How do you solve this trig problem?

Question

Find the value(s) of (inverse)Sec (5) the restriction is 0-2π

sorry but i wasn't able to type superscripts here

just had a test on this stuff and was wondering if I did it correctly

you have to use the unit circle....can't anyone do it?

Answer 1

Same as inverse cosine of 1/5

On the unit circle, it would be the angle between 0 and 90 where the x coordinate is 1/5.

Or just do cos inv of 1/5.

Answer 2

you need to memorize what applications are helpful wherein quadrants and that are adverse. which will help do those issues quickly. even nevertheless it is not too confusing to make certain in case you caricature the 4 quadrants and undergo in recommendations that the radius (the hypotenuse) is often helpful. interior the 1st question, you're given the sine, so which you recognize 2 facets of a triangle and you are going to be able to locate the third area via the tPythagorean theorem. that's 3(sqrt 5). you could now fill in different function values, remembering that any function that is composed of the adjoining area (which suits to a adverse x fee) would be adverse. interior the 2nd question, the sine is adverse, so which you recognize the attitude must be interior the third or fourth quadrant. The cotangent is helpful, so the tangent is helpful too (the cotangent is the reciprocal of the tangent). The cotangent is adverse interior the fourth quadrant, so the attitude won't be in a position to be there, so it is going to be interior the third quadrant. The cotangent is 4/3, and that's the adjoining over the different, so the hypotenuse must be 5 (a three-4-5 good triangle). you could now caricature the attitude. interior the third quadrant, the two the x and y values are adverse, so which you will fill interior the sign consistent with no rely if the fraction -- the ratio that defines the function -- has 2 adverse numbers or one helpful and one adverse.

Answer 3

INV SEC (5) over 0-2pi

well, that corresponds to INV COS (1/5)

which angles give the cosine value of 1/5? they occur in quadrants I and IV roughly around 80 degrees, more accurately, 78.5 degrees and 281.5 degrees.

Answer 4

If
secθ = 5
cosθ = 1/5 = 0.2
θ = 78.463°, 281.537°

In radians,
θ = 1.3694, 4.9137