Stuck On Some Trig Identities

**Hoopsaholic31** · June 5th 2007, 04:00 PM

I was having trouble with a few problems none of my friends could help me with, so I figured I'd give this forum a try. Thanks in advance, here are the problems:

Establishing the Identities of (1-sin^2x)/(1-cosx)=-cosx (I'm not sure how to make thetas) and (1-cotx^2)/(1+cotx^2)=1-2cosx

cos x = 3/5, 0 < x < 2pi, What is cos(2x)?

And lastly, find the exact value of these two:
cos(2tan^-1(4/3))
sec(tan^-1(4/3)+cot^-1(5/12))

Hopefully that wasn't too hard to read, any help on any problem would be VERY appreciated, thanks a lot!

**Soroban** · June 5th 2007, 05:02 PM

Hello, Hoopsaholic31!

Please check those two identities; there are typos.
. . As written, neither is true.

$\cos x = \frac{3}{5},\;0 < x < 2\pi$ . What is $\cos(2x)$ ?

You're expected to know the identity: . $\cos(2\theta) \:=\:2\cos^2\theta - 1$

Therefore: . $\cos(2x) \;=\;2\cos^2(x) - 1 \;=\;2\left(\frac{3}{5}\right)^2 - 1 \;=\;-\frac{7}{25}$

Find the exact value of these two:

. . $(a)\;\cos\left[2\tan^{-1}\!\!\left(\frac{4}{3}\right)\right]$

. . $(b)\;\sec\left[\tan^{-1}\!\!\left(\frac{4}{3}\right) + \cot^{-1}\!\!\left(\frac{5}{12}\right)\right]$

(a) Let: $\theta \:=\:\tan^{-1}\!\!\left(\frac{4}{3}\right)$

Then we have: . $\cos(2\theta) \;=\;2\cos^2\theta - 1$ .[1]

We have: . $\theta \:=\: \tan^{-1}\!\!\left(\frac{4}{3}\right) \quad\Rightarrow\quad \tan\theta \:=\: \frac{4}{3} \:=\: \frac{opp}{adj}$

$\theta$ is in a right triangle with: $opp = 4,\;adj = 3$
. . Then: . $hyp = 5$ . . . And hence: . $\cos\theta \,=\,\frac{3}{5}$

Substitute into [1]: . $\cos(2\theta) \;=\;2\left(\frac{3}{5}\right)^2 - 1$

. . Therefore: . $\cos\left[2\tan^{-1}\!\!\left(\frac{4}{3}\right)\right] \;=\;-\frac{7}{25}$

$(b)\;\sec\left[\tan^{-1}\left(\frac{4}{3}\right) + \cot^{-1}\left(\frac{5}{12}\right)\right]$

Let $\alpha \,=\,\tan^{-1}\left(\frac{4}{3}\right)$ . . . Let $\beta \,=\,\cot^{-1}\left(\frac{5}{12}\right)$

We have: . $\sec(\alpha + \beta) \;=\;\frac{1}{\cos(\alpha + \beta)} \;=\; \frac{1}{\cos\alpha\cos\beta - \sin\alpha\sin\beta}$ .[2]

We have: . $\alpha = \tan^{-1}\left(\frac{4}{3}\right)\quad\Rightarrow\quad \tan\alpha \,=\,\frac{4}{3} \,=\,\frac{opp}{adj}$
. . Then: . $hyp \,=\,5\quad\Rightarrow\quad \sin\alpha = \frac{4}{5},\;\cos\alpha = \frac{3}{5}$

We have: . $\beta \,=\,\cot^{-1}\left(\frac{5}{12}\right)\quad\Rightarrow\quad\c ot\beta \,=\,\frac{5}{12} \,=\,\frac{adj}{opp}$
. . Then: . $hyp = 13\quad\Rightarrow\quad\sin\beta \,=\,\frac{12}{13},\;\cos\beta \,=\,\frac{5}{13}$

