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Study Notes 033:
Trigonometry 7:
By William E. Steinman:
Following up on Study Note 032, we have more formulas. These are similar to those we have already encountered.
Products of sins and cosines.
sin a cos b = ½(sin (a + b) + sin(a - b))
cos a sin b = ½(sin (a + b) – sin(a – b))
cos a cos b = ½(cos (a + b) + cos(a – b))
sin a sin b = -½(cos (a + b) – cos(a – b))
Sum and differences.
sin a + sin b = 2 sin ½(a + b) cos ½(a - b)
sin a – sin b = 2 cos ½(a + b) sin ½(a – b)
cos a + cos b = 2cos ½(a + b) cos ½(a – b)
cos a – cos b = -2sin ½(a + b) sin ½(a – b)
Of course, these equations are derived from identities or equations we discovered earlier.
Let us look at our first product equation sin a cos b = ½(sin (a + b) + sin(a - b))
From our basic addition formulas, Study Notes 32, recall these formulas.
sin (a + b) = sin a cos b + cos a sin b
sin (a - b) = sin a cos b – cos a sin b
We proved the first one in the notes and, of course, you proved the other on by yourself.
Given these equations then:
sin (a + b) + sin(a - b) is just the sum of the two equations or,
(sin a cos b + cos a sin b) + (sin a cos b – cos a sin b)
When we reduce that we find it comes to 2 sin a cos b
Therefore, the formula becomes sin a cos b = ½(2sin a cos b) or
sin a cos b = cos a sin b
Okay, that’s one proof. I strongly urge you, if you want to become proficient at this, to prove the other seven on your own. You will need lots of scratch paper. You will find a good source of scratch paper in the unsolicited garbage you find every day in your US Mailbox. Quite often, those fools only use one side of the paper for their garbage. You can use the other side before you throw it away.
Up until now we have been dealing with right triangles. Well, we know the world is not all that neat. So, let us look at oblique triangles.
Oh, oh!
The basic definition is, an oblique triangle is one which does not contain a right angle. This means it will have at least two acute angles. The third angle could be acute or obtuse.
Let’s draw a oblique triangle.
We have angle A with its opposite side being a.
We have angle B with its opposite side being b.
We have angle C with its opposite side being c.
We have some laws about these kinds of triangles.
First, we have the laws of sins.
a/sin A = b/sin B = c/sin C
That is equivalent to sin A/a = sin B/b = sin C/c.
We also have the laws of cosines.
a2 = b2 + c2 – 2bc cos A
b2 = a2 + c2 – 2ac cos B
I bet you can guess the next one.
c2 = a2 + b2 – 2ab cos C
Of course, we should be able prove these.
Let’s try to prove a/sin A = b/sin B = c/sin C
Or ability to prove this stuff, depends on our ability to turn any triangle into some combination of right triangles.
That is easy with our triangle above.
We simply draw a perpendicular line h from side a to angle A and we have divided our oblique triangle into two right triangles.
We know from our basic relationships that the sin off an angle = side opposite/hypotenuse.
That means sin C = h/b or h = b sin C.
Also h = a sin B
So a sin B = b sin C
Or a/sin A = b/Sin B
We could also draw perpendiculars from side b to angle B or from side A to angle A e could derive the final relationship. You can see that we would finally get the law of sins.
a/sin A = b/sin B = c/sin C
You can use a similar technique to derive and thus prove the laws of cosines.
I leave it to you. Practice makes perfect, so I have been told.