0
$\begingroup$

I have this formula: $$\frac{\sin x + 3\cos x}{2\cos x}$$ and I'm supposed to transform it so that I'm only left with one function (sin, cos, tan etc, just one of these). It should not be based on any complicated formulas/equations since it was originally a high school task, yet I have no idea how to tackle it. I always end up having at least one sin/cos and unable to get rid of it. I'm pretty sure I'm missing something simple but I'd appreciate any help. Thanks

$\endgroup$
2
  • 3
    $\begingroup$ $\frac{1}{2}\tan(x) + \frac{3}{2}$ $\endgroup$
    – user335907
    Oct 7, 2017 at 19:55
  • $\begingroup$ ugh, thanks, can't believe i haven't noticed that before $\endgroup$
    – Jake
    Oct 7, 2017 at 19:59

2 Answers 2

Reset to default
0
$\begingroup$

$$ \frac{\sin x + 3 \cos x}{2 \cos x} = \frac{\sin x}{2 \cos x} + \frac{3 \cos x}{2 \cos x} = \frac{1}{2} \tan x + \frac{3}{2} $$

$\endgroup$
0
$\begingroup$

$$\dfrac{1}{2}\tan x+\dfrac{3}{2}$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .