aa asked in Science & MathematicsMathematics · 8 years ago

challenging trig identity please help!?


sin(x) + sin(x + 14π/3) + sin(x - 8π/3) = 0

thanks in advance! please show me your steps


hi Pope, do you mind explaining your answer a bit? :/ im not following the switch to cos and the negative sign out of the bracket thanks!

3 Answers

  • Pope
    Lv 7
    8 years ago
    Favorite Answer

    LHS = sin(x) + sin(x + 14π/3) + sin(x - 8π/3)

    = sin(x) + sin(x + 2π/3) + sin(x - 2π/3)

    = sin(x) + sin(x)cos(2π/3) + cos(x)sin(2π/3) + sin(x)cos(-2π/3) + cos(x)sin(-2π/3)

    = sin(x) + sin(x)cos(2π/3) + cos(x)sin(2π/3) + sin(x)cos(2π/3) - cos(x)sin(2π/3)

    = sin(x) + 2sin(x)cos(2π/3)

    = sin(x) + 2sin(x)(-1/2)

    = sin(x) - sin(x)

    = 0

    = RHS

    Here are a few identities that may address your followup questions:

    sin(A + B) = sin(A)cos(B) + cos(A)sin(B)

    sin(-A) = -sin(A)

    cos(-A) = cos(A)

  • Anonymous
    8 years ago

    sin(x) + sin(x + 14π/3) + sin(x - 8π/3) = 0

    First, you substitute the sins to the parentheses.

    sin(x)+ sin(x)+ sin(14π/3)+ sin(x)- sin(8π/3) = 0

    Next, you add all the single sins. There are 3 single sins with just an x

    3sin(x)+ sin(14π/3)- sin(8π/3) = 0

    The complex sins with what is in the parentheses have the same denominator. all you do is subtract what is in the parentheses. 14-8=6

    3sin(x)+ sin(6π/3)= 0

    Subtract sin(6π/3) from both sides

    3sin= -sin(6π/3)

    Divide both side by 3

    sin= sin(2π)

    After that, i don't know. (I hope i was able to help you a little)

  • 4 years ago

    i stumble at the ideal concern to do as soon as you could memorize distinctive formula is to make some flashcards. indoors of in simple terms some classes, you will possibly rather memorize 15-20 formula. I as soon as used that technique to memorize over a hundred thirty formula for an actuary examination. yet yet another trick: take the derivative of a number of your purposes. you could desire to have the technique of calculate the derivative of tanx by employing translating to sinx and cosx and employing the quotient rule. one greater trick: use your graphing calc. on your income. as soon as you have a TI-89 you could't have any problems as that calculator can compute highest integrals explicitly. yet, as soon as you have a TI-eighty 3 or TI-eighty 4 it extremely is easy to stick to numerical needed to check your strategies. i stumble on it mind-blowing what variety scholars do not use their graphing calculators to their complete income. I actually have had many suited exams, because of the actual fact I checked my paintings so heavily with a graphing calc. very final trick: as soon as you're given a group of tanx's and secx's you could continuously convert these to sinx and cosx, and likewise you could realize what the needed of sinx and cosx are, and function the technique of stick to a alleviation formula if mandatory.

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