Addition and Subtraction Trig. Identities Questions

[theonenonlynat]

In Trig. class we are learning Identities. I completely understand the multipication, dividing, factoring, etc problems. Except I was sick when my class learnt the addition and subtraction problems. She told me I would still have to take the test tomorrow, but she gave me a worksheet but it doesn't tell me how to do it so I am VERY confused. Could someone please give me a step-by-step instructions on how to do the following problems that I have spent hours on??

csc X - sin X = (cot X)(cos X)
another one like it:
sec X - cos X = (tan X)(sin X)

(cot X)(cos x) + sin X = csc X

(cos X)/(1 + sin X) - (1- sin X)/ (cos X) = 0

(1)/(1-sin X) + (1)/(1+ sin X) = 2 sec ^2 X

ANY help is appreciated. Thanks in advance.

 

[soroban]

[Hello, theonenonlynat!

I'll do a couple of them and hope you can do the others.

I'll assume you know the basic identities and can follow my substitutions.
\(\displaystyle \;\;\)I also hope you know how to add and subtract fractions . . .

\(\displaystyle \csc x\,-\,sin x\:=\\cot x)(\cos x)\)

The left side is:
\(\displaystyle \L\;\;\;\frac{1}{\sin x}\,-\,\sin x\;=\;\frac{1\,-\,\sin^2x}{\sin x}\;=\;\frac{\cos^2x}{\sin x}\;=\;\frac{\cos x}{\sin x}\cdot\cos x\;=\;\cot x\cdot\cos x\)

\(\displaystyle \frac{\cos x}{1\,+\,\sin x}\,-\,\frac{1\,-\.\sin x}{\cos x}\:=\:0\)

Get a common denominator and add on the left side:

\(\displaystyle \L\;\;\;\frac{\cos x}{\cos x}\cdot\frac{\cos x}{1\,+\,\sin x}\,-\,\frac{1\,+\,\sin x}{1\,+\,\sin x}\cdot\frac{1\,-\,\sin x}{\cos x}\;=\;\frac{\cos^2x\,-\,(1\,+\,\sin x)(1\,-\,\sin x)}{\cos x(1\,+\,\sin x)}\)

\(\displaystyle \L\;\;\;=\;\frac{\cos^2x\,-\,(1\,-\,\sin^2x)}{\cos x(1\,+\,\sin x)} \:=\:\frac{\cos^2x\,-\,1\,+\,\sin^2x}{\cos x(1\,+\,\sin x)} \;= \;\frac{(\sin^2x\,+\,\cos^2x)\,-\,1}{\cos x(1\,+\,\sin x)}\)

Since \(\displaystyle \,\sin^2x\,+\,\cos^2x\:=\:1\), we have: \(\displaystyle \L\:\frac{1\,-\,1}{\cos x(1\,+\,\sin x)} \:=\:\frac{0}{\cos x(1\,+\,\sin x)} \:=\:0\)

 

[theonenonlynat]

thank you but...

thank you so much for replying, but I am still confused...

when you made 1/(sin x) - sin x into 1- sin ^2 x/ sin x, i dont understand what you did to make it squared...wouldn't it be 1- sin x/ sin x ???

 

[galactus]

Soroban just cross-multiplied.

\(\displaystyle \frac{1}{sin(x)}-sin(x)=\frac{1-sin^{2}(x)}{sin(x)}\)

Since \(\displaystyle 1-sin^{2}(x)=cos^{2}(x)\), we have

\(\displaystyle cos(x)\frac{cos(x)}{sin(x)}= \frac{cos^{2}(x)}{sin(x)}\)

=\(\displaystyle cot(x)cos(x)\)

 

[theonenonlynat]

okay thank you so much, i did the rest of the worksheet using cross multiplication and i got them all right, and i understand them now.

i better do good on this test!

 

[soroban]

Re: thank you but...

Hello, theonenonlynat!

I was afraid of that . . . you don't know how to add/subtract fractions . . .

when you made \(\displaystyle \;\frac{1}{\sin x}\,-\,sin x\;\) into \(\displaystyle \;\frac{1\,-\,\sin^2 x}{\sin x}\),

i dont understand what you did to make it squared...

wouldn't it be \(\displaystyle \;\frac{1\,-\,\sin x}{\sin x}\) ?

Oh, sure . . . \(\displaystyle \frac{1}{2}\,+\,3\;=\;\frac{1\,+\,3}{2}\;=\,2\) . . . right?

To add or substract fraction, we need a common denominator.
If we don't have one (the denominators are different),
\(\displaystyle \;\;\)we must <u>force</u> them (legally) to be the same.


We have: \(\displaystyle \L\,\frac{1}{\sin x}\,-\,\frac{\sin x}{1}\) . . . different denominators.


Multiply the second fraction by \(\displaystyle \frac{\sin x}{\sin x}\)

\(\displaystyle \L\;\;\;\frac{1}{\sin x}\,-\,\frac{\sin x}{\sin x}\cdot\frac{\sin x}{1} \;= \;\frac{1}{\sin x}\,-\,\frac{\sin^2x}{\sin x}\) . . . same denominators!

Now we can subtract them . . .