Find the exact value without using a calculator (see problem)?

Question

sin[tan-1 (-3/square root of 5)]

Please show me how to solve this problem in detail. I would really appreciate it!

Answer 1

With these sort of problems, you need to draw a triangle.
Lets call tan x = -3/sqrt(5).
Now, usually with tan x = a/b, we make a the opposite side, and b the adjacent side. However, since we have a negative number, lets change something first:
Since tan x = -3/sqrt(5), we know that tan (-x) = 3/sqrt(5).

So, lets draw a triangle with one angle marked as -x, the opposite side having length 3, and the adjacent side having length sqrt(5). Using pythagoras, we can also work out the length of the other side:
y^2 = 3^2 + (sqrt(5))^2 = 9 + 5 = 14, so the other side is length sqrt(14).

Now, we want to work out sin(x). But we only have an angle of -x. By the same idea as before, we change it to: sin(x) = -sin(-x).

Now, sin(-x) is opposite/hypotenuse = 3/sqrt(14).

Thus sin(x) = -3/sqrt(14).

Answer 2

Draw a triangle to solve this problem. Since tan-1 always gives a value between -pi/2 and pi/2, and the tangent function isn't negative when x >=0, you'll want your triangle to have one side on the positive x-axis starting at the origin, one side connecting the origin to somewhere in the fourth quadrant (where x > 0, y < 0), and the third side parallel to the y-axis, joining the other sides.

Label the angle at the point at the origin "a". Say that a = tan-1(-3 / sqrt(5)). Then this means that the ratio of the side opposite the angle to the side adjacent to the angle is -3/(sqrt 5). The "opposite" side in this case is the one parallel to the y-axis; label it "3". (We don't have to worry about the negative sign; this is taken care of because we're in the fourth quadrant.) The "adjacent" side is the one on the x-axis; label it sqrt(5).

Now, we can compute the length of the third side (the slanty one) using the Pythagorean Theorem. If you do this, you'll get that it's sqrt(5 + 9) = sqrt(14). Now we know all three sides of our right triangle, so we can compute the sine of a. Looking at the diagram, the opposite side has length -3 (since it's in the fourth quadrant, the y-value is negative), and the adjacent side has length sqrt(14) (which we just computed). Thus, the sine of a is equal to -3/sqrt(14).

Therefore, sin (tan-1 (-3/sqrt(5))) = -3/sqrt(14).

Answer 3

The tangent is the opposite over the adjacent, right?

So, inverse tangent of -3/√5 would be the angle in the 4th quadrant formed by drawing a line through the point (√5, -3).

If you then draw a line straight up from that point to the x-axis, you have a right triangle with adjacent = √5 and opposite = -3.

The hypotenuse of that triangle would be √[(-3)^2 + (√5)^2] = √14

So the sin of that angle would be opposite over hypotenuse, or -3/√14 = -3√14/14

Answer 4

draw a picture where X is an angle, opposite side is -3, adjacent side is sqrt(5). then tan x = -3/sqrt(5). hypothenuse is sqrt(14). plug and chug, you get sin[tan-1 (-3/square root of 5)] = sin[x] and from your picture that is -3 / sqrt(14).

Answer 5

arctan(– ?3)= - pi / 3. revise your universal tables. oh, and yeah, the precise fee in radians won't be able to be stated as pi is irrational. even if the precise fee in degrees is - 60 degrees the fashion of the arctan function is purely ( - ninety , ninety ) so the linked fee is purely - 60 and under no circumstances 100 and twenty, or three hundred, or something else. ----------- the arctan function is a one-one function and can want to take purely one fee. yet in case you write tan theta = - rt 3, we would want to get a regular answer, 180n - 60.

Answer 6

Let α = tan^(-1) (3/√(5))
and let α' = tan^(-1) (3/√(5))
ie tan α' = 3/√(5)

So opposite = 3, adjacent = √(5) and thus hypotenuse = √(14)

Thus sin α' = 3/√(14)

Since in fact tan α = -3/√(5)

Then α either lies in the 2nd or 4th quadrants so sin α is either positive or negative.

So sin α = ±3/√(14)

ie sin (tan^(-1) (-3/√(5))) = ±3/√(14)

Answer 7

forget that it is impossible talk about math