Question 1171343
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QIV → cos, sec > 0; sin, csc, and tan < 0


cosine is the reciprocal of secant, so if *[tex \Large \sec(x)\ = 4] then *[tex \Large \cos(x)\ =\ \frac{1}{4}].


Then since *[tex \Large \sin(\varphi)\ =\ \pm\sqrt{1\,-\,\cos^2(\varphi)}], but since sine is negative in QIV,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(x)\ =\ -\sqrt{1\,-\,\frac{1}{16}}\ =\ -\frac{\sqrt{15}}{4}]


Then since tangent is the quotient of sine over cosine and cosecant is the reciprocal of sine, we have:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan(x)\ =\ -\sqrt{15}]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \csc(x)\ =\ -\frac{4\sqrt{15}}{15}]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(2x)\ =\ 2\sin(x)\cos(x)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(2x)\ =\ \cos^2(x)\ -\ \sin^2(x)]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan(2x)\ =\ \frac{2\tan(x)}{1\,-\,\tan^2(x)}]

The rest is just arithmetic -- that's on you.

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
*[illustration darwinfish.jpg]

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