Solving a Trigonometric Equation: Primary to Secondary Solutions

[cscott]

How do I get from the primary to the secondary solution of a trigonometric equation? This book tells me that the second angle is within $-\pi \leq \theta \leq \pi$, in a different quadrent, but I don't follow :\

Thanks.

Edit: I got it (I think!): I can pick the correct quadrent using the sign of the ratio and knowing where each trig function is positive or negative. Then, I know (as an example) $\sin \theta = +\sin \alpha$, where $\alpha = 180 - \theta$ if we're talking about first and second quadrents.

Please correct me if I'm wrong, but I seem to be getting correct answers

 

[Zurtex]

Depends what the question is. For example:

$$\cos \theta = 0$$

There are the 2 solutions within the range $-\pi \leq \theta \leq \pi$ namely:

$$\theta = \frac{\pi}{2} \quad \text{and} \quad \theta = - \frac{\pi}{2}$$

 

[cscott]

Edit: So there is an infinite set of angles for $\pm 1$ (for sine and cosine), $\pm \infty$ (for tangent), and zero if there is no angle range? For every other ratio there is two solutions?

 

[mathman]

For cos >0, there are two possible solutions for the same cos, first and fourth quadrant. For cos <0, second and third. For sin >0, first and second, for sin <0, third and fourth.

 

[whozum]

Depends what the question is. For example:

$$\cos \theta = 0$$

There are the 2 solutions within the range $-\pi \leq \theta \leq \pi$ namely:

$$\theta = \frac{\pi}{2} \quad \text{and} \quad \theta = - \frac{\pi}{2}$$

Doesn't that range encompass the entire unit circle?

It doesn't need to be zero, the cosine and sine function both return values within their range twice if the domain is the entire unit circle.

$$\cos \theta = \frac{\sqrt{3}}{2}$$ has two solutions,

$$\frac{\pi}{6} \ and \ \frac{5\pi}{6}$$

For the cosine function, the range is symmetric about the x axis, which means the values between x = 0 and x = Pi are repeated from x = Pi to x = 2Pi. For the sine function the symmetry is about the y axis, and the values from x = Pi/2 to x = 3Pi/2 repeat over x = -Pi/2 to Pi/2 .

 

[cscott]

So, I take it "primary solution" and "secondary solution" are just the fancy words and I can just say the function has two sloutions.

Is my method in my OP correct, though?

Also, I edited my last post if you want to try to make sense of it :tongue:

 

[cscott]

Err... I see now the function can have many solutions, not just two, within the specified ranged.

My revised method: find the primary and secondary solutions and cater them to meet the specified range.

 

[whozum]

Sin and cosine return values between -1 and 1 for ANY x, since they are repeating functions. They complete one cycle in a domain interval of 2pi, which means if you take sin(x) and sin(x+2pi) you will get the same result, the same holds for any multiple of 2pi, the rule:

$$\sin x = \sin ( x + 2n \pi)$$ for integer n [/tex]. The same holds for cosine. The problem you've been given is in the domain x = -Pi to x = Pi, which is one complete cycle of the sin function. That means that in that domain you will have 2 solutions.

In one complete cycle the sin and cosine function return the same value twice.

 

[cscott]

Sin and cosine return values between -1 and 1 for ANY x, since they are repeating functions. They complete one cycle in a domain interval of 2pi, which means if you take sin(x) and sin(x+2pi) you will get the same result, the same holds for any multiple of 2pi, the rule:

$$\sin x = \sin ( x + 2n \pi)$$ for integer n [/tex]. The same holds for cosine. The problem you've been given is in the domain x = -Pi to x = Pi, which is one complete cycle of the sin function. That means that in that domain you will have 2 solutions.

In one complete cycle the sin and cosine function return the same value twice.

Thank you, that clears it up!