# Thread: i Have tried this sum,,,but couldnt. can you help

2. ## Re: i Have tried this sum,,,but couldnt. can you help

this is a very simple sum

3. ## Re: i Have tried this sum,,,but couldnt. can you help

What have you tried? Did you start with the sum of angles formula? What about $\sin^2 x + \cos^2 x = 1$? If you use that, you should be able to find at least a possible $f(x)$.

4. ## Re: i Have tried this sum,,,but couldnt. can you help

This is what i got

5. ## Re: i Have tried this sum,,,but couldnt. can you help

Good start. You did not apply the sum of angles formula for sine correctly. The sum of angles formula is:

$$\sin(\alpha+\beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$$

Applying that to $\sin(3x)$ gives:

$$\sin(3x) = \sin(2x)\cos x + \cos(2x)\sin x$$

Now, you can apply the double angle formula for $\sin (2x) = 2\sin x \cos x$ and for $\cos(2x) = \cos^2 x - \sin^2 x = 2\cos^2 x - 1 = 1-2\sin^2 x$.

6. ## Re: i Have tried this sum,,,but couldnt. can you help

I used sin C- sin D

7. ## Re: i Have tried this sum,,,but couldnt. can you help

Originally Posted by IamSujith
I used sin C- sin D
Oh, my mistake. I missed that. Good point. In that case, I'm not seeing it. What makes you think this is possible?

8. ## Re: i Have tried this sum,,,but couldnt. can you help

Its a question from a final Advance level school paper. Then there must be a way.

9. ## Re: i Have tried this sum,,,but couldnt. can you help

Wolfram|Alpha: Computational Intelligence

According to Wolframalpha, there are no real solutions. Is it possible the problem statement has a typo? Maybe a plus should be a minus or vice versa?