The question was:
Solve the following on the interval . Give your answers to 3 sf.
I was able to solve this using the harmonic form, expressing the in terms of . I would like to know if it is possible, better, to solve this using identities, needless to say I can't see how and would appreciate a hint as how to start.
You would need the right trigonometric identity:
List of trigonometric identities - Linear combinations - Wikipedia
Edit: Oh never mind, I'm guessing that this is what you meant with the harmonic form.
As for finishing gives me and and gives me . The first and the last of these solutions I got using the harmonic form, which, since you mentioned them, leads me to believe that is an extraneous root, but looking at the graph of I don't see why. Could you please explain that.
Thank you very much.
Edit: I just checked the roots using a calculator and found that is not a solution. Is that how I check for extraneous roots, by seeing if they work?
To answer your question about extraneous roots, substitution into the original equation will in general show if the solution is extraneous or not.
I would have solved the equation much the same way you did:
Observing that in a right triangle, the two inverse trig. functions represent complementary angles, we have:
Adding to get the angle in the desired range, we get:
We could plug this into a calculator, but there is a slightly simpler representation.
Construct a right triangle with horizontal base 1, altitude 2 and hypotenuse with the right angle at the lower right, then from the lower left vertex draw a line segment to the vertical leg at the point 1/2 unit above the right angle, and we have the angle subtended by this segment and the hypotenuse as . Using the law of cosines, we find:
Which gave me and then
I will try and work through your method.
Thanks again for all your help
I just noticed a much simpler method to find . Rather than using the law of cosines, we may use the law of sines as well:
Multiplying through by 3/2 and recalling , we have:
All of this really points out that the method used by skeeter is the way to go.
Thanks again for your efforts.