more trigtastic fun!!?

Question

solve: sin(arctan (square root of 3)) + cos(arccot (square root of 3))
plz help me understand how to do this!!

Answer 1

sin(arctan(√3)) + cos(arccot(√3))

arctan(√3) = π/3
arccot(√3) = π/6

sin(arctan(√3)) + cos(arccot(√3))
= sin(π/3) + cos(π/6)
= √3/2 + √3/2 = √3

Answer 2

there are several ways to do this, but drawing right triangle is not such a bad way.

To find the sin(arctan(√3)), you are trying to find the sin of some angle. Call that angle x. So x = acrtan √3. That is, x is the angle whose tan is √3. Draw a right triangle with acute angle x. Since tan x = √3, put √3 as opposite and 1 as adjacent. Us√3/2e Pythagorus to help you find the hypotenuse. It's 2, right? So now we can compute sin x, just √3/2.

Go thru the same process to find cos(arccot√3). Again, you want the cos of some angle, say y. This means arccot √3 = y or, cot y = √3. In the right triangle now opposite is 1 and adjacent is √3. Still, hypotenuse is 2. Cos y is also √3/2.

So, the entire expression is is √3/2 + √3/2 = √3

You can do this strictly algebraically using sin^2 + cos^2 =1.

Answer 3

There's nothing to SOLVE, only something to EVALUATE. However, there are THREE possible values ! ( ### --- see my P.S.) They are:

+ sqrt(3), 0, and - sqrt(3).

[Later note: As of this writing, the 7 other responders have only given you the first possibility. That is because they have only considered acute angle solutions for the arctan and arccot functions. The problem, as set, contains no such limitation.]

I shall assume that you know how to draw a (30, 60, 90) deg. right-angled triangle with sides of length 1, sqrt(3) and 2 opposite these respective angles. The values of sine, cosine, and tangent of 30 or 60 deg. can then be read off easily from this triangle. The mnemonic SOHCAHTOA has been taight for generations to help one remember that sine = O/H, etc, where "O" means the "Opposite length" and "H" means the "Hypoteneuse." ("A" means the "Adjacent side.") The values of cotangents are the INVERSES of the corresponding tangents. Finally, you need to know how to obtain the values of these functions (including their signs) when the angles occupy any of the other three quadrants than just the 0 deg. to 90 deg. quadrant. Only then will you be equipped to obtain ALL THREE possible values for the sum posed!

O.K. that all understood, let's start:

arctan (square root of 3) = 60 deg. or 240 deg.

arccot (square root of 3) = 30 deg. or 210 deg.

So sin(arctan (square root of 3)) = sin 60 = [sqrt(3)]/2 or sin 240 = - [sqrt(3)]/2.

And cos(arccot (square root of 3)) = cos 30 = [sqrt(3)]/2 or cos 210 = - [sqrt(3)]/2.

Now any of the individual terms in sine or cosine are permitted --- there is NOTHING to say, for example, that the tangents or cotangents have to be in the SAME quadrant.

So the sum of these two terms COULD BE:

+ sqrt(3) (just from one combination), 0 (from two combionations) and - sqrt(3) (from one combination).

Live long and prosper.

### P.S. Notice that of the eight (8) answers so far, no-one else has given you anything but the rather limited evaluation for both angles being acute, i.e. the result : + sqrt(3). As I have shown you, if you consider all possible solutions for the arcctan and arccot functions in the range (0, 360) degrees, the values 0 and - sqrt(3) are also possible.

Answer 4

sin(arctan (square root of 3)) +
cos(arccot (square root of 3)) =
sin(60) + cos(30) =
(1/2)√3 + (1/2)√3 = √3

Answer 5

First, draw a right triangle.
Let one of the acute angles be arctan(sqrt(3)).
That means that the tangent of that angle is sqrt(3),
or that the opposite side could be sqrt(3) and the
adjacent side could be 1.
Use the pythagorean theorem to get the hypotenuse = 2.
Note that the other acute angle is the arccot(sqrt(3)).
So, sin(arctan(sqrt(3))) = sqrt(3)/2 and
cos(arccot(sqrt(3))) = sqrt(3)/2.
Add them together and you get sqrt(3)!

Answer 6

DRAW a TRIANGLE. arctan refers to the inverse of tan. So draw a right triangle... the tangent is root 3 / 1: so draw the two sides for tangent.. (SOH CAH TOA) opposite as root 3 and adjacent as 1. Solve the hypotenuse to be 2. Now take the sine in the same triangle. You get (root 3) / 2. Do the same for the other angle.. you get 1/2.

ANSWER: ( ( root 3) + 1)/2

(also, you can type this into your calculator: Any TI keystrokes would be [sin] [2nd][tan][root 3] ) ) + [cos] [2nd] [tan] (1/root3) ) )

Good luck!

Answer 7

sin( arctan(sqrt(3)) ) + cos ( arccot (sqrt(3)) )

Let's solve this individually.

Let t = arctan(sqrt(3)). Then
tan(t) = sqrt(3), which occurs at the point t = pi/3.

Therefore, sin(t) = sin(arctan(sqrt(3))) = sin(pi/3) = sqrt(3)/2.

For the second term, let
t = arccot(sqrt(3)). Then
cot(t) = sqrt(3), or
tan(t) = 1/sqrt(3), which occurs when
t = pi/6.

Therefore, cos(t) = cos(arccot(sqrt(3))) = cos(pi/6) = sqrt(3)/2.

sin( arctan(sqrt(3)) ) + cos ( arccot (sqrt(3)) ) =

sqrt(3)/2 + sqrt(3)/2 = 2sqrt(3)/2 = sqrt(3).

Answer 8

arctan (square root of 3) ...

consider the standard x-y plane

make a right triangle with one angle located at the origin and the right-angle located at (1,0) ...

[this is just for "scratch-paper visualization" .. don't worry ]

since tan(angle at origin) = opposite/adjacent
and you've already set the adjacent side = 1, then
the opposite side must be = sqrt(3)

[this is MUCH easier if you draw a little figure on paper]

in this triangle, the hypotenuse = sqrt(1^2 +
sqrt(3)^2) = sqrt (1 + 3) = sqrt(4) = 2
[[ Pythag theorem]]

so "sine" of this angle = oppsite / hypotenuse =
sqrt(3)/2 .... this completes the first part ..
no biggee

arccot (square root of 3)) ... draw anoth stick figure with "one angle at the origin" and the right angle at (0, sqrt(3) ) ... go vertical length "1"

since cot = adjacent/opposite = sqrt(3)/1

by [[ Pythag theorem]] again, the hypot = 2
so cos of this angle = adjacent/hypot =
sqrt(3)/2


add these two together:
sin(arctan (square root of 3)) + cos(arccot (square root of 3)) =

sqrt(3)/2 + sqrt(3)/2 = sqrt(3)

soooo,

sqrt(3) is your answer ....

hopefully it's clear if you draw the little scetches as suggest4ed ...

it's just easy triangles