# Solving $\tan x = \dfrac{\sin 10^\circ + \sin 40^\circ}{ \cos 10^\circ + \cos 40^\circ}$ [closed]

If $x$ satisfies the equation $$\tan x = \dfrac{\sin 10^\circ + \sin 40^\circ}{ \cos 10^\circ + \cos 40^\circ}$$ and $x$ is between $0^\circ$ and $90^\circ$, then $x$ is equal to what?

Is there an identity I can use here?

## closed as off-topic by Noah Schweber, астон вілла олоф мэллбэрг, Stefan Mesken, Daniel W. Farlow, user26857Nov 13 '16 at 20:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Noah Schweber, астон вілла олоф мэллбэрг, Stefan Mesken, Daniel W. Farlow, user26857
If this question can be reworded to fit the rules in the help center, please edit the question.

Try this in the numerator:

$$\sin a + \sin b = 2 \sin \frac{a+b}2 \;\cos \frac{a-b}2$$

And this in the denominator:

$$\cos a + \cos b = 2 \cos \frac{a+b}2\;\cos\frac{a-b}2$$

Divide the numerator by the denominator after applying these sum-product relations, cancel out common factors, and see what you get.

• Just enough hints to stimulate curiosity! +1. – blackpen Nov 13 '16 at 4:44
• (+1) The identity that ensues is one of my favorites. It also shows that $\tan\left(\frac a2\right)=\frac{\sin(a)}{1+\cos(a)}$ – robjohn Nov 13 '16 at 4:56

# Proof Without Words

$$\large\color{#8060A0}{\frac{\sin(a)+\sin(b)}{\cos(a)+\cos(b)}=\tan\left(\frac{a+b}2\right)}$$