Proving Sine Identities
Date: 05/16/98 at 13:02:36 From: Becka Susan Subject: Proofs of Trigonometry I need to prove the functions of the sum of two angles: sin(A + B)= sin(A)cos(B) + cos(A)sin(B) I can use this identity to find the sin(75). You put sin(45 + 30) into the equation. I know it works. I just can't figure out why it works. I also need to figure out why: sin(2A) = 2sin(A)cos(A) and why: sin(A/2) = +/- the square root of (1 - cos (A/2)) Maybe if you show me how to do my first question I could figure out the other ones using information from that. Thanks for your help!
Date: 06/05/98 at 13:10:09 From: Doctor Mateo Subject: Re: Proofs of Trigonometry Hello Becka, The key to all trigonometric proof is understanding the connection of the relations you are being asked to examine with respect to all other geometric relations you have studied so far. Trigonometric proof is rich in connecting geometric relations with algebraic results. In order to prove sin(A + B)= sin(A)cos(B) + cos(A)sin(B), you may want to consider understanding some of the identities available and their geometric understanding. A good identity to look at would be: cos(A - B) = cos(A)cos(B) + sin(A)sin(B) In the proof of cos(A - B) = cos(A)cos(B) + sin(A)sin(B), you begin by drawing a unit circle (a circle with radius 1). Place two distinct points, call them A and B, on the circle. These points will have coordinates (cos(A), sin(A)) and (cos(B), sin(B)). Do you see why? Then consider the angle AOB, formed with A, B, and the center of the circle O. From geometry, you should remember that two points form a line and that the distance between these two points can be calculated by using the distance formula. Remember the distance formula? d = sqrt((x2 - x1)^2 + (y2 - y1)^2) From trigonometry you also know of a second way to compute the length of a line segment - the law of cosines. The law of cosines can be written as: (AB)^2 = (A0)^2 + (B0)^2 - 2(AO)(BO) cos(angle O) when working with triangle AOB. By using the law of cosines, you have: (AB)^2 = 1^2 + 1^2 - 2*1*1*cos(A - B)= 2 - 2cos(A - B) From algebra you have a very useful property called the transitive property that can be applied now. That is, if A = B and A = C, then B = C. So if: (AB)^2 = 2- 2cos(A - B) and (AB)^2 = 2 - 2(cos(A)cos(B) + sin(A)sin(B)) then: 2 - 2 cos(A - B) = 2 - 2(cos(A)cos(B) + sin(A)sin(B)) which gives us: cos(A-B) = cos(A)cos(B) + sin(A)sin(B) After working through the algebra and geometry involved in the above identity, you need to think about other relations that relate sin(x) to cos(x) in some way. If you consider a right triangle, you should remember that the two acute angles of a right triangle are complementary. Two very useful relations are an immediate result of knowing that the two acute angles of a right triangle are complementary: (i) sin(x) = cos((pi/2) - x) and (ii) cos(x) = sin((pi/2) - x) So if you let x = A + B where A and B represent the measure of the two angles that you want to add, you can use the complementary function relation in (i) above so that after substituting A + B into (i) you have: sin(A + B) = cos ((pi/2) - (A + B)) or cos((pi/2) - A - B) = cos (((pi/2) - A) - B) Then by applying the identity for cos (A - B), you have: cos((pi/2) - A)cos(B) + sin((pi/2) - A)sin(B) Use the complementary function relations to conclude the proof. For the proof of sin(2A), you can use the result of the proof you have just completed. Think about another way to write 2A. For example, 3A = 2A + A. For the half-angle formula you will need to use the formula for the double angle of cosine (that is, cos(2A) = 1 - 2 (sin(A))^2 where A = x/2). Keep up the good work. -Doctor Mateo, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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