Verifying Trig Identities: 1-Sin^2 Theta/1-cos Theta=-cos th

V

Violagirl

Junior Member
Joined
Mar 9, 2008
Messages
87
Hi, I'm not sure how to go about doing these. I was wondering if anyone could help me on getting started with them? Thanks!

1. 1-Sin^2 Theta/1-cos Theta=-cos theta

2. 1 + sin theta/1-sin theta=(sec theta + tan theta)^2

3. tan theta+cot theta=sec theta csc theta.
 
Re: Verifying Trig Identities

Use the basic identities and definitions, e.g., sin⁡2(θ)+cos⁡2(θ)=1,tan⁡(θ)=sin⁡(θ)/cos⁡(θ)\displaystyle \sin^2(\theta)+\cos^2(\theta)=1, \tan(\theta) = \sin(\theta)/\cos(\theta), etc,
 
Re: Verifying Trig Identities

Hello, Violagirl!

1)    1−sin⁡2θ1−cos⁡θ = −cos⁡θ\displaystyle 1)\;\;\frac{1 - \sin^2\theta}{1-\cos\theta} \:=\:-\cos\theta
Click to expand...

There must be a typo . . . This is not an identity.


\(\displaystyle 2)\;\;\frac{1 +\sin\theta}{1-\sin\theta} \:=\:(\sec\theta + \tan\theta)^2\)
Click to expand...

The right side is:   (1cos⁡θ+sin⁡θcos⁡θ)2  =  (1+sin⁡θcos⁡θ)2  =  (1+sin⁡θ)2cos⁡2 ⁣θ  =  (1+sin⁡θ)21−sin⁡2 ⁣θ\displaystyle \text{The right side is: }\;\left(\frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta}\right)^2 \;=\;\left(\frac{1+\sin\theta}{\cos\theta}\right)^2 \;=\;\frac{(1+\sin\theta)^2}{\cos^2\!\theta} \;=\;\frac{(1+\sin\theta)^2}{1-\sin^2\!\theta}


. . Factor and reduce:   (1+sin⁡θ)(1+sin⁡θ)(1−sin⁡θ)(1+sin⁡θ)  =  1+sin⁡θ1−sin⁡θ\displaystyle \text{Factor and reduce: }\;\frac{(1+\sin\theta)(1 + \sin\theta)}{(1-\sin\theta)(1+\sin\theta)} \;=\;\frac{1+\sin\theta}{1-\sin\theta}


3)    tan⁡θ+cot⁡θ = sec⁡θcsc⁡θ\displaystyle 3)\;\;\tan\theta +\cot\theta\:=\:\sec\theta\csc\theta
Click to expand...

The left side is:   tan⁡θ+cot⁡θ  =  sin⁡θcos⁡θ+cos⁡θsin⁡θ\displaystyle \text{The left side is: }\;\tan\theta + \cot\theta \;=\;\frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta}

Get a common denominator and add:   sin⁡2 ⁣θ+cos⁡2 ⁣θ⏞This is 1cos⁡θsin⁡θ\displaystyle \text{Get a common denominator and add: }\;\frac{\overbrace{\sin^2\!\theta + \cos^2\!\theta}^{\text{This is 1}}}{\cos\theta\sin\theta}


. . So we have:  1cos⁡θsin⁡θ  =  1cos⁡θ⋅1sin⁡θ  =  sec⁡θcsc⁡θ\displaystyle \text{So we have: }\:\frac{1}{\cos\theta\sin\theta} \;=\;\frac{1}{\cos\theta}\cdot\frac{1}{\sin\theta} \;=\;\sec\theta\csc\theta

 
Thanks soroban! That did help a lot! I had a couple more that I'm stuck on:

1. 1-(sin^2 theta/1-cos theta)= -cos theta

(the first 1 is on the outside of the fraction.)

2. cos theta + 1/cos theta-1=1+sec theta/1-sed theta

3. sin(alpha+beta)/cos alpha cos beta=tan alpha+tan beta

Also is the third similar to using the regular properties? Or different from them?
 
