Makes complete sense thank you so much!!!
My only question is how come we respect domains only sometimes...
For example
now sin©¬/cos©¬ = (1/a)/(¡î3/a) = 1/¡î3
tan©¬ = 1/¡î3
©¬ = 30¨¬ or pi/6
then sin 30¨¬ = 1/a
1/2 = 1/a ---> a = 2
so our original ¡î3 sin x + cos x = 1 becomes
2sin(x+30¨¬) = 1
sin(x+30¨¬) = 1/2
x+30 = 0 or x+30 = 150¨¬
x = 0¨¬ or 120¨¬
the period of sin(x+30) is 360¨¬ so we can add 360 to any answer as long as that keeps us in our domain
so x = 0, 120 , or 360
just like I had before
Makes sense... perfect sense but...
notice when we decided on beta
tan©¬ = 1/¡î3
©¬ = 30¨¬ or pi/6
we used pi/6 not (7 pi)/6 because of the restriction of tangent is between (3 pi)/2 and pi/2
(7 pi)/6 is a perfect solution for beta but it fall out of the restrictios for tangent so we ignore it...
but what does not make sense
sin(x+30¨¬) = 1/2
x+30 = 0 or x+30 = 150¨¬
x = 0¨¬ or 120¨¬
the period of sin(x+30) is 360¨¬ so we can add 360 to any answer as long as that keeps us in our domain
so x = 0, 120 , or 360
makes sense makes sense...
but we did not respect the domain here and i don't know why but we still get the right answer...
how come???
see
sin(x+30¨¬) = 1/2
sin^-1 1/2
has two solutions pi/6 or(5 pi)/6 and the domain of sine is between (3 pi)/2 and pi/2 and as we can see the solution (5 pi)/6 falls out of this restriction...
x+30 = 30 or x+30 = 150¨¬
which gives us the right answers...
of 0 or (2 pi)/3
even though (5 pi)/6 was out of the domian of sin and we did not respect the domian here
but we respected the domain for tangent...
I just have to know why this is so...
I like how Jishka adds random symbols to my posts lol awsome :O
It seems like you have a good understanding of trigonometric functions and how to solve equations involving them. In this specific case, you're asking why the solution to sin(x+30°) = 1/2 includes values that fall outside the allowed domain for the sine function.
First, let's clarify that the domain of the sine function is defined for all real numbers. So technically, there are no restrictions on the domain of the sine function itself.
However, when we solve the equation sin(x+30°) = 1/2, we are looking for specific values of x that satisfy the equation. The solution to this equation requires finding the inverse sine (also known as arcsine) of 1/2. The inverse sine function, denoted as sin^(-1), returns angles within a specific range.
The standard range for inverse sine is between -pi/2 and pi/2 (or -90° and 90°). So when we calculate sin^(-1)(1/2), we get two possible angles: pi/6 (or 30°) and (5pi)/6 (or 150°). However, as you correctly pointed out, the sine function is restricted to values between -pi/2 and pi/2, so we discard the angle (5pi)/6 since it falls outside this range.
Now, let's discuss why the solutions x = 0 and x = (2pi)/3 are still valid, even though they include an angle outside the restricted range for the sine function.
When solving trigonometric equations, we often consider periodicity because the trigonometric functions repeat themselves after certain intervals. For the sine function, this interval is 2pi (or 360°). This means that if x is a solution to the equation sin(x+30°) = 1/2, then adding any multiple of 2pi to x will also be a solution.
In this case, the equation sin(x+30°) = 1/2 has a period of 2pi because of the added constant 30°. So we can add any multiple of 2pi (or 360°) to the solutions within the restricted range to find all possible solutions outside that range.
When you solved x+30 = 0°, you found x = 0° as a solution. Adding 360° to this gives us x = 360°, which is still a valid solution within the periodicity of the sine function.
Similarly, when you solved x+30 = 150°, you found x = 120° as a solution. Adding 360° to this gives us x = 480°, which is also a valid solution within the periodicity.
So, by adding multiples of 2pi (or 360°) to the valid solutions within the restricted range of the sine function, we still obtain valid solutions for the original equation sin(x+30°) = 1/2. This is why x = 0, 120°, and 360° are all valid solutions.
In summary, while we respect the domain restrictions when finding solutions using inverse trigonometric functions, we can still consider the periodic nature of trigonometric functions to find additional solutions outside the restricted range.