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  1. If $\theta$ is in quadrant $\text{I}$ and $\tan(\theta) = 0.6$ then $\sec(\theta) = $?
    This seems pretty easy to me:
    $\tan^2(\theta)-\sec^2(\theta)=1 \\ -\sec^2(\theta)=.64 \\ \sec(\theta)=8$

  2. Another one, $\cos(\theta)=\sin(2\theta)$
    Should that be $\cos^2(\theta)+\sin^2(\theta)=1/2$? Which would be 0.5

  3. For all angles $\theta$, $\cos(-\theta)$ = $\cos(\theta)$

$\ \ \ \ \ \ \ $I said false how can a negative angle be equal to a positive one?

  1. If $\sin^2(\theta)= 0.5$ then $\sin^2(\theta) = \cos^2(\theta)$
    I said false because $\sin$ and $\cos$ can never equal 0.5 together but it was wrong.
    This answer is wrong but I don't understand why.

I am seriously considering changing majors, I know most people here will think I am an idiot, lazy or whatever else for taking such simple high school level math courses in college but I really am having trouble with it. I keep making mistakes no matter what I do I can never get better than a D or C on a test. So I am trying to evaluate the mistakes I made on the test so I don't make them again, but inevitably I will.

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    Thank you, that stuff is very hard to figure out.2011-06-10

4 Answers 4

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part 1:

Identity is $\sec^2\theta -\tan^2\theta =1$ $\sec^2\theta -0.36 =1$ $\sec\theta =\sqrt{1.36}\implies\sec\theta=1.1662$

part 2:

$\cos \theta=\sin2\theta\implies\cos\theta=\cos\left(\dfrac\pi2-2\theta\right)\implies\theta=\dfrac\pi2-2\theta\implies\theta=\dfrac\pi6$ $\cos^2(\dfrac \pi6)+\sin^2(\dfrac \pi6)=1$

part3:

$\cos(-\theta)\implies\cos(0^\circ-\theta)\implies\cos 0^\circ\cdot\cos\theta+\sin 0^\circ\cdot\sin\theta\implies\cos\theta$

so: $\cos(-\theta)=\cos\theta$

part4:

if $\sin^2\theta=\dfrac12\implies\sin^2\theta=\left(\dfrac{1}{\sqrt{2}}\right)^2\implies\sin^2\theta=\sin^2\left(\dfrac{\pi}{4}\right)$

if

$\sin^2\theta=\sin^2\alpha$ then general solution of $\theta=n\pi\pm\alpha\;\;,n \in Z$.

so:

$\theta=n\pi\pm\dfrac{\pi}{4}\;\;,n \in Z$

so there are not $\sin\theta=\cos\theta=0.5\;\;while\;\;\sin^2\theta=\cos^2\theta=0.5$

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(2) If the problem asked you for the angle $\theta$ at which $\cos \theta = \sin(2\theta)$,

then think about what you know about a 30, 60, 90 right triangle: $\cos(30^\circ) = \sin(2\cdot 30^\circ) = \sin(60^\circ) = \frac {\sqrt{3}}{2}$

So theta would be $30^\circ$.

(4) If $\sin^2 \theta = .5$ then $\sin^2 \theta = \cos^2 \theta$.

This is true, since if $\sin^2 \theta = .5 = \frac 12$, then we know that $ \sin \theta = \pm\sqrt {\frac 12} = \pm \frac {\sqrt 2}{2} = \pm \frac {1}{\sqrt 2} = \pm \cos \theta$ (That means the reference angle is $45^\circ$.)

Hence $\cos^2 \theta = \sin^2 \theta = 0.5$.

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You need to read the questions more carefully. Checking the typing would make them easier to read, as there are many typos. Setting them in $\LaTeX$ (see the FAQ) would help a lot, too.

For the first, the identity is $1+tan^2\theta=sec^2\theta$. Your first equality is incorrect as $\tan^2\theta \neq 1$. The dash after $1$ looks like a minus sign, but it seems to be a separator. How about line feeds? Then you lost a decimal point at the end.

For the second, what is sin2theta? $\sin^2 \theta$ or $\sin 2\theta$?

Please review the question and make it legible.

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    @Adam: http://www.math.harvard.edu/texman/2011-06-10
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Draw a triangle in the first quadrant with $\tan(\theta) = 3/5$. It will have height 3, base 5 and hypoteneuse $\sqrt{34}$. When I compute $\sec(\theta)$ I get $\sqrt{34}/5$. Drawing the appropriate picture makes this worlds simpler.

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    @Adam: it might be labeled $\tan^{-1}$, but if you have $\tan$ you will have it.2011-06-10