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3 years ago

Rewrite with only sin x and cos x. sin 2x - cos 2x

Mathematics

2 answers:

Ymorist [56]3 years ago

4 0

Sin2x = 2sinxcosx
cos2x = (cosx)^2 - (sinx)^2 = 2(cosx)^2 -1 = 1- 2(sinx)^2

Any combination you want is here.

RoseWind [281]3 years ago

3 0

Sin2x-cos2x
2sinxcosx-2cos^2x-1
2cos^2x+2sinxcosx-1
I think that's how you do this one.

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3 years ago

Solve the following equation:

Rama09 [41]

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.

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

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

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

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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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2 years ago