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. $\endgroup$ Apr 2, 2013 at 15:40
  • $\begingroup$ Sorry I'm using a program that helps me to type these equations out $\endgroup$
    – seeker
    Apr 2, 2013 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! $\endgroup$ Apr 2, 2013 at 15:56

4 Answers 4


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

  • $\begingroup$ his angle relation was $0 <\alpha < 90^\circ$. So you can't take $90^\circ$ neither $0$. $\endgroup$ Feb 24 at 14:58
  • 1
    $\begingroup$ @BillyIstiak I think you have confused $\alpha$ with $\theta$. In this answer, I let $\theta$ be $0$. There is no restriction on $\theta$. $\endgroup$
    – 2'5 9'2
    Feb 24 at 15:27
  • $\begingroup$ Oops! I misread😑. Thanks! I will always take a look whatever I read from now. $\endgroup$ Feb 24 at 15:56

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. $\endgroup$ Apr 2, 2013 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. $\endgroup$ Apr 2, 2013 at 15:56
  • $\begingroup$ @RossMillikan, how about the current version $\endgroup$ Apr 2, 2013 at 15:59
  • $\begingroup$ See Cameron Buie's comment. I had missed the $\equiv$. Sorry $\endgroup$ Apr 2, 2013 at 16:29

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

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

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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.