Question 1176366
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos\varphi\ =\ \pm\sqrt{1\ -\ \sin^2\varphi}]


The sine is positive in QI and QII which means that there are two possible arguments for the function that have a positive sine value, but the cosine is positive in QI and negative in QII.  So, as posed the question has two answers as indicated by the *[tex \LARGE \pm] in the formula.


The formula is a consequence of the Pythagorean Identity: *[tex \LARGE \cos^2\varphi\ +\ \sin^2\varphi\ =\ 1]


Let's say that *[tex \LARGE \sin\varphi\ =\ x].  We square both sides of this equation to obtain:


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


Applying the Pythagorean Identity:


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


A little algebra music, Maestro:


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


Taking the square root (remembering to account for positive and negative roots):


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


But we know from earlier that *[tex \LARGE x^2\ =\ \sin^2\theta] we can substitute to get the desired formula.


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


Further, given the symmetrical nature of the Pythagorean Identity, it should be intuitively obvious that the following formula holds as well:


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


The following will help you choose which sign to affix based on your knowledge of the Quadrant location of the function argument:


<pre>
Quadrant  Sine Cosine Tangent

   I        +     +      +

   II       +     -      -

   III      -     -      +

   IV       -     +      -

</pre>


Also, secant has the same sign as cosine and cosecant has the same sign as sin.


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

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
I > Ø
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