a few problems having trouble with

**lsnyder** · April 21st 2009, 06:16 PM

okay I have like 50 problems to do and I have a few I need help with particularly. If someone can help, I would appreciate it so much.

Problem 1:
Prove the identity.

tan(t+(pi/2))=-cot(t).

Problem 2:
Rewrite sin(5t)-sin(3t).

Problem 3:
Solve 3 cos(theta)+3=2 sin^2(theta) where 0<=theta<2pi

Problem 4:
Solve cos(theta)+sin(theta)=1 where 0<=theta<2pi

**Soroban** · April 22nd 2009, 06:22 AM

Hello, lsnyder!

Here's some help . . .

(3) Solve: . $3\cos\theta+3\:=\:2\sin^2\!\theta\:\text{ where }0 \leq\theta < 2\pi$

Replace $\sin^2\!\theta$ with $1-\cos^2\!\theta$

$3\cos\theta + 3 \:=\:2(1-\cos^2\!\theta) \quad\Rightarrow\quad 3\cos\theta \:=\:2-2\cos^2\!\theta$

. . $2\cos^2\!\theta + 3\cos\theta + 1 \:=\:0 \quad\Rightarrow\quad (\cos\theta + 1)(2\cos\theta+1) \:=\:0$

Then: . $\begin{array}{cccccccccccc}<br /> \cos\theta + 1 \:=\:0 & \Rightarrow & \cos\theta \:=\:\text{-}1 & \Rightarrow & \theta \:=\:\pi \\<br /> 2\cos\theta + 1 \:=\:0 & \Rightarrow & \cos\theta \:=\:\text{-}\frac{1}{2} & \Rightarrow & \theta \:=\:\frac{2\pi}{3},\:\frac{4\pi}{3}\end{array}$

(4) Solve: . $\cos\theta+\sin\theta \:=\:1\:\text{ where }0 \leq \theta < 2\pi$

Square both sides: . $(\cos\theta + \sin\theta)^2 \:=\:1^2 \quad\Rightarrow\quad \cos^2\!\theta + 2\cos\theta\sin\theta + \sin^2\!\theta \:=\:1$

. . $\underbrace{2\sin\theta\cos\theta}_{\text{This is }\sin2\theta} + \underbrace{\sin^2\!\theta + \cos^2\!\theta}_{\text{This is 1}} \:=\:1 \quad\Rightarrow\quad \sin2\theta + 1 \:=\:1 \quad\Rightarrow\quad \sin2\theta \:=\:0$

. . $2\theta \:=\:0,\:\pi,\:2\pi,\;3\pi \quad\Rightarrow\quad \theta \;=\;0,\:\tfrac{\pi}{2},\;\pi,\;\tfrac{3\pi}{2}$

Since squaring an equation often introduces extraneous roots,
. . we must check our answers.

And we find that $\theta \:=\:\pi\,\text{ and }\,\tfrac{3\pi}{2}$ are extraneous.

Therefore, the answers are: . $\theta \;=\;0,\:\frac{\pi}{2}$

**lsnyder** · April 23rd 2009, 10:17 AM

okay, i got a question for number 1, which probably sounds so dumb but would it start this.....

tan(t)+tan(pi/2)=-cot (t)
tan(t)+tan(pi/2)=- 1/tan(t)

or....

sin/cos(t+(pi/2))=-cot(t)
sin/cos(t+(pi/2))=-1/tan(t)

or is that way off.

**lsnyder** · April 23rd 2009, 07:12 PM

i figured out that number 1 is an cofunction idenity

so it is set up as

tan(t+pi/2)=>(tan (t)-tan (pi/2)/(1+tan(t)tan(pi/2)

after that I am stuck.
does anyone know what to do

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