Serious trig. question, I need a step by step with the answers PLEASE!?

Question

I also need to know what number is representing each leg of the triangle. Is #x the smallest leg or #y, or vice versa. Here's the problem:

Consider an angle θ in standard position whose terminal side has the equation y = -5x, with x ≤ 0. Find the exact values of sin θ, cos θ, and tan θ.

I don't really need any explanations except for the leg part I noted above. All I need is to see the whole problem worked out and I'll get it. Our book is really vague, but I learn by seeing, so if you can help with the whole thing it'll be appreciated, and of course I'll reward you, serious people only please.

Answer 1

Here is what I would do. I hope you can adapt it to your way of understanding.

Graph the line y = -5x. It has slope -5 and goes through the origin. The part of the line for which x ≤ 0 forms the terminal side of θ. So it's the part of the line that is in Quadrant II. θ is greater than pi/2 but less than pi.

Pick a point on the line in Quadrant II; for example, choose x=-1, and then since y=-5x, y=5. Draw a vertical line segment from (-1,5) to the x-axis. This forms a triangle. Designate the acute angle opposite the vertical line segment as α. α is the "reference angle" for θ; it has the same trig values as θ except possibly for sign. Now, the triangle you've drawn has a horizontal leg of length 1, and a vertical leg of length 5. The hypotenuse (which is along the terminal side of θ) has length sqrt(26) (from Pythagorean theorem). So

cos(α) = adjacent/hypotenuse = 1/sqrt(26)
sin(α) = opposite/hypotenuse = 5/sqrt(26)

In Quadrant II, where the terminal side of θ is, x is negative and y is positive. Therefore

cos(θ) = -1/sqrt(26)
sin(θ) = 5/sqrt(26)

Finally, tan(θ) = sin(θ)/cos(θ) = -5

Alternatively, you could determine the point in Quadrant II where the terminal side of θ intersects the unit circle. The x-coordinate of that intersection point is cos(θ) and the y-coordinate of that intersection point is sin(θ). You would find this point by solving

x² + y² = 1 (the equation of the unit circle)
y = -5x

simultaneously, remembering that x < 0 in Quadrant II. You could solve this by substituting -5x for y in the circle equation:

x² + (-5x)² = 1
26x² = 1
x = ±1/sqrt(26)

But x < 0 in Quadrant II, so

x = -1/sqrt(26) = cos(θ)
y = -5x = 5/sqrt(26) = sin(θ)

the same (as it had better be!) as determined earlier.

Answer 2

a million) bypass the wide style to the right area of the equation: x^2 - 8x = 4 2) upload a time period to both area that makes the left area right into a acceptable sq.. the finest thanks to attempt it is to take the coefficient of the x time period (8), divide it by technique of two (8/2 = 4) and sq. it (4^2 = 16). hence, you should upload 16 to both area. x^2 - 8x + 16 = 4 + 16 3) aspect the appropriate sq.. (x - 4)^2 = 20 4) ensure for x. x - 4 = +/- sq. root 20 x = 4 +/- sq. root 20 The sq. root of 20 simplifies into 2 sq. root 5. hence, the answer is A.

Answer 3

Brick, all you need is a point on the terminal line for you to do the computations, Choose any negative value for x, compute y, and read off the appropriate values for your trig functions