Impossible Trigonometry Questions

**jhunt47** · November 16th 2009, 06:24 AM

K, so my teacher presented a set of problems and it is basically impossible to get all of them right (it was a competition to see who could do the most). There were a ton of problems and after going to class I still don't understand how to do some of them.

Disprove the following identity by counterexample:
#4
sin^2(x)+cos(x)=0

Everything that I plug in says that this isn't an identity. What am I doing wrong?

Prove the following identities:
#13
cotx-sinx=(cos^2(x)+cosx-1)csc(x)

I spent 2 hours trying to prove this, and I am coming with a nothing.

And the third one

Solve the following:
#14
1-2sin^2(2x)=3sin2x

I have a bunch of these that I didn't get right that I don't understand, but these 3 are really bugging me. I already know the answers, so I mostly care about how to get that answer because so far I haven't been able to figure it out. Any help will be great!!!

**pencil09** · November 16th 2009, 07:00 AM

Originally Posted by jhunt47

Prove the following identities:
#13
cotx-sinx=(cos^2(x)+cosx-1)csc(x)

Solve the following:
#14
1-2sin^2(2x)=3sin2x

i'll try "the lucky" #13

$cotx-sinx$
$=\frac{cosx}{sinx}-sinx$
$=\frac{cosx-sin^2x}{sinx}$

remember:
$sin^2x+cos^2x=1$ so,
$sin^2x=1-cos^2x$

$=\frac{cosx-(1-cos^2x)}{sinx}$
$=\frac{cosx-1+cos^2x}{sinx}$

remember:
$\frac{1}{sinx}=cscx$

$=(cosx-1+cos^2x)(cscx)$
---Q.E.D---

**Soroban** · November 16th 2009, 09:10 AM

Hello, jhunt47!

Here's #13 . . .

13. Prove: . $\cot x-\sin x\:=\:\left(\cos^2\!x+\cos x-1\right)(\csc x)$

The right side is: . $\bigg[\cos x - (1 - \cos^2\!x)\bigg]\cdot\frac{1}{\sin x}$

. . . . . . . . . . . $= \;\bigg[\cos x - \sin^2\!x\bigg]\cdot\frac{1}{\sin x}$

. . . . . . . . . . . $=\; \frac{\cos x}{\sin x} - \frac{\sin^2\!x}{\sin x}$

. . . . . . . . . . . $=\; \cot x - \sin x$

**Soroban** · November 16th 2009, 06:24 PM

Hello again, jhunt47!

Here's #14 . . . solved on the interval $[0,\:2\pi)$

Solve: . $(14)\;\;1-2\sin^2\!2x\:=\:3\sin2x$

We have: . $2\sin^2\!2x + 3\sin2x - 1 \:=\:0$

Quadratic Formula: . $\sin2x \:=\:\frac{-3 \pm\sqrt{3^2 - 4(2)(-1)}}{2(2)} \;=\;\frac{-3\pm\sqrt{17}}{4}$

We have two equations to solve:

$\sin2x \:=\:\frac{-3 - \sqrt{17}}{4} \:=\:-1.7807764067$ . . . no real solution

$\sin2x \:=\:\frac{-3 + \sqrt{17}}{4} \:=\:0.280776406 \quad\Rightarrow\quad 2x \:=\:\arcsin(0.280776406)$

. . $2x \:\approx\:\begin{Bmatrix}16.3^o \\ 163.7^o \\ 376.3^o \\ 523.7^o \end{Bmatrix} \quad\Rightarrow\quad x \;\approx\;\begin{Bmatrix}8.15^o \\ 81.85^o \\ 188.15^o \\ 261.85^o \end{Bmatrix}$

**jhunt47** · November 16th 2009, 06:53 PM

Thanks!!! You guys are the bomb-diggity!

I don't know if you guys will be able to help me, but I got halfway through on this problem and I think that I might be doing it wrong.

(16/81)^sin^2(x) + (16/81)^1-sin^2(x) = (26/27)

I know that you can substitute sin^2(x) for 1/4 or 3/4 and the problem will work, but the instructions say "Solve" and I don't think I am solving it very well.

**mr fantastic** · November 16th 2009, 06:58 PM

Originally Posted by jhunt47

K, so my teacher presented a set of problems and it is basically impossible to get all of them right (it was a competition to see who could do the most). There were a ton of problems and after going to class I still don't understand how to do some of them.

Disprove the following identity by counterexample:
#4
sin^2(x)+cos(x)=0

Everything that I plug in says that this isn't an identity. What am I doing wrong?

[snip]

Nothing. You've disproved it, as required.

Thread: Impossible Trigonometry Questions

LinkBack

Thread Tools

Display

Impossible Trigonometry Questions

RE

Similar Math Help Forum Discussions

Trigonometry Questions and identities.

Trigonometry Questions.

:(impossible factoring questions

3 Hard Trigonometry Questions

6 Trigonometry Questions needing help...

Search Tags