To find out sin(α+β), let’s build a diagram of
what we’re actually, trigonometrically doing.
Let’s make a triangle with angle β on it.
If its hypotenuse is some number H, then: opposite=H∙sin(β) adjacent=H∙cos(β)
Keep this in mind for later.
On top of its hypotenuse, let’s add a triangle with angle α on it.
Using the same intuition from △β,
we could figure out its sides, given its hypotenuse…
What’s this? Just by stacking triangles, we’ve made a
third triangle out of α+β! Let’s just draw
a dotted line so that we realize it’s there.
Remember the intuition we talked about, where all we
need is the hypotenuse to find the sides?
Let’s set the hypotenuse of △(α+β)
to 1. This way, if we find its sides, we’ll have
both:
1∙cos(α+β) 1∙sin(α+β)
Immaculate.
However, when we extend the logic to △β,
we notice that our goal, the sides of
△(α+β), is still out of reach.
Now, we have to get crafty. Let’s get the very top angle.
If we think back to geometry, we can draw a line
parallel to the bottom line to discover a secret 4th
right triangle. Parallel lines crossed by a
transversal!
Now all we do is hop between β and 90−β
to get to the top.
With ∠β and the hypotenuse sin(α),
we solve for the sides of our 4th triangle…
And just like that, we have the final things we
need to solve for △(α+β).
Because the cos double-angle formula can be expressed in both
a sin-only form and a cos-only form, we can use cos2α to solve
for the half-angle formulas.
cos2αcosα1−cosα21−cosα±21−cosαcos2αcosαcosα+12cosα+1±21+cosα=1−2sin2(α) Sine-friendly version=1−2sin2(2α)Solve for sin(2α)=2sin2(2α)=sin2(2α)=sin(2α)=2cos2α−1 Cosine-friendly version=2cos2(2α)−1Solve for cos(2α)=2cos2(2α)=cos2(2α)=cos(2α)
Hey, what does the plus-minus mean?
If a half-angle formula has a ± in it, we need to
consider the quadrant of where the new angle ends up, to
and then pick either + or −. Think of it this way:
cos(100°) is in quadrant II and thus negative
cos(2100°)=cos(50°) is in quadrant I
and thus positive. Although our formula says ±, we need to
recognize that a cosine of a quadrant I angle is positive.
For tangent, we once again apply tanθ=cosθsinθ:
tan(2α)tan(2α)=cos(2α)sin(2α)=±21+cosα±21−cosα=±1+cosα±1−cosα Rationalize the denominator!=±1+cosα±1−cosα∙1−cosα1−cosα=±12−cos2α±(1−cosα)2=±sin2α±(1−cosα)2 Pythag. Identity=sinα1−cosα
Interestingly, we get to remove the ± because of a very interesting
case-by-case basis that appears: instead of us needing to select
+ or −, the angle functions do it for us.
Look up at formula 12 again. No matter what angle α we have, the
numerator is greater than or equal to zero. (1−cosα)≥0
So, we only need to check if the signs of
sinα and tan2α match for any angle α.
And they do:
If 2α lands in quadrant I, then α is in quadrant I or II.
tan is positive in quadrant I
sin is positive in quadrant I and II
If 2α lands in quadrant II, then α is in quadrant III or IV.
tan is negative in quadrant II
sin is negative in quadrant III and IV
If 2α lands in quadrant III, then α is in quadrant I or II (wraps around).
tan is positive in quadrant III
sin is positive in quadrant I and II
If 2α lands in quadrant IV, then α is in quadrant III or IV (wraps around).