1. ## Trigonomic Identities

1. verify the following trigonometric identities and show all steps

let @=theta

1. cos^3@+sin^2@cos@=cos

2. cot@
---- = tan@
csc^2@-1

3. cot@csc@tan^2@=sec@

4. sin@ (1 + csc@)=sin@+1

5. cos@=sec@ - sin@tan@

6. 1+cos@ = cot@(sin@+tan@)

Answering any part of these would help a ton! Thanks.

2. Hello, daviskane!

Do you know any of the basic identities?

Verify the following trigonometric identities and show all steps.

$\displaystyle 1)\;\;\cos^3\theta + \sin^2\theta\cos\theta \:= \:\cos\theta$
$\displaystyle \text{Factor the left side: }\;\cos\theta\underbrace{(\cos^2\theta + \sin^2\theta)}_{\text{this is 1}} \;=\;\cos\theta\cdot1 \;=\;\cos\theta$

$\displaystyle 2)\;\frac{\cot\theta}{\csc^2\!\theta - 1} \:=\:\tan\theta$
$\displaystyle \text{We have: }\;\frac{\cot\theta}{\underbrace{\csc^2\!\theta-1}_{\text{this is }\cot^2\!\theta}} \;=\; \frac{\cot\theta}{\cot^2\theta} \;=\;\frac{1}{\cot\theta} \;=\;\tan\theta$

$\displaystyle 3)\; \cot\theta\csc\theta\tan^2\theta \:=\:\sec\theta$
The left side is: .$\displaystyle \frac{\cos\theta}{\sin\theta}\cdot\frac{1}{\sin\th eta}\cdot\frac{\sin^2\theta}{\cos^2\theta} \;=\;\frac{1}{\cos\theta} \;= \;\sec\theta$

$\displaystyle 4)\; \sin\theta (1 + \csc\theta)\:=\:\sin\theta + 1$
$\displaystyle \text{Multiply the left side: }$ .$\displaystyle \sin\theta\!\cdot\!1 + \underbrace{\sin\theta\cdot\csc\theta}_{\text{this is 1}} \;=\;\sin\theta + 1$

$\displaystyle 5)\; \cos\theta \:= \:\sec\theta - \sin\theta\tan\theta$
$\displaystyle \text{The right side is: }\;\frac{1}{\cos\theta} \,- \,\sin\theta\!\cdot\!\frac{\sin\theta}{\cos\theta} \;=\;\frac{1}{\cos\theta} - \frac{\sin^2\theta}{\cos\theta} \;= \frac{\overbrace{1 - \sin^2\theta}^{\text{this is }\cos^2\!\theta}}{\cos\theta}$

and we have: .$\displaystyle \frac{\cos^2\theta}{\cos\theta} \;=\;\cos\theta$

$\displaystyle 6)\; 1 + \cos\theta \: = \:\cot\theta(\sin\theta + \tan\theta)$

$\displaystyle \text{Multiply the right side: }\;\cot\theta\cdot\sin\theta \,+ \,\underbrace{\cot\theta\tan\theta}_{\text{this is 1}} \;=\;\frac{\cos\theta}{\sin\theta}\cdot\sin\theta \,+ \,1 \;= \;\cos\theta \,+ \,1$