Hello again, Violagirl!

1)    1−sin⁡2 ⁣θ1−cos⁡θ = −cos⁡θ\displaystyle 1)\;\;1-\frac{\sin^2\!\theta}{1-\cos\theta} \:=\:-\cos\theta
Click to expand...

On the left, we have:   1−sin⁡2 ⁣θ1−cos⁡θ    =    1−1−cos⁡2 ⁣θ1−cos⁡θ\displaystyle \text{On the left, we have: }\;1 - \frac{\sin^2\!\theta}{1 - \cos\theta} \;\;=\;\;1 - \frac{1-\cos^2\!\theta}{1-\cos\theta}

. . . =    1−(1−cos⁡θ(1+cos⁡θ)1−cos⁡θ    =    1−(1+cos⁡θ)    =    −cos⁡θ\displaystyle =\;\;1 - \frac{(1-\cos\theta(1+\cos\theta)}{1-\cos\theta} \;\;=\;\;1 - (1 + \cos\theta)\;\;=\;\;-\cos\theta



2)    cos⁡θ+1cos⁡θ−1 = 1+sec⁡θ1−sec⁡θ\displaystyle 2)\;\;\frac{\cos\theta + 1}{\cos\theta-1} \:=\:\frac{1+\sec\theta}{1-\sec\theta}
Click to expand...

On the right, we have:   1+sec⁡θ1−sec⁡θ    =    1+1cos⁡θ1−1cos⁡θ\displaystyle \text{On the right, we have: }\;\frac{1+\sec\theta}{1-\sec\theta} \;\;=\;\;\frac{1 + \frac{1}{\cos\theta}}{1 - \frac{1}{\cos\theta}}

Multiply by cos⁡θcos⁡θ ⁣:    cos⁡θ(1+1cos⁡θ)cos⁡θ(1−1cos⁡θ)    =    cos⁡θ+1cos⁡θ−1\displaystyle \text{Multiply by }\frac{\cos\theta}{\cos\theta} \!:\;\; \frac{\cos\theta\left(1 + \frac{1}{\cos\theta}\right)} {\cos\theta\left(1 - \frac{1}{\cos\theta}\right)} \;\;=\;\;\frac{\cos\theta+1}{\cos\theta-1}



3)    sin⁡(α+β)cos⁡αcos⁡β = tan⁡α+tan⁡β\displaystyle 3)\;\;\frac{\sin(\alpha+\beta)}{\cos\alpha\cos\beta} \:=\:\tan\alpha+\tan\beta
Click to expand...

This one is different . . .
We need a compound-angle identity: .sin⁡(A+B) = sin⁡Acos⁡B+sin⁡Bcos⁡A\displaystyle \sin(A + B) \:=\:\sin A\cos B + \sin B\cos A

On the left, we have:  sin⁡(α+β)cos⁡αcos⁡β   =   sin⁡αcos⁡β+sin⁡βcos⁡αcos⁡αcos⁡β\displaystyle \text{On the left, we have: }\:\frac{\sin(\alpha + \beta)}{\cos\alpha\cos\beta} \:\;=\;\:\frac{\sin\alpha\cos\beta + \sin\beta\cos\alpha}{\cos\alpha\cos\beta}

. . . =    sin⁡αcos⁡βcos⁡αcos⁡β+sin⁡βcos⁡αcos⁡αcos⁡β    =    sin⁡αcos⁡α+sin⁡βcos⁡β    =    tan⁡α+tan⁡β\displaystyle =\;\;\frac{\sin\alpha\cos\beta}{\cos\alpha\cos\beta} + \frac{\sin\beta\cos\alpha}{\cos\alpha\cos\beta} \;\;=\;\;\frac{\sin\alpha}{\cos\alpha} + \frac{\sin\beta}{\cos\beta} \;\;=\;\;\tan\alpha + \tan\beta

 
You must log in or register to reply here.
Top