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Let $\alpha$ be an acute angle. Solve:

$\log(\sin\alpha) + \log(\cos\alpha) + \log(\tan\alpha) + \log(\frac{1}{sin^2\alpha})$

I came to the following conclusions:

  • if $\alpha$ is an acute angle then $0<\sin\alpha<1$ and $0<\cos\alpha<1$

  • $\log(\frac{1}{sin^2\alpha}) = -2\log(\sin\alpha)$

  • if $\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$, then $\sin \alpha = \tan \alpha \cdot \cos \alpha$ and $\cos \alpha = \frac{\sin \alpha}{\cos \alpha}$, so

$$\log(\sin\alpha) + \log(\cos\alpha) + \log(\tan\alpha) + \log(\frac{1}{sin^2\alpha}) = \log(\sin \alpha) +\log(\frac{\sin \alpha}{\tan \alpha}) +\log(\frac{\sin \alpha}{\cos \alpha}) +\log(\frac{1}{\sin ^2\alpha}) = \log(\sin \alpha) + \log(\sin \alpha) - \log(\tan \alpha) +\log(\sin \alpha) - \log(\cos \alpha) + 0 - 2\log(\sin \alpha) = ???$$

But if I keep expanding the expression like this it will go on and on forever and I'll get nowhere.

How do I solve this?

My book says the solution is $0$.

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    $\begingroup$ $\log(a)+\log(b)=\log(ab)$ when $a, b\in(0,+\infty)$ $\endgroup$ – mrs Nov 20 '16 at 16:13
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The sum of logs is the log of a product. Thus:

$\log a + \log b + \log c + \log d = \log(abcd)$.

Letting $a$, $b$, $c$ and $d$ represent your four trigonometric expressions, try that, and see how it simplifies.

(I'm assuming you mean that you have to "simplify", not "solve".)

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  • $\begingroup$ I think it is really solve as the solution is supposed to be zero. But I'll try your suggestion $\endgroup$ – Mark Read Nov 20 '16 at 16:16
  • $\begingroup$ I got it. I will post the answer $\endgroup$ – Mark Read Nov 20 '16 at 16:18
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    $\begingroup$ It really is simplify, and it simplifies down to zero. "Solve" means to find the value of some variable, like to come away saying what $\alpha$ equals. Simplify means to find the simplest way of writing the expression. Some expressions, like this one, simplify to a number. $\endgroup$ – G Tony Jacobs Nov 20 '16 at 16:25
  • $\begingroup$ We simplify expressions; we solve equations. $\endgroup$ – G Tony Jacobs Nov 20 '16 at 16:25
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It is better to simply the expression as follows: $ log(sin\alpha)+log((sin(\alpha))^{-2}) + log(cos(\alpha))+log(tan(\alpha)) = \\ log((sin(\alpha)^{-1}) +log(cos(\alpha))+log(tan(\alpha)) = \\ log(cot(\alpha))+log(tan(\alpha)) = \\ log(cot(\alpha)tan(\alpha))=log(1)=0$

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Just multiply them together when amalgamating the logs.

You'll get $$ log\left(\sin \alpha \cos \alpha \frac{\sin\alpha}{\cos\alpha}\frac{1}{\sin^2\alpha} \right) = a$$

You then have $log(1) = a$. Thus $a=0$.

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$$\log(\sin\alpha) + \log(\cos\alpha) + \log(\tan\alpha) + \log(\frac{1}{sin^2\alpha}) = \log(\sin\alpha\cos\alpha\tan\alpha\frac{1}{\sin^2\alpha}) =\\ \log(\frac{\sin\alpha\cos\alpha\tan\alpha}{\sin^2\alpha}) = \log(\frac{\sin\alpha\sin\alpha}{\sin^2\alpha}) = \log(1) = 0$$

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