Substitute into [2]: . $\sec(\alpha +\beta) \;=\;\frac{1}{\left(\frac{3}{5}\right)\left(\frac{ 5}{13}\right) - \left(\frac{4}{5}\right)\left(\frac{12}{13}\right) } \;=\;\frac{1}{\frac{15}{65} - \frac{48}{65}} \;=\;-\frac{65}{33}$

**Hoopsaholic31** · June 5th 2007, 05:15 PM

Oh, sorry, for the identities, the one with the cotangents should equal 1-2cos^2(x) and the other should be 1-(sin^2(x)/(a-cos)) = -cos

The other 3 are great though, thanks.

**Jhevon** · June 5th 2007, 05:16 PM

Originally Posted by Hoopsaholic31

I was having trouble with a few problems none of my friends could help me with, so I figured I'd give this forum a try. Thanks in advance, here are the problems:

Establishing the Identities of (1-sin^2x)/(1-cosx)=-cosx (I'm not sure how to make thetas) and (1-cotx^2)/(1+cotx^2)=1-2cosx

i don't think either of these are true

cos x = 3/5, 0 < x < 2pi, What is cos(2x)?

$\cos(2x) = \cos^2 x - \sin^2 x$

$= 2 \cos^2 x - 1$

$= 2 \left( \cos x \right)^2 - 1$

Now just plug in the value for $\cos x$ and solve. your answer should be $- \frac {7}{25}$

cos(2tan^-1(4/3))

Let $\tan^{-1} \left( \frac {4}{3} \right) = \theta$

$\Rightarrow \tan \theta = \frac {4}{3} = \frac {opposite}{adjacent}$

See first triangle below, the theta triangle. we can use Pythagoras' theorem to find the missing side and then find all the trig ratios from the triangle. From the triangle, we see that $\cos \theta = \frac {3}{5}$

So, $\cos \left( 2 \tan^{-1} \left( \frac {4}{3} \right) \right) = \cos 2 \theta$

$= 2 \cos^2 \theta - 1$

Now just plug in the value of $\cos \theta$ and solve

Your answer should be $- \frac {7}{25}$

sec(tan^-1(4/3)+cot^-1(5/12))

Let $\tan^{-1} \left( \frac {4}{3} \right) = A$

$\Rightarrow \tan A = \frac {4}{3} = \frac {opposite}{adjacent}$

See the second triangle below, triangle A. We can find all corresponding trig ratios from this. We see from this triangle that $\sin A = \frac {4}{5} \mbox { and } \cos A = \frac {3}{5}$

Let $\cot^{-1} \left( \frac {5}{12} \right) = B$

$\Rightarrow \cot B = \frac {12}{13} = \frac {adjacent}{opposite}$

See the third triangle below, triangle B. From this triangle we see that $\sin B = \frac {12}{13} \mbox { and } \cos B = \frac {5}{13}$

So, $\sec \left( \tan^{-1} \left( \frac {4}{3} \right) + \cot^{-1} \left( \frac {5}{12} \right) \right) = \sec \left( A + B \right)$

$= \frac {1}{ \cos \left(A + B \right)}$

$= \frac {1}{ \cos A \cos B - \sin A \sin B}$

Now just plug in the values and solve. Your answer should be $- \frac {65}{33}$

EDIT: Beaten by Soroban again, and by far this time

**Hoopsaholic31** · June 5th 2007, 05:32 PM

Thanks a lot both of you, and there was one more thing I was having a little bit of a hard time with. It gave me the equation to solve of sin(2x)sin(x)=cos(x), and I managed to establish the identity using the double angle formula. Is there a way to solve this, or do you think it was just put in the wrong section of the review packet?

