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### Proving Sine Identities

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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.

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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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Associated Topics:
High School Trigonometry

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