right triangle trig review

Right Triangle Trig Review

Given the right triangle from the origin to the point (x, y) with the angle , we can find the following trig functions:

cos

cos

xr

x r

sin

sin

yr

y r

cos , sinr r

Replacing (x, y) with these new values, we get the point as:

cos , sinr r

Moving to the circle centered at the origin

Moving to the circle centered at the origin with radius “r”, we find two points A and B.

( ,0)r

( cos , sin )r r

We can use the distance formula to find the distance AB.

( ,0)r

( cos , sin )r r

2 2

2 1 2 1D x x y y

2 2cos sin 0AB r r r

Next, construct the angle in a circle with the same radius r. Using the SAS property, the triangle AOB in the previous example is congruent to the triangle COD in this example. Therefore, the length of segment AB must equal the length of segment CD.

It must also be true that u v

Finding points C and D and the length CD, we get:

( cos , sin )r u r u

( cos , sin )r v r v

2 2cos cos sin sinCD r u r v r u r v

By similar triangles, we know the length of AB = length of CD.

2 2 2 2cos sin 0 cos cos sin sin

AB CD

r r r r u r v r u r v

We can square both sides to get rid of the square roots.

2 22 2cos sin 0 cos cos sin sinr r r r u r v r u r v

Simplifying by squaring each group, we get:

2 2 2 2 2 2cos 2 cos sinr r r r 2 2 2 2 2 2 2 2 2 2cos 2 cos cos cos sin 2 sin sin sinr u r u v r v r u r u v r v

Every term has an r2. Divide each term by r2.

2 2cos 2cos 1 sin 2 2 2 2cos 2cos cos cos sin 2sin sin sinu u v v u u v v

Using the pythagorean identity, we know 2 2cos sin 1

2c2 2cos sos in1

2cos cos 2sin si2 2cos si2 2cos sin nnvu vu u vu v

1 2cos 1 1 2cos cos 1 2sin sinu v u v

Simplifying, we get:

2 2cos 2 2cos cos 2sin sinu v u v

Subtracting the 2’s from each side, we get:

2cos 2cos cos 2sin sinu v u v Each term has a -2, so divide out the -2.

cos cos cos sin sinu v u v

u v However, recall that

Replacing in the equation, we get:

cos cos cos sin sinu v u v

cos cos cos sin sinu v u v u v

To find a rule for , we replace v with –v. cos u v

cos cos cos sin sinu v u v u v

Simplifying with odd/even rules, we get:

cos cos cos sin sinu v u v u v

cos cos cos sin sinu v u v u v

To get the sum/difference rules for sin, we will use the co-function rule.

cos sin2 x x

Let’s use the cosine rule to find cos2u v

sin cos cos cos2 2 2u v u v u v u v

Using the cosine sum rule

cos cos cos sin sin2 2 2u v u v u v

Using the co-function rules, we get:

sin cos cos sinu v u v

Therefore:

sin sin cos cos sinu v u v u v

To get the sin(u+v) rule,

sin sin cos cos sinu v u v u v

Using the odd/even functions, we get:

sin sin cos cos sinu v u v u v