### 25th SCRIBE

Hi everyone, today's scribe is MARK.

today was a 1 period class and we did some questions. =P

first we had a little more practise on identities.

1) sec

*x*+ csc

*x*= (tan

*x*+ cot

*x*)(cos

*x*+ sin

*x*)

As we all know there are many ways to solve these problems. If you "correctly" worked one of the sides and then the other, you should have ended up with equal answers. One way to solve this problem is shown below:

__Right Side of "="__[(sin x/cos x) + (cosx/sinx)][cos x + sin x] find the common denominator for the first part, inside the yellow brackets

[( sin

^{2 }x + cos

^{2}x)/(cos x)(sin x)][cos x + sin x] multiply the common denominator to the numerator and denominator, you get the values shown in this line. You may also combine the two with the same denominator.

[1/(cos x sin x )][cos x +sin x] sin2 x + cos2x is replaced by 1 because sin2 x + cos2x= 1. multiply.

[cos x/(cos x sin x)] + [sin x/(cos x sin x)] if you notice in this line we multiplied and seperate the one fraction into two. It is still the same value, but we have split them apart.

(1/ sin x) + (1/ cos x) we know that these are the reciprocal functions...

csc x + sec x so we can name them appropriately

2) (sec y -1)/(sin^{2}y)= (sec^{2}y)(1+ sec y)

For this question we can simplify both sides. It would would be best to start with the right side since it has secant and the left side is partly simplified. ut i will show how to do the left side, instead of the right =).

__Left Side__

(1 cos y -1)/(sin ^{2} y)

[(1/cos y) - (cos y/cos y)] / [sin^{2 }y] turn everything into sine and cosine

[(1- cos y)/ cos y] ^{.} [1/sin^{2 }y] 1 = cos y/ cos y

[(1-cos y)/cos y] ^{.} [1/(1-cos^{2 }y)] multiply the first part of the expression by the reciprocal of the second(instead of dividing).

[(1-cos y)/cos y]** ^{.}** [1/((1- cos y)(cos y))] 1-cos2 y can e factored by the difference of squares

[(1-cos y)/cos y] ^{.} [1/(1+cos y)] final answer

you can use similar steps from this (left) side on the right side.

**NEXT**

Using these fractions or multiples of them (numerators only)

1/5, 1/4, 1/3, 1/2

find two who add or differnce is

a) 5/ 12 = (2/3) - (1/4)

b) 1/12 = (1/3) - (1/4)

c) 7/12 = (1/3) + (1/4)

d)11/ 12 = (5/4) - (1/3)

last thing of the day we learned some dance moves to an equation... it was fun and hilarious.

heres the song

sine sine cos cos sine

cos cos cos sign change sine sine

sine(alpha+ beta)= sine alpha cos beta __+__ cos alpha sine beta

cos (alpha + beta) - cos alpha cos beta __+__ sine alpha sine beta

We used these equations to solve one problem... I think the next scribe will post these equations ...so check out the next scribe post!!!.

hmm...Anh will be the next scribe =), YOu have three days =P...at least you don't have it on a week day =\

## 1 Comments:

Mark,

outstanding scribe!!Excellent use of colour and well organized. Your post is an example for all scribes who follow.

Hey! Y'know, you schould write a textbook. ;-)

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