3 years ago

Sin(x)-tan(x)cos(x)+cos(π/2-x)

Mathematics

1 answer:

Gala2k [10]3 years ago

7 0

Answer:

sin(x)

Step-by-step explanation:

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2 years ago

Solve the following equation:

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Complete the square.

$z^4 + z^2 - i\sqrt 3 = \left(z^2 + \dfrac12\right)^2 - \dfrac14 - i\sqrt3 = 0$

$\left(z^2 + \dfrac12\right)^2 = \dfrac{1 + 4\sqrt3\,i}4$

Use de Moivre's theorem to compute the square roots of the right side.

$w = \dfrac{1 + 4\sqrt3\,i}4 = \dfrac74 \exp\left(i \tan^{-1}(4\sqrt3)\right)$

$\implies w^{1/2} = \pm \dfrac{\sqrt7}2 \exp\left(\dfrac i2 \tan^{-1}(4\sqrt3)\right) = \pm \dfrac{2+\sqrt3\,i}2$

Now, taking square roots on both sides, we have

$z^2 + \dfrac12 = \pm w^{1/2}$

$z^2 = \dfrac{1+\sqrt3\,i}2 \text{ or } z^2 = -\dfrac{3+\sqrt3\,i}2$

Use de Moivre's theorem again to take square roots on both sides.

$w_1 = \dfrac{1+\sqrt3\,i}2 = \exp\left(i\dfrac\pi3\right)$

$\implies z = {w_1}^{1/2} = \pm \exp\left(i\dfrac\pi6\right) = \boxed{\pm \dfrac{\sqrt3 + i}2}$

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$\implies z = {w_2}^{1/2} = \boxed{\pm \sqrt[4]{3} \, \exp\left(-i\dfrac{5\pi}{12}\right)}$

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1 year ago