Anonymous
Anonymous asked in Science & MathematicsMathematics · 11 months ago

cscx - sinx = cotx*cosx?

with steps please? ive been trying to figure it out to no luck

9 Answers

Relevance
  • 11 months ago
    Best Answer

    To prove that csc(x) - sin(x) = cot(x) * cos(x), we begin evaluating from the LHS to show that it's equal to the RHS, i.e.

    LHS:

    csc(x) - sin(x)

    = [1/sin(x)] - sin(x) {from the property: csc(x) = 1/sin (x)}

    = [1 - sin²(x)]/sin(x)

    = cos²(x)/sin (x) {from the Pythagorean Property: sin²(x) + cos²(x) = 1}

    = [cos(x)/sin(x)] * cos(x) {from the concept: cos²(x) = cos(x) * cos(x)}

    = cot(x) * cos(x)

    = RHS

  • H
    Lv 7
    11 months ago

    Prove Cscx - sinx = cotx*cosx

    There really is no specific step for trig identities. You just have to toy with it using your knowledge. The point of practicing these is to get your familiar with trig identities by getting you to apply them.

    (1/sin(x))-(sin(x))=(cos(x)/sin(x))(cos(x))

    Finding common denominator for the left side

    (1/sin(x))-(sin^2(x)/sin(x))=(cos^2(x)/sin(x))

    The rest can be done simply by understanding cos^2(x)+sin^2(x)=1

    If you get cos^2(x) by itself you get cos^2(x)=1-sin^2(x)

    You have that as a numerator at the left side so just substitute and you get

    cos^2(x)/sin(x)=cos^2(x)/sin(x) and you are done there

    • H
      Lv 7
      11 months agoReport

      And those three dots seem to be blocking the full thing.

  • It’s all trigonometry to me.

  • 11 months ago

    csc(x) - sin(x) (left)

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

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

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

    = cos^2(x)/sin(x)

    = (cos(x)/sin(x))cos(x)

    = cot(x)cos(x) (right). Done.

    You got it right. :)

  • How do you think about the answers? You can sign in to vote the answer.
  • 11 months ago

    LHS:

    csc (x) - sin (x)

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

    = 1/2 (csc(x) + cos(2 x) csc(x))

    = 1/2 (cos(2 x) + 1) csc(x)

    = cos(x) cot(x)

    = RHS

  • 11 months ago

    your answer is correct:

    cscx= 1/sinx

    1/sinx -sinx= (1-sin^2x)/ sinx = cos^2x/sinx

    we know that cotx= cosx/sinx

    cotx * cosx = (cosx/sinx) cosx = cos^2x/sinx

    Hence, proved

  • Como
    Lv 7
    11 months ago

    1/sin x - sin x

    1 - sin² x

    ---------------------

    sin x

    cos²x

    ------------ = cot x cos x

    sin x

  • Huh
    Lv 6
    11 months ago

    RHS:

    cot(x) * cos(x)

    = [cos(x) / sin(x)] * cos(x)

    = cos^2(x) / sin(x)

    = [1 - sin^2(x)] / sin(x) # Pythagorean Trigonometric identity:

    = [1 / sin(x)] - [sin^2(x) / sin(x)]

    = csc(x) - sin(x)

  • Pope
    Lv 7
    11 months ago

    LHS

    = csc(x) - sin(x)

    = 1/sin(x) - sin²(x)/sin(x)

    = [1 - sin²(x)]/sin(x)

    = cos²(x)/sin(x)

    = [cos(x)/sin(x)]cos(x)

    = cot(x)cos(x)

    = RHS

Still have questions? Get your answers by asking now.