jonnburton
Junior Member
- Joined
- Dec 16, 2012
- Messages
- 155
I've spend a good deal of time today trying to solve some problems in this section of my book, with little success. The easier examples are fairly straightforward for me to see the identity in question and reach a solution but as they get more difficult I often don't really know how to proceed.
This is an example of a question I've been working on and how I approached it. To be honest, I think I am just 'stabbing in the dark' when I begin these problems and just start re-arranging things in the hope that an identity becomes apparent. But no matter how much I fiddled around with this, I simply can't see how to deal with it.
Solve the equation \(\displaystyle cos x sin \frac{\pi}{3} + sin x cos\frac {\pi}{3} = \frac{1}{4}\)
(as an aside, my first thought was this appears to be of the form \(\displaystyle sin (A + B) = sinAcosB+ cosA sin B\) but this doesn't seem to lead anywhere).
The other thing I noticed was that \(\displaystyle \frac{1}{4} = sin \frac{\frac{\pi}{3}}{2}\)
\(\displaystyle cos x sin \frac{\pi}{3} + sin x cos\frac {\pi}{3} = sin \frac{\frac{\pi}{3}}{2}\)
Then I thought dividing all terms by \(\displaystyle cos\frac{\pi}{3} sin\frac{\pi}{3}\) would help bring the x's to one side:
\(\displaystyle \frac{cos x}{cos\frac{\pi}{3}} + \frac{sin x}{cos\frac{\pi}{3}} = \frac{sin\frac{\pi}{3}}{2 (cos\frac{\pi}{3} sin\frac{\pi}{3})}\)
\(\displaystyle \frac{cosx + sinx}{cos\frac{\pi}{3}sin\frac{\pi}{3}} = \frac{1}{2} \frac{sin \frac{\pi}{3}}{(cos\frac{\pi}{3} sin\frac{\pi}{3})}\)
Multiplying both sides by 2:
\(\displaystyle \frac{2 cos x + 2 sin x}{ cos\frac{\pi}{3} sin\frac{\pi}{3}} = \frac{sin \frac{\pi}{3}}{(cos\frac{\pi}{3} sin\frac{\pi}{3})}\)
Now the denominator is the same so:
\(\displaystyle 2 cos x + 2 sin x = sin{\pi}{3}\)
If this is correct, I still can't see how the equation can be solved (however, it is the closest I've managed to get to a solution).
I would be very grateful for any pointers anyone might be able to give on this.
This is an example of a question I've been working on and how I approached it. To be honest, I think I am just 'stabbing in the dark' when I begin these problems and just start re-arranging things in the hope that an identity becomes apparent. But no matter how much I fiddled around with this, I simply can't see how to deal with it.
Solve the equation \(\displaystyle cos x sin \frac{\pi}{3} + sin x cos\frac {\pi}{3} = \frac{1}{4}\)
(as an aside, my first thought was this appears to be of the form \(\displaystyle sin (A + B) = sinAcosB+ cosA sin B\) but this doesn't seem to lead anywhere).
The other thing I noticed was that \(\displaystyle \frac{1}{4} = sin \frac{\frac{\pi}{3}}{2}\)
\(\displaystyle cos x sin \frac{\pi}{3} + sin x cos\frac {\pi}{3} = sin \frac{\frac{\pi}{3}}{2}\)
Then I thought dividing all terms by \(\displaystyle cos\frac{\pi}{3} sin\frac{\pi}{3}\) would help bring the x's to one side:
\(\displaystyle \frac{cos x}{cos\frac{\pi}{3}} + \frac{sin x}{cos\frac{\pi}{3}} = \frac{sin\frac{\pi}{3}}{2 (cos\frac{\pi}{3} sin\frac{\pi}{3})}\)
\(\displaystyle \frac{cosx + sinx}{cos\frac{\pi}{3}sin\frac{\pi}{3}} = \frac{1}{2} \frac{sin \frac{\pi}{3}}{(cos\frac{\pi}{3} sin\frac{\pi}{3})}\)
Multiplying both sides by 2:
\(\displaystyle \frac{2 cos x + 2 sin x}{ cos\frac{\pi}{3} sin\frac{\pi}{3}} = \frac{sin \frac{\pi}{3}}{(cos\frac{\pi}{3} sin\frac{\pi}{3})}\)
Now the denominator is the same so:
\(\displaystyle 2 cos x + 2 sin x = sin{\pi}{3}\)
If this is correct, I still can't see how the equation can be solved (however, it is the closest I've managed to get to a solution).
I would be very grateful for any pointers anyone might be able to give on this.