Trigonometric Identities
Installing wardrobes, AKA rsin(θ+α)
Although you may be told this
is for calculating how wardrobes fit through doors, it’s actually about triangles
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that’s simple
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this is the same as above but with...
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& 
…this substituted in
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soh
cah toa, tanθ = b/a
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that’s
just Pythagoras’ theorem
What happens?
You get a question like “solve
this: 6sinθ + 9cosθ = 7 for 0 ≤ θ ≤ 360”.
- First
find a value for r.
- Second
find a value for α.
- Third
sub r and α into rsin(θ+α)=y and rearrange for the value of
θ that maps onto y.
- Finally
sketch your value of θ on the graph y = sin x, then calculate the other
values of θ in the given range that map onto y.
Stage one is filling in some gaps. Generally it’s asinθ + bcosθ = y. Here it’s
6sinθ + 9cosθ = 7.
6 equals side a.
9 equals side b.
Now you can fill in r = √(a2
+ b2) with a and b:
r = √(62 + 92)
= 10.82.
Secondly, work out angle α. α = tan-1(b/a) = tan-1(9/6)
= 56.31°.
The third and final step is to sub r, α and y into the
general equation rsin(θ+α)
. 10.82 sin (θ + 56.31) = 7.
As George W would say, rearrangify! To isolate θ I mean.
sin
(θ + 56.31) = 7 / 10.82 = 0.647
sin-1
0.647 = θ + 56.31
(sin-1
0.647) – 56.31 = θ
θ = -16 (to
2 s.f.)
If we superimpose the graphs of y = sin (θ + 56.31) and
y = 0.647, the lines cross where θ = -16.
But it they also cross a few times between θ = 0 and θ =
360.
To work out where, we have to compare our value of θ to
a peak or trough on the graph. sin (θ + 56.31) shifts the graph y = sinθ to the
left by 56.31°.
A peak that would normally be at 90° is now at (90-56.31)°, or 33.69°. -16°
is 49.69° less than this. So sin (θ
+ 56.31)° must be equal to 0.647 where θ is ± 49.69° from a positive peak,
i.e. at 33.69° + 49.69°, and at (270-56.31)° or 213.69° + 49.69°.
These values of θ are 83.38° and 263.38°.
Addition formulae
These formulae tell you how
to add up the cosine and sine of two angles
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You get given them in the
exam, in this format
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Double angle formulae
However, these ones you gotta
learn
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Life changing formula
The Swiss army knife of maths
Useful triangles
Proving that three sides are
better than one. Erm… use these to find
the surd value for irrational values of y for sines and cosines of common
angles in the function f(x) = sinx.
