How do I solve this trigonometric equation?


2 sin^2a +sin⁡ a cos ⁡a(cot⁡ a - tan ⁡a)

8 Answers

  • 1 month ago

    CotA - Tan A = CosA / SinA - SinA/CosA 


    Cos^2A - Sin^2A /(CosASinA) 

    Re inserting 

    2Sin^2A + SinACosA( Cos^2A - Sin^2A) / (SinACosA)

    Cancel down 'SinACosA

    2Sin^2A + ( Cos^A - Sin^2A) 

    Use the Trig. Identity Cos^2A = 1 - Sin^2A 


    2Sin^2A + ( 1 - Sin^2A - Sin^2A) 

    2Sin^2A + 1 - 2Sin^2A = '1'    NB ( 2Sin^2A add to 'zero'. 

  • 1 month ago

    Mistake: what you posted was not an equation, after

    making correction & re-post it.

  • 1 month ago

    i feel some thing is rong

  • 2 months ago

    Step 1 is to realize that it's not an equation.  It's an expression.  There's nothing to solve for.  The value of a can be anythign that makes both (tan a) and (cot a) defined.

    You can simplify, though.  If I don't see an identity I can use right away, I find it helps to write everything in terms of sines and cosines:

    2 sin² a + (sin a)(cos a)(cot a - tan a) = 2 sin² a + (sin a)(cos a)[(cos a)/(sin a) - (sin a)/(cos a)]

          = 2 sin² a + (cos² a - sin² a)

    That last step multiplied (sin a)(cos a) into each term in [] brackets and canceled denominators.  Finish with:

        = sin² a + cos² a

        = 1

    That's only true when both cos a and sin a are nonzero, making both the original expression and the cancellation of (sin a)/(sin a) and (cos a)/(cos a) valid.

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  • fcas80
    Lv 7
    2 months ago

    It's not an equation.

    However, if you set it to zero, i think there are no solutions.

  • 2 months ago

    work out the second term

    sin(a)cos(a)(cot(a) - tan(a)) = sin(a)cos(a)( cos(a)/sin(a) - sin(a)/cos(a))

                                                = cos^2(a) - sin^2(a)


    2 sin^2a +sin⁡ a cos ⁡a(cot⁡ a - tan ⁡a) = 2*sin^2(a) + cos^2(a) - sin^2(a) = sin^2(a) +cos^2(a) = 1

  • 2 months ago

    I meant "calculate"

  • rotchm
    Lv 7
    2 months ago

    You can't solve it because there's nothing to solve. To "solve" an equation you must have an equals sign.

    If you meant to find the roots then here just factor out sin(a). 

    What does this imply?

    To your update. Still your question doesn't make sense. Calculate what? What you type is like 5 + x, then you say calculate 5+x... well, its 5+x.  

    Type your question exactly as is it is posed to you. Provide a pic if need be.

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