### UNIT III – Trigonometric Identities

This is John's scribe post. He had technical difficulties so I'm posting it for him.

Scribe Post:

October 13, 2006

Hi, I’m John, Scribe Post for today.

At the beginning of class, Mr. K asked the class to sketch the graph and

solve for the exact value of the following equation.

f (x) = 2 sin (π/2)x) - 3

g (x) = -2 cos [π/2 (x+1)] -3

Note: blue graph is the g (x)

Black graph is the f (x)

Finding the period:

Period = 2π/B

2 π over the reciprocal of B which is 2/π. The pi; will be

reduce

So, the period is 4.

Did you notice something? The graph is the same. Right? So, we can say

f(x)=g(x).

It looks different because they have different equation but still the graph

is identical.

Find the exact values of the following.

1. sin2 (π /6) + cos2 (π/6)

(1/2)2 + (√3/2)2 = 1

¼ + ¾ = 1

2. cos2 (5π/3) + sin2 (5π/3)

(½)2 + (-√3/2)2 =1

¼ + ¾ = 1

3. sin2 (-11π /4) + cos2 (-11π /4)

(-√2/2)2 + (-√2/2)2

(2/4) + (2/4) or

(1/2) + (1/2) = 1

Blue = sin π/6

Black = cos π/6

Therefore, we can say that the coordinates of the points are (sin π/6 ,

cos π /6). Which can also be written in their exact value (√3/2 ,

1/2).

We know that sine are equivalent to opposite over hypotenuse.

In the unit circle they can all be divided by 1.

Facts: There was an old ugly guy, Pythagoras, who invented the

Pythagorean theorem, a2 + b2 = 1, in the unit circle, we can say that

hypotenuse is always equal to 1.

ALGEBRAICALLY:

Sin2θ + cos2θ = 1

By Subtraction:

Sin2 θ = 1 – cos2 θ They are

still identical.

Cos2 θ = 1 – sin2 θ

Divide by sin2 θ

By Division: sin2 θ + cos2 θ = 1

Divide by cos2 θ sin2 θ

sin2 θ sin2 θ

Sin2 θ + cos2 θ = 1 __

Cos2 θ cos2θ cos2 θ 1 + cot2 θ = csc2

θ

Tan2 θ + 1 = sec2 θ

All in colored equations are call the Pythagoras Identities.

Note: All trigonometric function can be reduced only in sine and cosine.

The building blocks of All the trigonometric functions Sin2 θ = 1 –

cos2 θ

Cos2 θ = 1 – sin2 θ

The fundamental identities Tan2 θ = sin2 θ / cos2 θ

Csc2 θ = 1 / sin2 θ

Sec2 θ = 1 / cos2 θ

Cot2 θ = cos2 θ / sin2 θ

Therefore: tan θ * cos θ

= Sin θ * cos θ

Cos θ

= sin θ

In 2nd period, Mr. K., just gave us exercises about the trigonometric

identities.

1. sin θ sec θ

Sin θ * 1_

cos θ

= sin θ

Cos θ

= tan θ

2. csc x - cot x

Sin x tan x

1 - cos x

Sin x sin x

Sin x sin x

cos x

1 1 _ cos x cos x

Sin x sin x sin x sin x

1 _ cos2 x

Sin2 x sin2 x

1- cos2 x

Sin2 x

= sin2 x

Sin2 x

= 1

3. sec β - sec β cos β

cos β

1

= cos β - 1 * cos β

cos β cos β

= 1 - 1 - 1

Cos β cos β

= 1 __ - 1

Cos2 β

= sec2 β – 1

= tan2 β

4. 1 + tan2 β

Tan2 β

= sec2 β

Tan2 β

_ 1_

= Cos2 β

Sin2 β

Cos2 β

1 * cos2 β

Cos2 β sin2 β

= __ 1

Sin2 β

= csc2 β

PROVING IDENTITIES:

-2 cos (π/2 (x+1)) -3 = 2 sin (π/2)x)

– 3

These are the same graph but different equations.

Prove this;

?

Sec τ – cos τ =

_ 1 _ cos τ

Cos τ

= _ 1 _ cos2 τ

Cos τ cos τ

= 1- cos2 τ

Cos τ

= Sin2 τ

Cos τ

= sin τ * sin τ

cos τ

QED

Sin τ tan

τ

Sin τ * sin τ

cos τ

= sin2 τ

cos τ

= 1 - cos2 τ

Cos τ

= 1 _ cos2 τ

Cos τ cos τ

= sec τ - cos τ

QED

Q = QUOD

E = ERAT

D = DEMONS

RHS = LHS

5 CLUES:

1. Look at the expression that has complication for you.

2. Rewrite the equation into sine and cosine form.

3. If there are fraction, simplify the compound fraction into simple

fraction.

4. Use the Pythagorean substitution.

5. Where possible, try to factor the equation.

Sin4 θ - cos4 θ

(sin2θ – cos2θ)(sin2θ + cos2θ)

= Sin2θ - cos2θ Sin2 θ – cos2 θ

QED

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