yielding: cos(u - h) = cosu cosh + sinu sinh cos[(/2) - x - h] = cos[(/2) - x] cosh + sin[(/2)- x] sinh cos[(/2) - (x + h)] = cos[(/2) - x] cosh +sin[(/2) - x] sinh recall the identities: cos[(/2) - ] = sin sin[(/2) - ] = cos thus, in conclusion: sin(x + h) = sinx cosh + cosx sinh your identity being proved I hope it helps... Bye!!
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