Given: $$2\sin\theta -\sqrt 5 \cos \theta \equiv - 3\cos (\theta + \alpha ),$$ where $$0 <\alpha < 90^\circ, $$ find $α.$

The issue I have with this question is the $-3$ on the right hand side, it really complicates things and I don't know how to deal with it.

When I usually come across equations of the form $$a\sin \theta + b\cos \theta $$ it's a relatively straight forward process of converting them into another equation of the form $$\sqrt {{a^2} + {b^2}} \left({a \over {\sqrt {{a^2} + {b^2}} }}\sin \theta - {b \over {\sqrt {{a^2} + {b^2}} }}\cos \theta \right)$$ where $$\cos \alpha = {a \over {\sqrt {{a^2} + {b^2}} }}$$ and $$\sin \alpha = {b \over {\sqrt {{a^2} + {b^2}} }}$$

(or some other variant of this).

I cant wrap my head around this, specifically. If $$2\sin\theta -\sqrt 5 \cos \theta \equiv - 3\cos (\theta + \alpha ),$$ this must mean the square root expression prefixing the cosine addition identity must be $-3$, as $$\sqrt {{a^2} + {b^2}} \cos(\theta + \alpha ) \equiv - 3\cos(\theta + \alpha ).$$

I am aware the square root operation outputs negative values I just dont know how to deal with that in this instance.

I hope this doesn't read too convoluted, thank you.

  • $\begingroup$ The use of displaystyle in the title is strongly depreciated. $\endgroup$ – Sangchul Lee Apr 2 '13 at 15:40
  • $\begingroup$ Sorry I'm using a program that helps me to type these equations out $\endgroup$ – seeker Apr 2 '13 at 15:42
  • $\begingroup$ That is understandable, but we also encourage you to learn a preliminary level of $\LaTeX$. That is, it will be helpful to know that using one dollar sign (\$) both at the opening and the closing results in textstyle mode, while using two dollar signs (\$\$) results in displaystyle mode. Thus even if you are aided with a software that produces displaystyle formulas, you can easily convert them to textstyle by erasing some dollar signs! $\endgroup$ – Sangchul Lee Apr 2 '13 at 15:56

$$\Rightarrow {\rm{2sin}}\theta {\rm{ - }}\sqrt 5 \cos \theta = - 3\cos (\theta + \alpha )$$ $$\Rightarrow -\frac{2}{3}\sin \theta+\frac{\sqrt 5 }{3}\cos \theta= \cos (\theta + \alpha ) $$ $$\Rightarrow -\frac{2}{3}\sin \theta+\frac{\sqrt 5 }{3}\cos \theta= \cos \theta \cos \alpha - \sin \theta \sin \alpha$$ $$\Rightarrow \frac{\sqrt 5 }{3}\cos \theta -\frac{2}{3}\sin \theta = \cos \theta \cos \alpha - \sin \theta \sin \alpha$$

Equating both sides and considering $0<\alpha < 90^0$ i.e. both $\cos \alpha$ and $\sin \alpha$ lie on the first quadrant. $$\cos \alpha = \frac{\sqrt 5 }{3}\tag1$$ $$\sin \alpha = \frac{2}{3}\tag2$$

Dividing $(2)$ by $(1)$

$$\frac{\sin \alpha}{\cos \alpha} =\frac{\frac{2}{3}}{\frac{\sqrt 5 }{3}}=\frac{2}{sqrt(5)}$$ $$\tan \alpha = \frac{2}{\sqrt5}$$ $$\alpha = tan^{-1}\frac{2}{\sqrt5}$$

Note, the reason you went wrong because of your formulation,


$$\sqrt {{a^2} + {b^2}} \left({a \over {\sqrt {{a^2} + {b^2}} }}\sin \theta - {b \over {\sqrt {{a^2} + {b^2}} }}\cos \theta \right) \ne \sqrt {{a^2} + {b^2}} \cos(\theta + \alpha )$$

But rather

$$-\sqrt {{a^2} + {b^2}} \left({b \over {\sqrt {{a^2} + {b^2}} }}\cos - {a \over {\sqrt {{a^2} + {b^2}} }}\sin \theta \theta \right) = - \sqrt {{a^2} + {b^2}} \cos(\theta + \alpha )$$

Which will derive to

$$-\sqrt {{a^2} + {b^2}} \cos(\theta + \alpha ) = - 3\cos(\theta + \alpha ).$$


Your assumption is that the relation holds for all $\theta$. If that's really the assumption, then you could let $\theta$ be any angle, say $0$.

Then $$\frac{\sqrt{5}}{3}=\cos(\alpha)$$ and $$\alpha=\arccos\left(\frac{\sqrt{5}}{3}\right)$$


Let $2=r\sin\beta,\sqrt5=r\cos\beta$ where $r\ge0$

Squaring and adding we get $r^2=2^2+5=9\implies r=3$

and $\sin\beta=\frac23,\cos\beta=\frac{\sqrt5}3\implies \beta=\arccos \frac{\sqrt5}3=\arcsin \frac23$


$$\implies 3\cos(\theta+\beta)=3\cos(\theta+\alpha)$$

$$\implies \cos(\theta+\beta)=\cos(\theta+\alpha)$$

So, $$\theta+\beta=2n\pi\pm (\theta+\alpha)$$

Taking the '+' sign, $\theta+\beta=2n\pi+ (\theta+\alpha)$ $\implies \alpha=\beta-2n\pi\equiv \beta\pmod{2\pi}$

Taking the '-' sign, $\theta+\beta=2n\pi- (\theta+\alpha)$ $\implies \alpha=-\beta+2n\pi-2\theta\equiv-(\beta+2\theta)\pmod{2\pi}$

  • 1
    $\begingroup$ I don't think it is fair to equate the coefficients of $\sin \theta$ and $\cos \theta$ separately. This is the same as saying this has to work for all $\theta$ There will be some more solutions. $\endgroup$ – Ross Millikan Apr 2 '13 at 15:43
  • $\begingroup$ @Ross: The assumption is that it does work for all $\theta$--that's what the $\equiv$ in the OP is intended to denote. $\endgroup$ – Cameron Buie Apr 2 '13 at 15:56
  • $\begingroup$ @RossMillikan, how about the current version $\endgroup$ – lab bhattacharjee Apr 2 '13 at 15:59
  • $\begingroup$ See Cameron Buie's comment. I had missed the $\equiv$. Sorry $\endgroup$ – Ross Millikan Apr 2 '13 at 16:29

You can use the so-called double angle formula. Notice that

$$-3\cos(\theta+\alpha) \equiv -3\cos\theta\cos\alpha + 3 \sin\theta\sin\alpha $$

Now you can compare coefficients. You want $-3\cos(\theta+\alpha) \equiv 2\sin\theta - \sqrt{5}\cos\theta$, and so you need to find an $\alpha$ for which $3\sin\alpha = 2$ and $3\cos\alpha = \sqrt{5}$. It follows that

$$\frac{3\sin\alpha}{3\cos\alpha} \equiv \tan\alpha = \frac{2}{\sqrt{5}}$$

You $\alpha$ is then $\arctan(2/\sqrt{5}) \approx 41.8^{\circ}$.


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