Trigonometric equation help?
1) cos(2x) - sin^2(x/2) + 3/4 = 0 substitute cos(2x) = 2cos^2(x) - 1 and sin^2(x/2) = (1-cosx)/2 2cos^2(x) - 1 - 1/2 + (1/2)cosx + 3/4 = 0 multiply with 4 8cos^2(x) - 4 - 2 + 2cosx + 3 = 0 simplify 8cos^2(x) + 2cosx -3 = 0 factorize 8cos^2(x) - 4cosx + 6cosx - 3 = 0 4cosx(2cosx - 1) + 3(2cosx - 1) = 0 (2cosx - 1)(4cosx + 3) = 0 cosx = 1/2 or -3/4 x = pi/3, 5pi/3, 77pi/100 , 123pi/100 in the interval [0 - 2pi] 2) cos(4x) = sin(2x) 1 - 2sin^2(2x) = sin(2x) 2sin^2(2x) + sin(2x) - 1 = 0 2sin^2(2x) + 2sin(2x) - sin(2x) - 1 = 0 2sin(2x)(sin(2x) + 1) - 1(sin(2x) + 1) = 0 (sin(2x) + 1)(2sin(2x) - 1) = 0 sin(2x) = -1 or 1/2 2x = 3pi/2, 7pi/2, pi/6, 5pi/6, 13pi/6, 17pi/6 x = 3pi/4, 7pi/4, pi/12, 5pi/12, 13pi/12, 17pi/12
Trigonometric equations help?
1) 3sinx - 4cosx = 2 3sinx = 4cosx + 2 squaring 9sin^2(x) = 16cos^2(x) + 16cosx + 4 9 - 9cos^2(x) = 16cos^2(x) + 16cosx + 4 25cos^2(x) + 16cosx - 5 = 0 cosx = [-16 +/- sqrt(256 + 500)]/50 cosx = [-16 +/- 27.5]/50 cosx = 0.23 or -0.87 x = 76.7, 283.3, 150.5 , 209.5 degrees 2) tan(x+15) = 3tanx (tanx + tan15) /(1 - tanx tan15) = 3tanx tanx + tan15 = 3tanx - 3tan^2(x)tan15 2tanx = tan15(3tan^2(x) + 1) 2tanx = (3tan^2(x) + 1)*tan(45-30) 2tanx = (3tan^2(x) + 1)* (√3 -1)/(√3+1) 2tanx = (3tan^2(x) + 1)(2-√3) [2/(2-√3)]tanx = 3tan^2(x) + 1 (4 + 2√3)tanx = 3tan^2(x) + 1 3tan^2(x) - (4 + 2√3)tanx + 1 = 0 use quadratic formula [a= 3, b =-(4 + 2√3), c = 1]to solve tanx
I Need help solving some trigonometric equations.?
1) sin x = cos x when x = pi/4 and 5pi/4 2) sin u = 1/2 when u = pi/6, 11pi/6, 13pi/6, and 23pi/6 (because it's 2x, you need to go around the circle twice) so... 2x = pi/6, 11pi/6, 13pi/6, and 23pi/6 x = pi/12, 11pi/12, 13pi/12, and 23pi/12 3) sin u = sqrt 3 / 2 when u = pi/3, 5pi/3, 7pi/3, 11pi/3, 13pi/3, 17pi/3 So... 3x = pi/3, 5pi/3, 7pi/3, 11pi/3, 13pi/3, 17pi/3 x = pi/9, 5pi/9, 7pi/9, 11pi/9, 13pi/9, 17pi/9 4) cos u = -1/2 when u = 5pi/6, 7pi/6, 17pi/6, 19pi/6, 29pi/6, 31pi/6, 41pi/6, 43pi/6 So, 4x = 5pi/6, 7pi/6, 17pi/6, 19pi/6, 29pi/6, 31pi/6, 41pi/6, 43pi/6 divide all by 4 to get what x = 5) tan u = 0 when u = pi and 3pi so 2x = pi and 3pi x = pi/2, 3pi/2
How do I solve this trigonometric equation for alpha?
just use the standard technique of solving the trigonometric equation of the form a cos A + b sin A=cdivide the whole equation with (a^2+b^2)^(1/2)assume the coefficient of cosine term as Sin Band the coefficient of sine term will automatically be Cos B where tan B=(a/b)now you have SinB CosA + CosB SinA=c replace the left hand side with Sin(A+B)now you have Sin(A+B)=c which you can solve easily.
Need help for solving trigonometric equation of ward problems. (T .T)?
Here are solutions to your problems... https://drive.google.com/file/d/0B93wLOq... Hope this helps
Trigonometric equations, radicals and identities, help needed?
