# If $\sin\theta + \sin\phi = a$ and $\cos\theta + \cos\phi = b$, then $\sin(\theta+\phi) = ???$ [closed]

If \begin{align} \sin\theta + \sin\phi &= a \\ \cos\theta + \cos\phi &= b \end{align} then find the value of $\sin(\theta + \phi)$.

## closed as off-topic by Nikunj, Travis, Watson, C. Falcon, user 170039Jun 20 '16 at 13:35

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." – Nikunj, Travis, Watson, C. Falcon, user 170039
If this question can be reworded to fit the rules in the help center, please edit the question.

• @DietrichBurde, Can you please show how it is duplicate? – lab bhattacharjee Jun 20 '16 at 11:50
• Why off-topic? What IS wrong? – Jamil Ahmed Jun 20 '16 at 16:53
• @JamilAhmed: As the "put on hold" note suggests, we like (and often demand) one's thoughts and/or attempts at a problem. This helps us help you, and it helps us avoid duplicating your effort. (It also helps us think that we're not just doing your homework for you.) I notice under another question, in response to a comment to use the $\tan(a-b)$ formula, you replied "I did. But [...]." While a comment isn't such a big deal, it's very annoying to spend time on an answer only to be told: "Yeah, I already tried that." Anyway ... Please tell us what you know, up-front, and everyone will be happy. – Blue Jun 20 '16 at 18:53
• Incidentally, I posted a trigonographic solution to trigonography.com. – Blue Jun 21 '16 at 3:42

HINT:

Use Prosthaphaeresis Formula and divide one by the other to find $$\tan\dfrac{\theta+\phi}2$$

Now use Weierstrass Substitution, $$\sin2A=\dfrac{2\tan A}{1+\tan^2A}$$

Multiplying both equations, we get $$\sin \theta\cos\theta+\sin\phi\cos\phi+\sin(\theta+\phi)=ab$$ $$\implies \frac{1}{2}(\sin 2\theta+\sin 2\phi)+\sin(\theta+\phi)=ab$$ Using the identity $\sin a+\sin b=2\sin\frac{a+b}{2}\cos\frac{a-b}{2}$ and re-arranging, we get $$\sin(\theta+\phi)(1+\cos(\theta-\phi))=ab$$

You can easily calculate $\cos(\theta-\phi)$ by squaring and adding the original two equations.

• I was going to add this solution! Anyways (+1) :) – Max Payne Jun 20 '16 at 12:25

Multiply the given equations to get $\sin2\theta+2\sin(\theta+\phi)+\sin2\phi=2ab,$ or$$[2+2\cos(\theta-\phi)]\sin(\theta+\phi)=2ab.$$Squaring and adding the original equations gives$$2+2\cos(\theta-\phi)=b^2+a^2.$$Dividing these two results yields$$\sin(\theta+\phi)=\frac{2ab}{b^2+a^2}.$$