donderdag 19 april 2018
zondag 15 april 2018
Dat is dan nog best een aardige opgave
Naar aanleiding van twee goniometrische vergelijkingen. Die '2' gooit roet in het eten... inderdaad. Zonder die '2' was het best een aardig sommetje:
\( \eqalign{ & \cos ^2 (x) - \sin ^2 (x) = 0 \cr & 1 - \sin ^2 (x) - \sin ^2 (x) = 0 \cr & 1 - 2\sin ^2 (x) = 0 \cr & 2\sin ^2 (x) = 1 \cr & \sin ^2 (x) = \frac{1} {2} \cr & \sin (x) = - \sqrt {\frac{1} {2}} \vee \sin (x) = \sqrt {\frac{1} {2}} \cr & \sin (x) = - \frac{1} {2}\sqrt 2 \vee \sin (x) = \frac{1} {2}\sqrt 2 \cr & x = \frac{3} {4}\pi + k \cdot 2\pi \vee x = 1\frac{1} {4}\pi + k \cdot 2\pi \vee x = \frac{1} {4}\pi + k \cdot 2\pi \vee x = 1\frac{3} {4}\pi + k \cdot 2\pi \cr & x = \frac{1} {4}\pi + k \cdot \frac{1} {2}\pi \cr} \)
Nou ja... 't idee was prima...:-)
\( \eqalign{ & \cos ^2 (x) - \sin ^2 (x) = 0 \cr & 1 - \sin ^2 (x) - \sin ^2 (x) = 0 \cr & 1 - 2\sin ^2 (x) = 0 \cr & 2\sin ^2 (x) = 1 \cr & \sin ^2 (x) = \frac{1} {2} \cr & \sin (x) = - \sqrt {\frac{1} {2}} \vee \sin (x) = \sqrt {\frac{1} {2}} \cr & \sin (x) = - \frac{1} {2}\sqrt 2 \vee \sin (x) = \frac{1} {2}\sqrt 2 \cr & x = \frac{3} {4}\pi + k \cdot 2\pi \vee x = 1\frac{1} {4}\pi + k \cdot 2\pi \vee x = \frac{1} {4}\pi + k \cdot 2\pi \vee x = 1\frac{3} {4}\pi + k \cdot 2\pi \cr & x = \frac{1} {4}\pi + k \cdot \frac{1} {2}\pi \cr} \)
Nou ja... 't idee was prima...:-)
zaterdag 14 april 2018
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