posted by .

ok so I'm learning to solve problems like




I've wrapped my head around the process required to do so and I must say my brain is like @_@ mosh right now. Whenever I ask this question I never get a direct answer. I understand it's important to be able to define such angles and that this field of math is a valid field and worth studying but why??? Why would I ever need to know how to do this? Like it really hurts my head and I don't see when I would ever need to be able to solve such a problem in real life other than a text book...

  • MATH -


  • MATH -

    I agree with you... I will not be able to solve for x either, when I am given
    cos(x)=2, or sin(x)=5.

    The reason is because for any real value of x, cos(x) and sin(x) vary between -1 and +1. This means that the solutions to the two given equations do not exist!

    I hope you have not made a typo, unless your teacher is training you to think outside of the box.

    As to whether you'll ever going to need to know how to solve these problems, it all depends on what you want to do later in life. If you want to flip burgers for the rest of your life, you will never need to know this stuff. If you will be doing renovations, landscaping, and a whole lot of "ordinary guy" jobs, this skill will definitely help you.

  • MATH -

    By the way George, you said you have done the programming exercises for the whole computer programming textbook before classes even started, you could not possibly have this cos(x) problem bother you, could you?

    May I also add computer programming as one of the areas where trigonometry would be a definite asset?

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Trigonometry

    I need to prove that the following is true. Thanks. csc^2(A/2)=2secA/secA-1 Right Side=(2/cosA)/(1/cosA - 1) = (2/cosA)/[(1-cosA)/cosA] =2/cosA x (cosA)/(1-cosA) =2/(1-cosA) now recall cos 2X = cos^2 X - sin^2 X and we could say cos …
  2. Cos(pi/4 + h) = 2^(-½)[1 - h - h^2/2 + h^3/6 +..]

    This reply to some student got deleted, so I'm reposting it. Cos(pi/4 + h) = Cos(pi/4)Cos(h) - Sin(pi/4)Sin(h) = 1/sqrt(2)[1 - h - h^2/2 + h^3/6 + h^4/24 - h^5/120 - h^6/720 + ...] Note that the way students like you are asked to solve …
  3. Trigonometry

    I need help with I just can't seem to get anywhere. this is as far as I have got: Solve for b arcsin(b)+ 2arctan(b)=pi arcsin(b)=pi-2arctan(b) b=sin(pi-2arctan(b)) Sub in Sin difference identity let 2U=(2arctan(b)) sin(a-b)=sinacosb-cosasinb …
  4. Mathematics - Trigonometric Identities

    Let y represent theta Prove: 1 + 1/tan^2y = 1/sin^2y My Answer: LS: = 1 + 1/tan^2y = (sin^2y + cos^2y) + 1 /(sin^2y/cos^2y) = (sin^2y + cos^2y) + 1 x (cos^2y/sin^2y) = (sin^2y + cos^2y) + (sin^2y + cos^2y) (cos^2y/sin^2y) = (sin^2y …
  5. Trigonometry

    Hello all, In our math class, we are practicing the trigonometric identities (i.e., sin^2(x)+cos^2(x)=1 or cot(x)=cos(x)/sin(x). Now, we are working on proofs that two sides of an equation are equal (for example, sin(x)*csc(x)=1; sin(x)csc(x)=sin(x)/sin(x)=1; …
  6. TRIG!

    Posted by hayden on Monday, February 23, 2009 at 4:05pm. sin^6 x + cos^6 x=1 - (3/4)sin^2 2x work on one side only! Responses Trig please help! - Reiny, Monday, February 23, 2009 at 4:27pm LS looks like the sum of cubes sin^6 x + cos^6 …
  7. Math

    Solve this equation algebraically: (1-sin x)/cos x = cos x/(1+sin x) --- I know the answer is an identity, and when graphed, it looks like cot x. I just don't know how to get there. I tried multiplying each side by its conjugate, but …
  8. Precal

    I do not understand how to do this problem ((sin^3 A + cos^3 A)/(sin A + cos A) ) = 1 - sin A cos A note that all the trig terms are closed right after there A's example sin A cos A = sin (A) cos (A) I wrote it out like this 0 = - …
  9. Calculus - MathMate Please help

    ok, i tried to do what you told me but i cant solve it for c because they cancel each others out! the integral for the first one i got is [sin(c)cos(x)-cos(c)sin(x)+sin(x)+c] and the integral for the 2nd one i got is [-sin(c)cos(x)+cos(c)sin(x)-sin(x)+c] …
  10. english

    Conservatives love to teach and you decide. Liberals love to teach and enforce their views on you. They even make it a law. There is one thing I am learning about liberals, they must attend a school of learning in how to talk over …

More Similar Questions