How do I solve these trigonometry questions?
inverse Trigonometric functionsIn trigonometry, if an angle is given then using any of the 6 trigonometric functions we find its trigonometric function value. But if the trigonometric function value is given, then finding the angle (working backwards) using a function is called Inverse trigonometric function.Example:Since “radian” is the standard unit for angle measure, we know:The above example shows that inverse trigonometric functions are called as: arcsine(x), arccos(x), arctan(x), arccsc(x), arcsec(x), arccot(x). Here ‘x’ is the value of the function and inverse is used to find the angle measure!Example:Examples:1) What is arcsin(0)? -> ‘0’ since sin(0) = 02) arctan(θ) = p if and only if tan(p) = θ3) arccsc(u) = m -> csc(m) = uExample: Evaluate the smallest angle measure for arctan(-1/√3)Need more help with Math homework help ,connect with expert tutors at now!!Source:http://blog.tutorsonnet.com/2017...
Can someone please help me solve these Trigonometry questions?
1) I think this first one should read: 2cos²(x/2) = sin²(x) / (1 - cos(x)) To prove this, let's work with the right hand side, and turn everything into sines and cosines, with argument x/2. That is, sin(x) = 2sin(x/2)cos(x/2), cos(x) = 1-2sin²(x/2) So, the right hand side becomes (2sin(x/2)cos(x/2))² / (2sin²(x/2) ) =4sin²(x/2)cos²(x/2) / 2sin²(x/2) =2cos²(x/2) 2) a) We will need to know tan(π/12). Given that sin(π/6) = 1/2, cos(π/6)=sqrt(3)/2, and the half angle formula tan(t/2) = sin(t)/(1+cos(t)) we get that tan(π/12) = 1/( 2+sqrt(3) ), and so 8tan(π /12) / 3 - 3tan²(π /12) =8/(6+3sqrt(3)) - 3/( 2+sqrt(3) )² b) cos(13π/12) = cos(π + π/12) = cos(π)cos(π/12) - sin(π)sin(π/12) = -cos(π/12) - 0 Now, the half-angle formula for cos(x) is cos(x/2) = sqrt( (1+cos(x))/2 ) so cos(13π/12) = -sqrt( (1+cos(π/6))/2 ) =-sqrt ( (1+sqrt(3)/2)/2 ) =-1/2*sqrt( 2 + sqrt(3) )
Can you solve these questions for me? (trigonometry)?
Number 6 quickly took my eye, but it turned out to be easy. It's obvious that x is not zero. So divide the whole equation by x. You'll get one term as a fraction and you should see that it can take on only two values, as it has to be an integer, because all the other terms are integers. You should be able to compute those two ordered pairs. EDIT: Alright, I'm biting, piecemeal, but it's bedtime very soon. Number 1: Please fill in the blanks. Equilateral triangle with QR = 30. Therefore, PR = __ ? QA bisects PR, so PA = RA = __ ? Now use Pythagoras in right-angled triangle QAR : QA^2 = QR^2 - RA^2, so QA = __ ? QA is bisected by B, so AB = QB = __ ? Use Pythagoras again in right-angled triangle PBA : PB^2 = PA^2 + AB^2, so PB = __ ? (I got 15√7/2). EDIT: Number 7 : (1) 2xy = 204 (2) x^2 - y^2 = 253 (3) x^2 + y^2 = 325 Add (2) + (3) to get : 2x^2 = 578. Therefore, x = __ ? Substitute this x into (1) to get : y = 204/(2x) = __ ? EDIT: Number 5 : 3x - 5y + 4 = 0. Therefore, y = (3/5)x + 4/5 2x + ay - 11 = 0. Therefore, y = (-2/a)x + 11/a Remember the formula for perpendicular lines : m2 = -1/m1. Let slopes, m1 = 3/5 and m2 = -2/a. Therefore, 3/5 = -1/(-2/a). Thus, a = 6/5. EDIT: Number 10 : Each angle of an equilateral triangle = 60º. Bisecting them yields 6 angles, each = 30º. The 3 bisectors are equal in length. Each of the 3 central angles around P = 120º. Thus, in say, triangle APB, by the sine rule, AP / sin(30º) = X / sin(120º) But sin(120º) = sin(60º). Therefore, AP / sin(30º) = X / sin(60º) so, AP = X * sin(30º) / sin(60º) = X * (1/2) / (√(3)/2) = X√(3)/3.
