The original question was solve for $\theta $ in $65\mathrm{cos}(2\theta )-56\mathrm{sin}(2\theta )-55=0$

Jaqueline Kirby
2022-06-30
Answered

The original question was solve for $\theta $ in $65\mathrm{cos}(2\theta )-56\mathrm{sin}(2\theta )-55=0$

You can still ask an expert for help

Layla Love

Answered 2022-07-01
Author has **29** answers

Remember the formulas

$\mathrm{cos}2\theta =\frac{1-{\mathrm{tan}}^{2}\theta}{1+{\mathrm{tan}}^{2}\theta},\phantom{\rule{2em}{0ex}}\mathrm{sin}2\theta =\frac{2\mathrm{tan}\theta}{1+{\mathrm{tan}}^{2}\theta}$

but first examine the cases $\theta =\pi /2$ and $\theta =-\pi /2$ that would invalidate the substitution.

We have

$65\mathrm{cos}\pi -56\mathrm{sin}\pi -55=-120\ne 0\phantom{\rule{0ex}{0ex}}65\mathrm{cos}(-\pi )-56\mathrm{sin}(-\pi )-55=-120\ne 0$

so the substitution is good and doesn't discard solutions.

Set $t=\mathrm{tan}\theta $ for simplicity, so you get

$65\frac{1-{t}^{2}}{1+{t}^{2}}-56\frac{2t}{1+{t}^{2}}-55=0$

that becomes

$60{t}^{2}+56t-5=0$

and the quadratic has roots

$\frac{-14+\sqrt{271}}{30}\phantom{\rule{2em}{0ex}}\frac{-14-\sqrt{271}}{30}$

so you get

$\theta =\mathrm{arctan}\frac{-14+\sqrt{271}}{30}+k\pi \phantom{\rule{2em}{0ex}}\text{or}\phantom{\rule{2em}{0ex}}\theta =\mathrm{arctan}\frac{-14-\sqrt{271}}{30}+k\pi $

Note also that your transformation to

$\mathrm{cos}\theta (130\mathrm{cos}\theta -122\mathrm{sin}\theta )-10=0$

does not reduce the equation to $130\mathrm{cos}\theta -122\mathrm{sin}\theta -10=0$. From

$a(b+c)-d=0$

you can't deduce

$b+c-d=0$

$\mathrm{cos}2\theta =\frac{1-{\mathrm{tan}}^{2}\theta}{1+{\mathrm{tan}}^{2}\theta},\phantom{\rule{2em}{0ex}}\mathrm{sin}2\theta =\frac{2\mathrm{tan}\theta}{1+{\mathrm{tan}}^{2}\theta}$

but first examine the cases $\theta =\pi /2$ and $\theta =-\pi /2$ that would invalidate the substitution.

We have

$65\mathrm{cos}\pi -56\mathrm{sin}\pi -55=-120\ne 0\phantom{\rule{0ex}{0ex}}65\mathrm{cos}(-\pi )-56\mathrm{sin}(-\pi )-55=-120\ne 0$

so the substitution is good and doesn't discard solutions.

Set $t=\mathrm{tan}\theta $ for simplicity, so you get

$65\frac{1-{t}^{2}}{1+{t}^{2}}-56\frac{2t}{1+{t}^{2}}-55=0$

that becomes

$60{t}^{2}+56t-5=0$

and the quadratic has roots

$\frac{-14+\sqrt{271}}{30}\phantom{\rule{2em}{0ex}}\frac{-14-\sqrt{271}}{30}$

so you get

$\theta =\mathrm{arctan}\frac{-14+\sqrt{271}}{30}+k\pi \phantom{\rule{2em}{0ex}}\text{or}\phantom{\rule{2em}{0ex}}\theta =\mathrm{arctan}\frac{-14-\sqrt{271}}{30}+k\pi $

Note also that your transformation to

$\mathrm{cos}\theta (130\mathrm{cos}\theta -122\mathrm{sin}\theta )-10=0$

does not reduce the equation to $130\mathrm{cos}\theta -122\mathrm{sin}\theta -10=0$. From

$a(b+c)-d=0$

you can't deduce

$b+c-d=0$

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