12th Grade

{"voice_prompt": "","manuscript": {"title": {"text": "How do you prove the addition and subtraction formulas for sine?","audio": "How do you prove the addition and subtraction formulas for sine?" },"description": {"text": "To prove the addition and subtraction formulas for sine, you construct geometric figures using right triangles and the unit circle. By applying trigonometric definitions and angle relationships, you show that sine of alpha plus beta equals sine of alpha times cosine of beta plus cosine of alpha times sine of beta, and sine of alpha minus beta equals sine of alpha times cosine of beta minus cosine of alpha times sine of beta.","audio": "To prove the addition and subtraction formulas for sine, you construct geometric figures using right triangles and the unit circle. By applying trigonometric definitions and angle relationships, you show that sine of alpha plus beta equals sine of alpha times cosine of beta plus cosine of alpha times sine of beta, and sine of alpha minus beta equals sine of alpha times cosine of beta minus cosine of alpha times sine of beta." },"scenes": [ {"text": "Draw a horizontal line (the x-axis); mark an origin O. Draw a line from O at an angle alpha above the horizontal line and a second line at an angle beta above that; the angle between the second line and the x-axis is alpha + beta","latex": "\\text{Tegn opp tre linjer som alle krysser i origo, hvorav en er vannrett}" }, {"text": "Place P on the line defined by alpha + beta at a unit distance from the origin","latex": "\\text{Tegn et punkt P på øverste linje}" }, {"text": "Let PQ be a line perpendicular to line OQ defined by angle alpha, drawn from point Q on this line to point P. OQP is a right angle.","latex":"\\text{Tegn en linje fra P normalt ned midterste linje. Dette punktet er Q. Marker at vinkelen i Q er 90 grader}" }, {"text": "Let QA be a perpendicular from point A on the x-axis to Q and PB be a perpendicular from point B on the x-axis to P. OAQ and OBP are right angles.","latex": "\\text{Tegn to linjer opp fra nederste linje. En opp til P og en opp til Q. \\\\Punktene linjene går opp fra er henholdsvis B og A. Marker at vinklene i B og A er 90 grader}" }, {"text": "From evaluating the angles in the figure, you can find the angle BPQ, to equal alpha. This is because the point of intersection formed by the lines BP and OQ form vertical angles. As both OQP and OBP are right angles. you know that all triangles has a angle sum of 180 degrees. Meaning angle BPQ must be equal to alfa.","latex": "\\text{Marker trekanten mellom O B og krysningspunktet mellom OQ og PB, vis at vinkelen i krysningspunket har en toppvinkel. Highlight at det er en 90 graders graders vinkel i både B og Q, Marker at vinkelen BPQ er alfa}" }, {"text":"Now, you can try to find sine of alpha plus beta. The formula for sine is opposite side divided by hypotenuse. By applying this formula to the figure you get sine of alpha plus beta equals PB over OP which equals PB because P is a unit away from Origo.","latex":"\\text{Highlight OP og PB samt vinkelen alfa+beta.} \\sin(\\alpha+\\beta)= \\frac{\\text{PB}}{\\text{OP}}=\\text{PB}" }, {"text": "Lets split PB into two parts by drawing a point R such that QR is parallel to the x-axis. Sine of alfa plus beta is equal to PB which is equal to PR + BR","latex":"\\text{Tegn en linje fra Q til PB som er parallell med x-aksen. Behold den figuren vi har laget i bildet fra og med nå} \\\\\\text{Hold følgende ligning på skjermen:}\\sin(\\alpha+\\beta)=\\text{PB}" }, {"text":"Sine of alfa plus beta is equal to PB which is equal to PR + BR","latex":"\\text{Marker PR og BR i to forskjellige farger og bytt ut PB slik at:} \\sin(\\alpha+\\beta)=\\text{PR+BR} \\text{(Kan kanskje skrive PR og BR texten i samme farge som markeringsfargen i figuren.)