1a. The reference angle for 210 is 30 (it is found by subtracting 180 from the angle). cos(30) = √3/2 Because 210 is in the third quadrant and cosine is negative in the third quadrant, then cos(210) = -cos(30) = -√3/2 So 1/3 + 2cos(210) = 1/3 - 2√3/2 = 1/3 - √3 = (1-3√3)/3 1b. The reference angle for 225 is 45. cos(45) = √2/2 cos(225) = -cos(45) = -√2/2 The reference angle for 240 is 60. sin(240) = -sin(60) = -√3/2 cos(60) = 1/2 cos(225)/sin(240) + cos(60) = (-√2/2)/(-√3/2) + 1/2 = (√2/2)(2/√3) + 1/2 = √(2/3) + 1/2 = √6/3 + 1/2 = (2√6+3)/6 2. Check out these trig identities: http://www.clarku.edu/~djoyce/trig/ident... 8. sinx = -√2/2 We are looking for the angle(s) whose sine is -√2/2. We know that sin(45) = √2/2. But since it's negative, we need the corresponding angle in the third and fourth quadrant. 180+45 = 225 360-45 = 315 Answer: 225 and 315 degrees. 9. Use the trig identities from the link above. Start with the left side and transform it into the right side. 10. This is similar to #1. You try it. 11. 2 cosx - √3 = 0 Add √3 to each side. 2cosx = √3 Divide both sides by 2. cosx = √3/2 You need to find the angle whose cosine is √3/2. You will have one angle in the first quadrant and one angle in the fourth quadrant.
Which is a good book for trigonometric equations, inverse trigonometry for iit jee?
Trigonometry requires a lot of practice to get a good grasp of. And it’s very tricky of course. I’ll suggest a few books to help you out:Skills in Mathematics-Trigonometry for JEE Main and Advanced-Amit M Agarwal-The best book for JEE. It has loads of examples,subjective questions and MCQ. It should suffice.Plane Trigonometry Part 1-SL Loney(optional)-The godbook for trigonometry. So;ve the height and distance chapter and solution and properties of triangles part. Also try the miscellaneous section of the book. The earlier chapters are easy,so you can skip them.Arihant’s Skills in Mathematics Play with Graphs-Amit M Agarwal-It has all the possible type of questions you can solve by drawing a graph. It’s really helping in solving problems. It also helps you to find the solutions of trigonometric inequalities.For more rigorous practice,you may go through the following MCQ books:Mathematics MCQ-Bharati Bhawan publishers-A Dasgupta(optional)-It has loads of MCQs,Practice tests-a very good practice book indeed.Comprehensive Mathematics for JEE Advanced-Tata McGraw Hill-The best book for JEE Maths-it is suggested by almost all JEE aspirants,toppers and mentors. It has a huge collection of solved examples based on the current JEE Advanced format of questions asked. It also has past year questions at the end of each chapter.Also,do solve the previous years’ questions. It helps you boost your confidence and you’ll get to know what type of questions are asked.Hope this helps.
I need help around trigonometric functions. How do you solve this?
You've got a right triangle with a 36° angle in it. The hypotenuse is 50 and you want the opposite side. [math]\sin 36^\circ=\frac{\mbox{opp}}{\mbox{hyp}}[/math]The opposite side has length [math]50\sin 36^\circ.[/math]The exact value of [math]\sin 36^\circ[/math] is [math]\frac{\sqrt{10-2\sqrt 5}}4.[/math] See Exact Value of sin 36° for an explanation why.
How do I solve this trigonometric function?
The question is ambiguous.But let’s assume you want to know where this function’s zeroes are.Always remember that sine and cosine are closely related functions, and this applies to the multiple angles.In this case, we need two identities, valid for all values of x:[math]\cos{2x} = \cos^2{x}-\sin^2{x}[/math][math]\sin^2{x} + \cos^2{x} = 1[/math]The second identity allows us to express the first identity as[math]\cos{2x} =1 - 2\sin^2{x}[/math]It also allows this version:[math]\cos{2x}=2\cos^2{x}-1[/math]We don’t need this for the present problem but it’s very useful to keep all these facts in mind.For your question, set [math]S=\sin{x}[/math]In view of the identities above, we have the following equation in S[math]2S+(1–2S^2)=0[/math]There are 2 solutions of this quadratic equation, but only one meets the condition of[math]\left | S \right | \le 1[/math]The solutions for [math]x[/math] are the inverse sines of suitable value(s) of [math]S[/math]
Does anyone know how to solve Trig Equations? I need help on 2 problems!?
I got the first part. just get the cosx by itself. then you wind up with the cosx = -(the square root of 2/2) which is 225+360n and 315 +360n(in degrees) Sorry i can't help you on the other one. Good luck, just keep trying new things with it, would be my advice.