Can you help to solve this trigonometry question?
i hope this helps you!!
Can somebody please solve these trig questions using the sum and difference formula?
a) Use the formula, Sin (A ± B) = SinACosB ± CosASinB In this problem we're provided with the right hand side of the formula. A = π/18, B = 5π/18 sin(π/18)cos(5π/18) + cos(π/18)sin(5π/18) = Sin(π/18 + 5π/18) = Sin(6π/18) = Sin(π/3)= √3/2 = 0.866 (3dp). b) If Tana = 5/12 then we regard this measure as being two sides of a right angled triangle with the hypotenuse², h² = 5² + 12² = 169 , thus h = 13 and therefore Sina = 5/13 and Cosa = 12/13 assuming that angle a was in the first quadrant. However, a lies somewhere in the second and third quadrants. For the tangent to be positive then a lies in the third quadrant where both Sine and Cosine ratios are negative. Thus, Sina = -5/13 and Cosa = -12/13. Similarly, as Sinb = -½, then we have a triangle with a hypotenuse length of 2 and sides of 1 and √3. Sines are negative in the third quadrant so Cosb is also negative and equals -√3/2 and Tanb is positive and equals 1/√3. Cos(a + b) = cosacosb - sinasinb = (-12/13)(-√3/2) - (-5/13)(-½) = 12√3/26 - 5/26 = 0.6071 (4dp). Sin(a + b) = sinacosb + cosasinb = (-5/13)(-√3/2) + (-12/13)(-½) = 5√3/26 + 12/26 = 0.7946 (4dp) Tan(a + b) = (tana + tanb) ÷ [1 - tanatanb] = [(5/12) + (1/√3)] ÷ [1 - (5/12)(1/√3)] = [(5√3 + 12) / 12√3] ÷ [(12√3 - 5) / 12√3] = (5√3 + 12) x 12√3 / 12√3 x (12√3 - 5) = (5√3 + 12) / (12√3 - 5) = 1.3089. Alternatively, Tan(a + b) = Sin(a + b / Cos(a + b) = 0.7946 / 0.6071 = 1.3089.
How do I solve this question related to trigonometry?
first recall that sin(30º) = 1/2 and 30º is splitting the rect 90º into 3 partsyou know that pi/2 was 90º so 30º would be pi/(2*3) = pi/6angles x between A and B are those whose sin > 1/2 in the first turnin the second turn A is 2 pi + pi/6 = 13 pi/6 and B is 2 pi+5 pi/6 =17pi/6the third turn does not interest us because is greater than 4 pisolution : (pi/6, 5pi/6) U ( 13pi/6, 17pi/6)
How do I do these trig questions?
cos3xsin4x = (1/2)*2cos3xsin4x = 1/2[sin7x + sinx] cos2x + 7sinx = 4 => 1 -- 2sin^2x + 7sinx = 4 => 2sin^2x - 7sinx + 3 = 0 => 2sinx(sinx -- 3) -- 1(sinx -- 3) = 0 => (sinx -- 3)(2sinx -- 1) = 0 giving sinx = 3 (absurd), sinx = 1/2 = sin(pi/6) x = n(pi) + (--1)^n(pi/6) 0 < x < 360 => x = 30deg, 150deg
How can I solve this Inverse Trigonometric Function question?
Ans. x^2 = sin(2*alpha)I am reminded of NCERT problem just like that . It was in the miscellaneous exercise .Well , the deal is to approach an Trigonometric problem , you should remember the formulas , which provide the base to think further .OK I have assumed x^2 = cos(2theta)This simplifies to (cos(2theta)-sin(2theta))/(cos(2theta))+sin(2theta) Dividing by cos(2theta)Then put 1 = tan(pi/4)Hope this helps .P.S. Excuse my handwriting .
Can someone please solve these trig identities: [(sin4x)/(1-cos4x)][(1-cos2x)/... = tanx?
When you see stuff like 2x and 4x, start thinking about the double angle formulas. For example::: cos 2x=1 - 2sin^2 x and sin 2x= 2 cosx sinx.