Angle Sum Trigonometric Identities

SAT Questions that focus on Angle Sum Trigonometric Identities require knowledge of the following topics.

Summary of the most important Angle Sum and Difference Identities

$sin(a+b)=sin(a)cos(b)+sin(b)cos(a)$

$sin(2a)=2sin(a)cos(a)$

$sin(a-b)=sin(a)cos(b)-sin(b)cos(a)$

$cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$

$cos(a-b)=cos(a)cos(b)+sin(a)sin(b)$

$tan(a+b)=\frac{tan(a)+tan(b)}{1-tan(a)tan(b)}$

$tan(2a)=\frac{2tan(a)}{1-tan^2(a)}$

$tan(a-b)=\frac{tan(a)-tan(b)}{1+tan(a)tan(b)}$

Angle Sum Formula for the Sine

Consider the Unit Circle in the following figure, where $(CD)=sin(a+b)$:

Triangles AOB, ODB and ABC are all right triangles.

In triangle OBD:
$sin(a)=(BD)/(OB)$
$sin(a)=(BD)/cos(b)$
$(BD)=sin(a)cos(b)$

In triangle ABC:
$cos(a)=(BC)/(AB)$
$cos(a)=(BC)/sin(b)$
$(BC)=sin(b)cos(a)$

Since:
$(CD)=(BD)+(BC)$
$(CD)=sin(a)cos(b)+sin(b)cos(a)$, that is

$sin(a+b)=sin(a)cos(b)+sin(b)cos(a)$

If angles "a" and "b" are equal:

$sin(a+b)=sin(a)cos(b)+sin(b)cos(a)$
$sin(a+a)=sin(a)cos(a)+sin(a1)cos(a)$
$sin(2a)=2sin(a)cos(a)$

Now let's find the angle difference formula for the sine:

$sin(a-b)=sin(a+(-b))$
$sin(a-b)=sin(a)cos(-b)+sin(-b)cos(a)$

Given that $sin(-b)=-sin(b)$ and $cos(-b)=cos(b)$,

$sin(a-b)=sin(a)cos(b)-sin(b)cos(a)$

Angle Sum Formula for the Cosine

Consider the Unit Circle in the following figure, where $(OE)=cos(a+b)$:

Triangles AOB, ODB, ABC and OEA are all right triangles.

In triangle ABC:
$sin(a)=(AC)/(AB)$
$sin(a)=n/sin(b)$
$n=sin(a)sin(b)$

In triangle ODB:
$cos(a)=(OD)/(OB)$
$cos(a)=(cos(a+b)+n)/cos(b)$
$cos(a+b)+n=cos(a)cos(b)$
$cos(a+b)=cos(a)cos(b)-n$

Since $n=sin(a)sin(b)$:

$cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$

Now let's find the angle difference formula for the cosine:

$cos(a-b)=cos(a+(-b))$
$cos(a-b)=cos(a)cos(-b)-sin(a)sin(-b)$

Since $sin(-b)=-sin(b)$ e $cos(-b)=cos(b)$:

$cos(a-b)=cos(a)cos(b)+sin(a)sin(b)$

Angle Sum Formula for the Tangent

The tangent of an angle is given by: $tan(x)=sin(x)/cos(x)$

Thus, $tan(a+b)=sin(a+b)/cos(a+b)$

$tan(a+b)=[sin(a)cos(b)+cos(a).sin(b)]/[cos(a)cos(b)-sin(a)sin(b)]$

Dividing the nominator and the denominator of the fraction by $cos(a)cos(b)$:

$tan(a+b)=[sin(a)/cos(a)+sin(b)/cos(b)]/[1-(sin(a)/cos(a))(sin(b)/cos(b))]$

$tan(a+b)=\frac{tan(a)+tan(b)}{1-tan(a)tan(b)}$

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