**Jhevon** · June 5th 2007, 05:54 PM

Originally Posted by Hoopsaholic31

Prove (1 - cot^2 x)/(1 + cot^2 x) = 1 - 2cos^2x

Prove $\frac {1 - \cot^2 x}{1 + \cot^2 x} = 1 - 2 \cos^2 x$

Consider LHS:

$\frac {1 - \cot^2 x}{1 + \cot^2 x} = \frac {1 - \frac {\cos^2 x}{\sin^2 x}}{1 + \frac {\cos^2 x}{\sin^2 x}}$

$= \frac { \frac {\sin^2 x - \cos^2 x}{\sin^2 x}}{ \frac {\sin^2 x + \cos^2 x}{\sin^2 x}}$

$= \frac { \frac { \sin^2 x - \cos^2 x}{\sin^2 x}}{\frac {1}{\sin^2 x}}$

$= \frac {\sin^2 x - \cos^2 x}{\sin^2 x} \cdot \frac {\sin^2 x}{1}$

$= \sin^2 x - \cos^2 x$

$= - \cos 2x$

$= - \left( 2 \cos^2 x - 1 \right)$

$= 1 - 2 \cos^2 x$

$= RHS$

Therefore, we see $LHS \equiv RHS$

1-(sin^2(x)/(a-cos)) = -cos

this makes no sense

**Soroban** · June 5th 2007, 06:04 PM

Hello, Hoopsaholic31!

This is an equation . . . not an identity.

Solve: . $\sin(2x)\sin(x)\:=\:\cos(x)$

I'll assume that: $0 \leq x < 2\pi$

Using $\sin2\theta \:=\:2\!\cdot\!\sin\theta\!\cdot\!\cos\theta$

. . we have: . $[2\!\cdot\!\sin(x)\!\cdot\!\cos(x)]\cdot\sin(x) \:=\:\cos(x)\quad\Rightarrow\quad 2\!\cdot\sin^2(x)\!\cdot\!\cos(x) - \cos(x) \;=\;0$

Factor: . $\cos(x)\cdot[2\!\cdot\!\sin^2(x) - 1] \;=\;0$

And we have two equations to solve:

. . $\cos(x) \:=\:0\quad\Rightarrow\quad\boxed{x \:=\:\frac{\pi}{2},\:\frac{3\pi}{2}}$
. . $2\sin^2(x) - 1 \:=\:0\quad\Rightarrow\quad\sin(x) \:=\:\pm\frac{1}{\sqrt{2}}\quad\Rightarrow\quad\bo xed{x \:=\:\frac{\pi}{4},\;\frac{3\pi}{4},\:\frac{5\pi}{ 4},\:\frac{7\pi}{4}}$

**Jhevon** · June 5th 2007, 06:09 PM

Originally Posted by Hoopsaholic31

Thanks a lot both of you, and there was one more thing I was having a little bit of a hard time with. It gave me the equation to solve of sin(2x)sin(x)=cos(x), and I managed to establish the identity using the double angle formula. Is there a way to solve this, or do you think it was just put in the wrong section of the review packet?

$\sin 2x \sin x = \cos x$

$\Rightarrow (2 \sin x \cos x) \sin x - \cos x = 0$

$\Rightarrow 2 \sin^2 x \cos x - \cos x = 0$

$\Rightarrow \cos x (2 \sin^2 x - 1) = 0$

$\Rightarrow \cos x = 0 \mbox { or } 2 \sin^2 x - 1 = 0$

$\Rightarrow \cos x = 0 \mbox { or } \sin x = \pm \frac {1}{\sqrt {2}}$

$\Rightarrow x = \frac {\pi}{2} + k \pi \mbox { or } x = \pm \frac {\pi}{4} + n \pi$ for integers $k \mbox { and } n$

EDIT: And yet again. Ok, I will leave any further questions in this thread to you Soroban.

**Hoopsaholic31** · June 5th 2007, 06:30 PM

Okay, ugh, I'm sorry, for the LAST time on that identity...

1 - (sin^2(x)/(1-cos(x))) = -cos(x)

I feel like I'm being a total bonehead and this is a really easy one, but I've had a really high fever the last couple days and I'm kinda out of it. Thanks for all the work you've put in for me though guys, it'll be a huge help on my test tomorrow, I appreciate it.