Trig questions?

These are the two questions I do not understand

If tan A =  sqrt3/3 find tan 2A.

Simplify the following.

tan (22.5° )/1 − tan^2 (22.5° )

Please explain and show as many steps as possible, thank you!

Update:

Thank you all who replied, I now understand this much better!

5 Answers

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  • 1 month ago
    Favourite answer

    There's a bunch of formulas known as "trigonometric identities". One of these formulas states that:

    tan(2x) = 2*tan(x) / (1 - tan^2(x))

    When we apply that formula to our original problem:

    If tan(A) = sqrt(3)/3

    Then tan(2A) = 2*(sqrt(3)/3) / (1 - (sqrt(3)/3)^2)

    = (2*sqrt(3)/3) / (1 - (3/9))

    = (2*sqrt(3)/3) / (1 - 1/3)

    = (2*sqrt(3)/3) / (2/3)

    Cancel out the 2's and 3's:

    = sqrt(3) ---> ANSWER

    Next question: this is actually kind of related to the first question:

    22.5 is exactly half of 45

    So tan(22.5 degrees) / (1 - tan^2(22.5 degrees)) is of the form:

    (1/2) * 2*tan(x) / (1 - tan^2(x))

    Which we know is equal to (1/2)*tan(2x) from the trig identity above.

    So thus:

    tan(22.5 degrees) / (1 - tan^2(22.5 degrees)) = (1/2)*tan(2 * 22.5 degrees) = (1/2)*tan(45 degrees)

    We know that tan(45) = 1

    So thus:

    tan(22.5 degrees) / (1 - tan^2(22.5 degrees)) = 1/2 ---> ANSWER

  • Philip
    Lv 6
    1 month ago

    1) tan A = (1/3)rt3. tan(2A) = 2tanA/[1-tan^2(A)] = (2/3)rt3/[1 - (1/3)] = rt3

    2) Put (s,c) = [sin(22.5°) , cos(22.5°)];

    tan(22.5°)/[1 - tan^2(22.5°)] becomes (s/c)/[1 -(s/c)^2] = (s/c)c^2/(c^2-s^2) =;

    (1/2)2sc/(c^2-s^2) = (1/2)sin(45°)/cos(45°) = (1/2)tan(45°) = (1/2)(1) =  (1/2).

  • Ian H
    Lv 7
    1 month ago

    a) Let t = tan A = √(3)/3

    tan 2A = 2t/(1 – t^2) ........................(1)

    tan 2A = [2√(3)/3]/(2/3) = √(3)

    b) Simplify tan (22.5°)/[1 − tan^2 (22.5°)]

    Immediately recognise this from (1) as

    (1/2)tan (45) = 1/2

    For questions like this and for future tests you need to learn the

    sin, cos and tan values of 0, 30, 45, 60, 90 and 180, and similar values

    around the unit circle, (with certain sign changes), and those trig identities.

    So, take the good advice to print and learn

    https://tutorial.math.lamar.edu/pdf/Trig_Cheat_She...

  • 1 month ago

    Remember the Trig. Identity

    Tan(2A) = 2Tan(A) / (1 - Tan^2(A)). 

    Hence 

    Tan(2A) = (2(sqrt(3)/3)) / ( 1 - (sqrt(3)/3)^2)

    Tan(2A) = (2/3)Sqrt(3) / ( 1 - (3/9)

    Tan(2A) = (2/3)sqrt(3) /( 6/9)

    Tan(2A) = (2/3)sqrt(3) X 9/6 

    Cancel down by '2'  

    Tan(2A) = (sqrt(3)/3 X 9/3 

    Cancel down by '3' (('9')  

    Tan(2A) = sqrt(3) 

    2A = 60 degrees 

    A = 30 degrees. 

    tan (22.5° )/1 − tan^2 (22.5° )

    0.4142... / (1 - (0.4142...)^2)

    0.4142.... / 1 - 0.17151...) 

    0.4142... / 0.82842.... 

    1/2 = 0.5 

     

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  • 1 month ago

    First, print out: 

    https://tutorial.math.lamar.edu/pdf/Trig_Cheat_She...

    Now, find the identity for tan(2x)

    tan(2x) = 2tan(x)/(1-tan^2(x)

    so tan(2A) 

    = 2tan(A)/(1-tan^2(A)

    You know tan(A), so insert the values and do the calculations

    for the other: look at the formation: and compare to tan(2x) above...

    so tan(2*22.5) = 2tan(22.5)/(1-tan^2(22.5))

    so tan(22.5)/(1-tan^2(22.5) = tan(2*22,5)/2 = tan(45) / 2

    We can go a little farther:

    tan(45) = 1 

    ... something that you should know are 

    ... the sin, cos and tan values of 0, 30, 45, 60, 90 and 180

    ... you can look them up on the unit circle in the PDF above

    so the final simplification is:

    tan(22.5)/(1-tan^2(22.5) = tan(45)/2 = 1/2

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