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Apr 8, 2016 · tan(π 6) = √3 3. Using a calculator, tan(π 6) ≈ 0.577. tan (pi/6)=1/sqrt (3)=sqrt (3)/3 (see image below) This is one of the standard trigonometric triangles. sqrt (3) has been determined using Pythagorean Theorem. Sum of interior angles of a triangle is always pi radians.
Apr 19, 2016 · Answer link. sqrt3 If you know the values of sin (pi/3) and cos (pi/3), you can write that tan (pi/3)=sin (pi/3)/cos (pi/3)= (sqrt3/2)/ (1/2)=sqrt3/2 (2/1)=sqrt3 Alternatively, you could think of this as tan (60˚), and then draw a 30˚-60˚-90˚ triangle: tan (60˚) will be equal to "opposite"/"adjacent" in reference to the 60˚ angle, so we ...
Mar 13, 2018 · Explanation: Use special right triangles to find the value. Since π 4 = 45 degrees, use the special right triangle on the left. If you do the tan(π 4), you do the opposite adjacent. According to the triangle, the ratio is just one. 1 Use special right triangles to find the value.
Nov 5, 2015 · Arctan(− 1) = − π 4. Answer link. The answer is -pi/4 Alright, archtan / tan^-1 (x) is the inverse of tangent. Tan is sin/cos. Like the inverse of sin, the inverse of tan is also restricted to quadrants 1 and 4. Knowing this we are solving for the inverse of tan -1. We are basically being asked the question what angle/radian does tan (-1 ...
Mar 15, 2017 · There are 2 real roots: tan(π 12) = − b 2a ± d 2a = − 2√3 2 ± 4 2 = −√3 ± 2. Since tan (pi/12) is positive, there fore: tan(π 12) = 2 −√3. Answer link. tan (pi/12) = 2 - sqrt3 Use trig identity: tan 2a = (2tan a)/ (1 - tan^2 a) In this case, trig table gives: tan (pi/6) = (2tan (pi/12))/ (1 - tan^2 (pi/12)) = 1/sqrt3 Cross ...
Feb 24, 2017 · Explanation: Using the sum identity: tan(π +θ) = tanπ+ tanθ 1 −tanπ ⋅ tanθ. tan(π) = sinπ cosπ = 0 −1 = 0. Therefore: tan(π +θ) = 0 + tanθ 1 −0 = tanθ. Answer link. tan (pi+theta) = tan theta I don't know the tan addition formulae of the top, so I'd go with the sine/cosine ones: tan (pi+theta) = ( sin (pi+theta) )/ (cos (pi ...
Jun 21, 2016 · Check by calculator. tan(π 12) = tan15∘ = 0.27. 2 − √3 = 2 −1.73 = 0.27. OK. Answer link. 2 - sqrt3 Use the trig identity: tan 2a = (2tan a)/ (1 - tan^2 a) Trig table --> tan 2a = tan (pi/6) = 1/sqrt3 Call tan (pi/12) = t, we get: 1/sqrt3 = (2t)/ (1 - t^2). Cross multiply --> (1 - t^2) = 2sqrt3 t t^2 + 2sqr3t - 1 = 0 Solve this ...
Aug 4, 2016 · θ = tan−1(tan(5π 6)) by definition satisfies both of the conditions: Xtanθ = tan(5π 6) X − π 2 <θ <π 2. Note that tan has period π, so for any integer n: tan(5π 6 + nπ) = tan(5π 6) When n = −1, we have: 5π 6 + nπ = 5π 6 − π = − π 6. which lies in the range (− π 2, π 2), so satisfies the second condition for tan−1 ...
Apr 4, 2016 · Verify trig expression Two ways. 1. First way. Using the trig unit circle. tan (pi - x) = - tan x 2. Second way. Apply the identity: tan (a - b) = (tan a - tan b)/ (1 - tan a.tan b) tan (pi - x) = (tan pi - tan x)/ (1 - tan (pi).tan x) Since tan (pi) = 0, therefor: tan (pi - x) = - tan x.
Nov 7, 2015 · Find exact value of tan (-pi/12) Ans: (sqrt3 - 2) Call tan ((-pi)/12) = tan x. tan 2x = tan ((-2pi)/12) = - tan (pi/6) = - sqrt3/3 = - 1/sqrt3 Apply the trig identity ...
