Trig identities, etc.

[gecko5068]

Ive done around 80 of these, but these last ones are a bit tricky. Also, how do you plug in cos^2x, versus cosx^2 in a calculator (ti-86 btw)

1)cscx/sin^2x+secx+cos^2x-1 I think that one is cotx

2)cot^2x/cscx-1 try factoring and using fundamental identities, according to my teacher

3)prove sec^2x/sin^2x+cos^2x=1+tan^2x (I think both simplify to sec^2x

4)2sin^2xcosx=cosx solve by factoring and/or extracting square roots?

5)2sin^2x+3sinx-2=0 find all EXACT solutions in [0,2pi)

6)lcotxl <or= square root of 3, find all exact solutions in [0,2pi]

7)tan 15deg, using sum/difference identities, exact anwser

8)which identity is true?
a)sin^2x-cosx=1=2sinx
b)sin^3x-sinx=-cos^2xsinx

any help is GREATLY appreciated

 

[soroban]

Hello, gecko50681

Here are a few of them . . .

For the "solve" problems, I assume the domain is \(\displaystyle \,[0,\,2\pi)\)

\(\displaystyle 4)\;\;2\cdot\sin^2x\cdot\cos x\:=\:\cos x\)

We have: \(\displaystyle \,2\cdot\sin^2x\cdot\cos x\,-\,\cos x\:=\:0\)

Factor: \(\displaystyle \,\cos x(2\cdot\sin^2x\,-\,1)\:=\:0\)

Then: \(\displaystyle \,\cos x\:=\:0\;\;\Rightarrow\;\;x\,=\,\frac{\pi}{2},\;\frac{3\pi}{2}\)

And: \(\displaystyle \,2\cdot\sin^2x\,-\,1\:=\:0\;\;\Rightarrow\;\;\sin x\,=\,\pm\frac{1}{\sqrt{2}}\;\;\Rightarrow\;\;x\:=\:\frac{\pi}{4},\;\frac{3\pi}{4},\;\frac{5\pi}{4},\;\frac{7\pi}{4}\)

\(\displaystyle 5)\;\;2\sin^2x\,+\,3\sin x\,-\,2\;=\;0\)

Factor: \(\displaystyle \,(2\sin x\,-\,1)(\sin x\,+\,2)\:=\:0\)

Then: \(\displaystyle \,2\sin x\,-\,1\:=\:0\;\;\Rightarrow\;\;\sin x\,=\,\frac{1}{2}\;\;\Rightarrow\;\;x\,=\,\frac{\pi}{6},\;\frac{5\pi}{6}\)

And: \(\displaystyle \,\sin x\,+\,2\:=\:0\;\;\Rightarrow\;\;\sin x\,=\,-2\) . . . no solutions

\(\displaystyle 6)\;\;|\cot x|\:\leq\:\sqrt{3}\)

We have: \(\displaystyle \,-\sqrt{3} \:\leq\:\cot x\:\leq\:\sqrt{3}\)

Therefore: \(\displaystyle \,x\) is in \(\displaystyle \,\left[\frac{\pi}{6},\,\frac{5\pi}{6}\right]\,\) or \(\displaystyle \,\left[\frac{7\pi}{6},\,\frac{11\pi}{6}\right]\)

\(\displaystyle 7)\;\;\text{Find }\tan15^o\)

Formula: \(\displaystyle \,\tan(A\,-\,B)\;=\;\L\frac{\tan A \,-\,\tan B}{1\,+\,\tan A\cdot\tan B}\)

We have: \(\displaystyle \tan(45^o\,-\,30^o) \;= \;\L\frac{\tan45^o\,-\,\tan30^o}{1\,+\,\tan45^o\cdot\tan30^o} \;= \;\frac{1\,-\,\frac{1}{\sqrt{3}}}{1\,+\,1\cdot\frac{1}{\sqrt{3}}}\)

Multiply top and bottom by \(\displaystyle \sqrt{3}:\L\;\;\frac{\sqrt{3}\,-\,1}{\sqrt{3}\,+\,1}\)

Rationalize: \(\displaystyle \,\L\frac{\sqrt{3}\,-\,1}{\sqrt{3}\,+\,1}\,\cdot\,\frac{\sqrt{3}\,-\,1}{\sqrt{3}\,-\,1} \;= \;\frac{3\,-\,2\sqrt{3}\,+\,1}{3\,-\,1} \;= \;\frac{4\,-\,2\sqrt{3}}{2}\)\(\displaystyle \;=\;2\,-\,\sqrt{3}\)

 

[gecko5068]

1 and 3 I have solved, number two I still havent confirmed sinx^2 is right, and soroban thanks alot for the help. I hate asking for anwsers, so ill make sure to study up how you did it =)

thanks a ton

 

[gecko5068]

could anyone get me started one number two? the rest I understand pretty well now, thanks

 

[stapel]

could anyone get me started one number two?

What are the instructions for this exercise?

Thank you.

Eliz.

 

[gecko5068]

oh im sorry about that

just simplify

 

[stapel]

To "get started", try applying the suggestions. What identity, involving cot²(x), have you learned? Plug that in. Factor the resulting numerator. Cancel.

Eliz.

 

[gecko5068]

cot^2x is involved in 1+cot^2x=csc^2x

so cot^2x=csc^2x-1
and
csc^2x-1/cscx-1

numerator factored to csc(cscx-1)

so csc(cscx-1)/(cscx-1) is csc, doesnt the x get canceled out?

I must be making a stupid error.

 

[stapel]

How are you getting that csc²(x) - 1 is equal to csc(csc(x) - 1)? They're quite unrelated.

Eliz.

 

[gecko5068]

I dont know, ive been dealing with this since 5 o clock and im getting tired of it, maybe it would be best to just sleep on it and come back tomorrow morning.

Thanks for the help everyone