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ijabbott · (This post was last modified: 11-22-2018 05:35 PM by ijabbott.)

(11-04-2018 04:57 PM)Dieter Wrote:
(11-04-2018 03:32 PM)Thomas Klemm Wrote: We don't really need trigonometric functions here.
Good old Pythagoras is good enough:

Great! This way it can also be done on the 12C and other calculators without trigs or polar/rectangular conversion:

Code:

01 X<>Y 02 - 03 LstX 04 ENTER 05 x 06 EEX 07 4 08 + 09 √x 10 ÷ 11 × 12 GTO 00

Since no →P is required this may even run slightly faster than Thomas' original version. If available, replace "ENTER x" with x².

(11-04-2018 03:32 PM)Thomas Klemm Wrote: Examples:

56 ENTER
20 ENTER
40 A
10.9825

56 ENTER
-20 ENTER
40 A
32.9475

Same for the above version. Press [R/S] instead of [A]. ;-)

That's a neat solution! It's also worth mentioning that if you know the tangent, sine or cosine of an angle between 0 and 90 degrees, you can derive the others with standard arithmetic and the square root function.

\( \tan(x) = \frac{\sqrt{1-\cos^2(x)}}{\cos(x)} \), or: \( \tan(x) = \sqrt{\frac{1}{\cos^2(x)} - 1} \)

Code:

ENTER

×

1/x

1

-

√x

\( \cos(x) = \frac{1}{\sqrt{1 + \tan^2(x)}} \)

Code:

ENTER

×

1

+

√x

1/x

\( \tan(x) = \frac{\sin(x)}{\sqrt{1 - \sin^2(x)}} \)

Code:

ENTER

ENTER

×

1

X<>Y

-

√x

÷

\( \sin(x) = \frac{\tan(x)}{\sqrt{1 + \tan^2(x)}} \)

Code:

ENTER

ENTER

×

1

+

√x

÷

\( \sin(x) = \sqrt{1 - \cos^2(x)} \), and: \( \cos(x) = \sqrt{1 - \sin^2(x)} \)

Code:

ENTER

×

1

X<>Y

-

√x

Of course, "ENTER", "×" can be replaced by "x²" in all of the above, if available.