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### Sine and Cosine Addition Formulas

```
Date: 11/29/2000 at 00:55:14
From: Phara
Subject: Deriving addition formulas for trig

How do you prove this?

sin(x+y) = sin(x)cos(y) + cos(x)sin(y)

If sin(2Q) = a, then what is sin(Q) + cos(Q) in terms of a?
```

```
Date: 11/29/2000 at 12:14:08
From: Doctor Rob
Subject: Re: Deriving addition formulas for trig

Thanks for writing to Ask Dr. Math, Phara.

To derive the above identity, draw a diagram as follows. Start with a
Cartesian rectangular coordinate system with origin O. Pick a point A
on the positive x-axis. Construct ray OB so that <AOB has measure x.
Construct ray OC so that <BOC has measure y. Then <AOC has measure
x + y. Erect a perpendicular to OA at A intersecting OB at D. Erect a
perpendicular to OB at D intersecting OC at E. Drop a perpendicular
from E to OA, intersecting it at F, and intersecting OD at G. Now you
have several right triangles, OAD, ODE, OFE, OFG, and GDE, each
containing one of the angles with measure x, y, and x + y:

^             /C
|           E/
|           /\
|          /|x\
|         / |  \
|        /  |   \
|       /   |    \             B
|      /    |     \       _,-'
|     /     |      \D _,-'
|    /      |     _,o'  |
|   /      G| _,-'  |
|  /      _,+'      |
| /y  _,-'  |       |
|/_,-'x     |       |
--o-----------o-------o--->
O|           F       A

Now let OE have length r. Then, using triangle ODE and the definitions
of sine and cosine,

OD = r*cos(y)
DE = r*sin(y)

Next, using triangle GDE,

EG = r*sin(y)/cos(x)
DG = r*sin(y)*tan(x)

That implies that

OG = OD - DG
= r*cos(y) - r*sin(y)*tan(x)
= r*[cos(y)-sin(y)*tan(x)]

Now using triangle OFG,

FG = OG*sin(x)
= r*[sin(x)*cos(y)-sin(y)*tan(x)*sin(x)]

That tells you the length

EF = EG + FG
= r*sin(y)/cos(x) + r*[sin(x)*cos(y)-sin(y)*tan(x)*sin(x)]
= r*[sin(x)*cos(y)-sin(y)*sin^2(x)/cos(x)+sin(y)/cos(x)]
= r*[sin(x)*cos(y)+sin(y)*[1-sin^2(x)]/cos(x)]
= r*[sin(x)*cos(y)+sin(y)*cos^2(x)/cos(x)]
= r*[sin(x)*cos(y)+cos(x)*sin(y)]

On the other hand, using triangle OFE,

EF = r*sin(x+y)

Setting these last two equal and dividing out the r gives

sin(x+y) = sin(x)*cos(y) + cos(x)*sin(y)

To derive the cosine identity, consider

cos(x+y) = sin(pi/2-[x+y])
= sin([pi/2-x]+[-y])

Now use the sine identity proved above:

cos(x+y) = sin(Pi/2-x)*cos(-y) + cos(Pi/2-x)*sin(-y)
= cos(x)*cos(y) - sin(x)*sin(y)

sin(2*Q) = a
2*sin(Q)*cos(Q) = a
1 + 2*sin(Q)*cos(Q) = 1 + a
sin^2(Q) + 2*sin(Q)*cos(Q) + cos^2(Q) = 1 + a
[sin(Q)+cos(Q)]^2 = 1 + a
sin(Q) + cos(Q) = +-sqrt(1+a)

The sign to use is determined by which quadrant Q + pi/4 might lie in.
If Q + pi/4 is in the first or second quadrant, then the positive sign
is correct, and otherwise the negative sign is the one to use. This is
because

sin(Q) + cos(Q) = sqrt(2)*sin(Q+Pi/4)

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Trigonometry

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