Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
The use of displaystyle in the title is strongly depreciated. – Sangchul Lee Apr 2 '13 at 15:40
Sorry I'm using a program that helps me to type these equations out – seeker Apr 2 '13 at 15:42
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! – Sangchul Lee Apr 2 '13 at 15:56
up vote 3 down vote accepted

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

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

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

share|cite|improve this answer

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

share|cite|improve this answer

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

share|cite|improve this answer
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. – Ross Millikan Apr 2 '13 at 15:43
@Ross: The assumption is that it does work for all $\theta$--that's what the $\equiv$ in the OP is intended to denote. – Cameron Buie Apr 2 '13 at 15:56
@RossMillikan, how about the current version – lab bhattacharjee Apr 2 '13 at 15:59
See Cameron Buie's comment. I had missed the $\equiv$. Sorry – 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 $\arctan(2/\sqrt{5}) \approx 41.8^{\circ}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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