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

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 $0$.

• $\log(a)+\log(b)=\log(ab)$ when $a, b\in(0,+\infty)$ – mrs Nov 20 '16 at 16:13

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".)

• I think it is really solve as the solution is supposed to be zero. But I'll try your suggestion – Mark Read Nov 20 '16 at 16:16
• I got it. I will post the answer – Mark Read Nov 20 '16 at 16:18
• 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. – G Tony Jacobs Nov 20 '16 at 16:25
• We simplify expressions; we solve equations. – G Tony Jacobs Nov 20 '16 at 16:25

It is better to simply the expression as follows: $log(sin\alpha)+log((sin(\alpha))^{-2}) + log(cos(\alpha))+log(tan(\alpha)) = \\ log((sin(\alpha)^{-1}) +log(cos(\alpha))+log(tan(\alpha)) = \\ log(cot(\alpha))+log(tan(\alpha)) = \\ log(cot(\alpha)tan(\alpha))=log(1)=0$

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 $a=0$.

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