12th Grade

{"voice_prompt": "",

"manuscript":

{"title": {"text": "How do you prove the addition and subtraction formulas for cosine?","audio": "How do you prove the addition and subtraction formulas for cosine?" },"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 cosine of alpha plus beta equals cosine of alpha times cosine of beta minus sine of alpha times sine of beta, and cosine of alpha minus beta equals cosine of alpha times cosine of beta plus sine 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 cosine of alpha plus beta equals cosine of alpha times cosine of beta minus sine of alpha times sine of beta, and cosine of alpha minus beta equals cosine of alpha times cosine of beta plus sine 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. the logarithm with base 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":"Draw a line from Q to point R such that QR is parallel to the x-axis","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":"Now, you can try to find cossine of alpha plus beta. The formula for cosine is adjacent side divided by hypotenuse. By applying this formula to the figure you get cosine of alpha plus beta equals OB over OP which equals OB because P is a unit away from Origo.","latex":"\\text{Highlight OB og OP og vinkelen alfa+beta.} \\cos(\\alpha+\\beta)= \\frac{\\text{OB}}{\\text{OP}}=\\text{OB}" }, {"text": "Point B is a part of the line OA, this means you can find the distance OB by calculating the distance OA and subtracting AB","latex":"\\text{Hold følgende ligning på skjermen:}\\cos(\\alpha+\\beta)=\\text{OB}" }, {"text":"Cosine of alfa plus beta is equal to OB which is equal to OA - AB","latex":"\\text{Marker OA og AB i to forskjellige farger og bytt ut OB slik at:} \\cos(\\alpha+\\beta)=\\text{OA-AB} \\text{(Kan kanskje skrive OA og AB texten i samme farge som markeringsfargen i figuren.)}" }, {"text": "OA is a part of the right triangle, AOQ. This implies that cosine of alfa must equal OA over OQ. Rearange the equation to get OA equals OQ times cosine 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 OA markert. Marker OQ og vinkelen AOQ=alfa. Sett oppligningen:} \\cos(\\alpha)=\\frac{OA}{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 OA you get. OA equal to cosine of alfa times cosine of beta.","latex": "\\text{Skriv om utrykket for OA til: OA}=\\cos(\\alpha)\\cos(\\beta)" }, {"text": "AB is equal to the length of QR. QR is the oposite side to the angle alfa in the triangle OQR. This means AB is equal to PQ sine 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 QR. Marker vinkelen RPQ=alfa, og siden QR. Skriv opp:} \\sin(\\alpha)= \\frac{AB}{PQ} \\text{og skriv om til AB} = PQ\\sin(\\alpha)" }, {"text":"PB 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:} AB = PQ\\sin(\\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 AB gives AB equal to sine of beta times sine of alfa.","latex":"\\text{Skriv om utrykket for AB til: AB}= \\sin(\\beta)\\cos(\\alpha)" }, {"text":"Now you can insert your expressions for OA and AB into the formula for cosine of alfa plus beta. Cosine of alfa plus beta equals OB, OB equals OA minus AB. Cosine of alfa plus beta thus equals cosine of beta times cosine of alfa minus sine of beta times sine of alfa. This is the addition formula for cosines","latex":"\\text{Her kan figuren fjernes} \\cos(\\alpha+ \\beta)= \\cos(\\alpha)\\cos(\\beta) + \\sin(\\alpha)\\sin(\\beta)" }, {"text":"To find the subtraction formula for cosines start with: cosine of alfa plus minus beta. Using the adition formula you get: Cosine of alfa times cosine of minus beta minus sine 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":"Cosine of alfa minus beta then equals cosine of alpha times cosine of beta plus sine 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 cosine formulas. This leads to the conclusion that cosine of alpha plus beta and cosine of alpha minus beta can be expressed as cosine of alpha times cosine of beta minus sine of alpha times sine of beta, and cosine of alpha minus beta equals cosine of alpha times cosine of beta plus sine 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 cosine formulas. This leads to the conclusion that cosine of alpha plus beta and cosine of alpha minus beta can be expressed as cosine of alpha times cosine of beta minus sine of alpha times sine of beta, and cosine of alpha minus beta equals cosine of alpha times cosine of beta plus sine of alpha times sine of beta." } }}