Lahenda S=2cos(2/3t-pi/9) | Microsofti matemaatika lahendaja

Lahendage ja leidke S (complex solution)

S=e^{\frac{2it}{3}-\frac{\pi i}{9}}+e^{-\frac{2it}{3}+\frac{\pi i}{9}}

Lahendage ja leidke t (complex solution)

t=3\pi n_{1}+\frac{1}{6}\pi +\left(-\frac{3}{2}i\right)\ln(\frac{1}{2}S+\left(-\frac{1}{2}\right)\left(S^{2}-4\right)^{\frac{1}{2}})\text{, }n_{1}\in \mathrm{Z}

t=3\pi n_{2}+\frac{1}{6}\pi +\left(-\frac{3}{2}i\right)\ln(\frac{1}{2}S+\frac{1}{2}\left(S^{2}-4\right)^{\frac{1}{2}})\text{, }n_{2}\in \mathrm{Z}

Lahendage ja leidke S

S=2\cos(\frac{6t-\pi }{9})

Lahendage ja leidke t

t=\frac{1}{6}\pi +3\pi n_{49}+\frac{3}{2}ArcCosI(\frac{1}{2}S)\text{, }n_{49}\in \mathrm{Z}

t=\frac{1}{6}\pi +\left(-\frac{3}{2}\right)ArcCosI(\frac{1}{2}S)+3\pi n_{48}\text{, }n_{48}\in \mathrm{Z}

Viktoriin

Trigonometry

5 probleemid, mis on sarnased:

S = 2 \cos ( \frac { 2 } { 3 } t - \frac { \pi } { 9 } )

Sarnased probleemid veebiotsingust

Show a regular n-sided polygon is constructible, using only ruler and compasses, iff the number \alpha = 2 \cos(2\pi/n) is constructible.

https://math.stackexchange.com/questions/2963133/show-a-regular-n-sided-polygon-is-constructible-using-only-ruler-and-compasses

If \cos(2\pi/n) is constructible, then make a unit circle C intersect with the vertical line through \cos(2\pi/n) . The point of intersection gives \zeta_n (treating the Euclidean plane ...

How do you find the exact value of \displaystyle{\cos{{\left({\left({2}\frac{\pi}{{3}}\right)}-{\left(\frac{\pi}{{6}}\right)}\right)}}} ?

https://socratic.org/questions/how-do-you-find-the-exact-value-of-cos-2pi-3-pi-6

Zero Explanation: Apply the trig identity: cos (a - b) = cos a.cos b + sin a.sin b \displaystyle{\cos{{\left(\frac{{{2}\pi}}{{3}}-\frac{\pi}{{6}}\right)}}}={\cos{{\left(\frac{{{2}\pi}}{{3}}\right)}}}.{\cos{{\left(\frac{\pi}{{6}}\right)}}}+{\sin{{\left(\frac{{{2}\pi}}{{3}}\right)}}}.{\sin{{\left(\frac{\pi}{{6}}\right)}}} ...

How do you evaluate \displaystyle{\cos{{\left(\frac{{{2}\pi}}{{9}}-\frac{\pi}{{18}}\right)}}} ?

https://socratic.org/questions/how-do-you-evaluate-cos-2pi-9-pi-18

\displaystyle\frac{\sqrt{{3}}}{{2}} Explanation: \displaystyle{\cos{{\left(\frac{{{2}\pi}}{{9}}-\frac{\pi}{{18}}\right)}}}={\cos{{\left(\frac{{{4}\pi}}{{18}}-\frac{\pi}{{18}}\right)}}}={\cos{{\left(\frac{{{3}\pi}}{{18}}\right)}}}= ...

How do you find the exact value of \displaystyle{\cos{{\left(\frac{\pi}{{4}}\right)}}}+{\cos{{\left(\frac{\pi}{{3}}\right)}}} ?

https://socratic.org/questions/how-do-you-find-the-exact-value-of-cos-pi-4-cos-pi-3

You can get the exact values of these functions from the unit circle. According to the unit circle; Explanation: cos( \displaystyle\frac{\pi}{{4}} ) = \displaystyle\frac{\sqrt{{{2}}}}{\text{2}} ...

How do you solve \displaystyle{\cos{{\left({\left[\frac{θ}{{3}}\right]}-{\left[\frac{\pi}{{4}}\right]}\right)}}}=\frac{{1}}{{2}} ?

https://socratic.org/questions/how-do-you-solve-cos-3-pi-4-1-2

\displaystyle\theta=-{45}^{{o}} Explanation: \displaystyle\text{please remember that }\ {\cos{{\left({a}-{b}\right)}}}={\cos{{a}}}\cdot{\cos{{b}}}+{\sin{{a}}}\cdot{\sin{{b}}} \displaystyle{\cos{{\left(\frac{\theta}{{3}}-\frac{\pi}{{4}}\right)}}}={\cos{\theta}}\cdot\frac{{\cos{\pi}}}{{4}}+{\sin{\theta}}\cdot\frac{{\sin{\pi}}}{{4}} ...

How do you evaluate \displaystyle{\cos{{\left(\frac{\pi}{{3}}-\frac{\pi}{{6}}\right)}}} ?

https://socratic.org/questions/how-do-you-evaluate-cos-pi-3-pi-6

mason m Jun 3, 2016 If we momentarily forgot how to subtract fractions, we could use the cosine subtraction formula: \displaystyle{\cos{{\left({A}-{B}\right)}}}={\cos{{\left({A}\right)}}}{\cos{{\left({B}\right)}}}+{\sin{{\left({A}\right)}}}{\sin{{\left({B}\right)}}} ...

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