}" }, {"text": "BR is equal to AQ, which is a part of the right triangle, AOQ. This implies that sine of alfa must equal BR over OQ. Rearange the equation to get BR equals OQ times sine of alfa. OQ is the adjacent side of angle beta in the triangle OQP,OP has a length of one, meaning cosine of beta equals OP","latex": "\\text{Hold BR markert. Marker opp AQ og fjern markeringen av BR. Marker AQ og OQ og vinkelen AOQ=alfa. Sett oppligningen:} \\sin(\\alpha)=\\frac{AQ}{OQ} \\text{, på skjermen og hold den der/}\\\\\\text{Marker opp OP, OQ og vinkelen POQ. Sett opp formelen }\\cos(beta)=\\frac{\\text{OQ}}{\\text{OP}}=\\text{OQ}" }, {"text":"Exchanging the formula for OQ into the formula for BR you get. BR equal to sine of alfa times cosine of beta.","latex": "\\text{Skriv om utrykket for BR til: BR}=\\sin(\\alpha)\\cos(\\beta)" }, {"text": "PR is the adjacent to the angle alfa in the triangle OQR. This means PR is equal to PQ cosine of alfa. PQ is the oposide side of angle beta in the triangle OQP. OP has a length of one, meaning sine of beta equals PQ.","latex":"\\text{Highlight PR. Marker vinkelen RPQ=alfa, og siden PQ. Skriv opp:} \\cos(\\alpha)= \\frac{PR}{PQ} \\text{og skriv om til PR} = PQ\\cos(\\alpha)" }, {"text":"PQ is the oposide side of angle beta in the triangle OQP. OP has a length of one, meaning sine of beta equals PQ.","latex": "\\text{Hold:} = PQ\\cos(\\alpha) \\text{på skjermen, marker opp OP, PQ og POQ. Skriv opp: PQ=OP}\\sin(\\beta)" }, {"text":"Exchanging PQ with sine of beta in the equation for PR gives PR equal to sine of beta times cosine of alfa.","latex":"\\text{Skriv om utrykket for PR til: PR}= \\sin(\\beta)\\cos(\\alpha)" }, {"text":"Now you can insert your expressions for PR and PB into the formula for sine of alfa plus beta. Sine of alfa plus beta equals PB, PB equals PR + BR. Sine of alfa plus beta thus equals sine of beta times cosine of alfa plus cosine of beta times sine of alfa. This is the addition formula for sines","latex":"\\text{Her kan figuren fjernes} \\sine(\\alpha+ \\beta)= \\sin(\\alpha)\\cos(\\beta) + \\cos(\\alpha)\\sin(\\beta)" }, {"text":"To find the subtraction formula for sines start with: sine of alfa plus minus beta. Using the adition formula you get: Sine of alfa times cosine of minus beta plus cosine of alfa times sine of minus beta.","latex":"\\sine(\\alpha+(-\\beta))= \\sin(\\alpha)\\cos(\\beta) + \\cos(\\alpha)\\sin(\\-beta)" }, {"text":"To simplify this expression you have to use the following identities. Sine of minus theta= minus theta and cosine of minus theta equals cosine of theta.","latex":"\\sin(-\\theta)=- \\sin(\\theta) \\text{and} \\cos(-\\theta) = \\cos(\\theta)" }, {"text":"Sine of alfa minus beta then equals sine of alpha times cosine of beta minus cosine of alpha times sine of beta.","latex":"\\sine(\\alpha-\\beta))= \\sin(\\alpha)\\cos(\\beta) - \\cos(\\alpha)\\sin(\\beta)" }
],"outro": {"text": "In summary, by using right triangles and the unit circle, you can visually break down the angle relationships needed for the sine formulas. This leads to the conclusion that sine of alpha plus beta and sine of alpha minus beta can be expressed as sine of alpha times cosine of beta plus cosine of alpha times sine of beta, and sine of alpha minus beta equals sine of alpha times cosine of beta minus cosine of alpha times sine of beta.","audio": "In summary, by using right triangles and the unit circle, you can visually break down the angle relationships needed for the sine formulas. This leads to the conclusion that sine of alpha plus beta and sine of alpha minus beta can be expressed as sine of alpha times cosine of beta plus cosine of alpha times sine of beta, and sine of alpha minus beta equals sine of alpha times cosine of beta minus cosine of alpha times sine of beta." } }}