Question 1161053
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\varphi\ =\ x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2\varphi\ =\ x^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ -\ \cos^2\varphi\ =\ x^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2\varphi\ =\ 1\ -\ x^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos\varphi\ =\ \pm\sqrt{1\,-\,x^2}]

and then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan\varphi\ =\ \frac{\sin\varphi}{\cos\varphi}]


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Similarly, given


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos\phi\ =\ x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\phi\ =\ \pm\sqrt{1\,-\,x^2}]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan\phi\ =\ \frac{\sin\phi}{\cos\phi}]



==================================


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan\theta\ =\ x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\sin\theta}{\cos\theta}\ =\ x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\theta\ =\ x\cos\theta]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2\theta\ =\ x^2\cos^2\theta]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2\theta\ =\ x^2\ -\ x^2\sin^2\theta]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2\theta\ +\ x^2\sin^2\theta\ =\ x^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2\theta\(1\ +\ x^2)\ =\ x^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\theta\ =\ \pm\sqrt{\frac{x^2}{1\ +\ x^2}}]

or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2\theta\ =\ x^2\cos^2\theta]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ -\ \cos^2\theta\ =\ x^2\cos^2\theta]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\cos^2\theta\ +\ \cos^2\theta\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2\theta(x^2\ +\ 1)\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos\theta\ =\ \pm\sqrt{\frac{1}{x^2\ +\ 1}]


In all cases, select the sign based on the quadrant of the terminal ray of the angle.


Sine is positive in QI and QII


Cosine is positive in QI and QIV


Tangent is positive in QI and QIII
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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