trigonometry1 Questions and Answers

trigonometry1

To solve the equation 5sinx = 4cosx, first, bring all terms to one side of the equations, so that the other side becomes 0: 5sinx - 4cosx = 0. Then, factor out 4cosx from the right hand side:...

Latest answer posted January 20, 2019, 5:17 am (UTC)

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trigonometry1

We have to prove that (1+ sin x + cos x) / (1+ sin x - cos x) = cot x/2 Now sin 2x = 2 sin x * cos x and cos 2x = 2 (cos x)^2 - 1 (1+ sin x + cos x) / (1+ sin x - cos x) => [1+2*(sin x/2)*(cos...

Latest answer posted January 21, 2011, 1:20 am (UTC)

1 educator answer

trigonometry1

We have to prove : cos x + cos 2x + cos 3x = cos 2x(1 + 2cos x) We know that cos A + cos B = 2 * cos (A + B)/2 * cos (A − B)/2 Now cos x + cos 2x + cos 3x => 2 * cos (4x / 2) * cos (2x / 2) +...

Latest answer posted January 21, 2011, 1:08 am (UTC)

1 educator answer

trigonometry1

I think by expansion you mean sin 3x in terms of sin x. We start with the relations sin (x + y) = sin x* cos y + cos x*sin y and cos (x + y) = cos x* cos y – sin x*sin y sin 3x = sin (2x + x) = sin...

Latest answer posted January 27, 2011, 3:45 pm (UTC)

1 educator answer

trigonometry1

We are given that sin a + sin b =1 and cos a + cos b = 1/2. We have to determine cos(a - b). cos(a - b) = (cos a)(cos b) + (sin a)(sin b) sin a + sin b =1 square both the sides => (sin a)^2 +...

Latest answer posted May 3, 2011, 4:39 pm (UTC)

1 educator answer

trigonometry1

We have to solve 2*sin 2x + sqrt 2 = 0 2*sin 2x + sqrt 2 = 0 => 2*sin 2x = -sqrt 2 => sin 2x = -1/sqrt 2 2x = arc sin (-1/sqrt 2) => 2x = -45 and -135 x = -45/2 + n*180 and x = -135/2 + n*180

Latest answer posted March 21, 2011, 10:47 am (UTC)

1 educator answer

trigonometry1

We are given 4 cos x + 2 sin x = 0, we have to find tan 2x 4 cos x + 2 sin x = 0 => 4 cos x = - 2sin x => sin x/cos x = 4/-2 => tan x = -2 tan 2x = 2*tan x/[1-(tan x)^2] => 2*(-2)/(1 -...

Latest answer posted March 8, 2011, 12:48 am (UTC)

1 educator answer

trigonometry1

We have cos x = -4/5. Also 180< x< 270. For these values of x sin x is negative and so is cos x. We use the relation (cos x)^2 + (sin x)^2 = 1 => (-4/5)^2 + (sin x)^2 = 1 => (sin x)^2 =...

Latest answer posted January 28, 2011, 3:48 pm (UTC)

1 educator answer

trigonometry1

To find the limit of sin x + cos x for x--> pi/3, we first substitute x with pi/3 to see if we get a valid solution. sin x + cos x for x--> pi/3 => sin (pi/3) + cos (pi/3) we see that both...

Latest answer posted January 17, 2011, 3:44 pm (UTC)

1 educator answer

trigonometry1

To show that sin 15 = (6^1/2 - 2^1/2)/4, let's start with the values of the functions of a number that is well known, sin 30 = 1/2 and cos 30 = (sqrt 3)/2 sin 15 can be written as sqrt [(1 - cos...

Latest answer posted May 6, 2011, 11:41 pm (UTC)

1 educator answer

trigonometry1

We have to prove that (sin x)^4 + (cos x)^4 + (sin 2x)^2/2 = 1 (sin x)^4 + (cos x)^4 + (sin 2x)^2/2 => (sin x)^4 + (cos x)^4 + (2*sin x * cos x)^2/2 => (sin x)^4 + (cos x)^4 + 4*(sin...

Latest answer posted June 2, 2011, 12:17 am (UTC)

1 educator answer

trigonometry1

We have to solve cos^2(x/2)-2cos^2x=(3/2)*square root2(1+cosx)+2sin^2x for x. Rewriting the equation in a simpler form: (cos x/2)^2 - 2 ( cos x)^2 = (3/2)sqrt 2 ( 1 + cos x) + 2 (sin x)^2 we know...

Latest answer posted January 21, 2011, 9:01 pm (UTC)

1 educator answer

trigonometry1

We have to prove that (3 sec x+3 cos x)(3 sec x - 3 cos x) = 9[(tan x)^2 + (sin x)^2] (3 sec x+3 cos x)(3 sec x - 3 cos x) => 9[ (1/cos x) + cos x][ (1/cos x) - cos x] => 9[ (1/cos x)^2 -...

Latest answer posted February 25, 2011, 5:37 pm (UTC)

1 educator answer

trigonometry1

tan x = -2/3 We need to find cosx such that x is in the second quadrant. We know that: sinx/cosx = tanx = -2/3 ==> sinx = (-2/3) cosx Also, sin^2 x + cos^2 x= 1 ==> (-2/3)cosx)^2 + cos^2 x =...

Latest answer posted February 4, 2011, 11:58 am (UTC)

1 educator answer

trigonometry1

We have to evaluate cos 13pi/6 and sin 37pi/4 Now cos (x+ 2*pi) = cos x cos 13pi/6 = cos ( pi/6 + 2*pi) = cos (pi/6) = sqrt 3 / 2 sin (x + pi) = -sin x sin ( 37*pi/4) = sin (36*pi/4 + pi/4) = sin...

Latest answer posted January 29, 2011, 2:33 am (UTC)

1 educator answer

trigonometry1

We have to prove the identity: (tan^4)t + (tan^2)t + 1 = [(1 - (sin^2)t * (cos^2) t] / (cos^4) t First we write it in a more regular way: (tan t)^4 + (tan t)^2 + 1 = [(1 - (sin t)^2 * (cos t)^2] /...

Latest answer posted January 24, 2011, 11:40 pm (UTC)

1 educator answer

trigonometry1

We have to solve the equation: (tan x)^2 - 8 tan x + 12 = 0 let tan x = y (tan x)^2 - 8 tan x + 12 = 0 => y^2 - 8y + 12 = 0 => y^2 - 6y - 2y + 12 = 0 => y(y -6) - 2( y - 6) = 0 => ( y -...

Latest answer posted January 26, 2011, 4:59 pm (UTC)

1 educator answer

trigonometry1

We have to calculate sin x + sqrt ( 1 - (sin x)^2) sin x + sqrt ( 1 - (sin x)^2) => sin x + sqrt ((cos x)^2) => sin x + cos x This sum can be changed to many other forms but the basic...

Latest answer posted January 26, 2011, 4:43 pm (UTC)

1 educator answer

trigonometry1

We have to prove that cos 4x - sin 4x * cot 2x = -1 cos 4x - sin 4x * cot 2x = -1 use cos 2x = (cos x)^2 - (sin x)^2 and sin 2x = 2 sin x cos x and cot x = (cos x)/(sin x) => (cos 2x)^2 - (sin...

Latest answer posted January 21, 2011, 1:26 am (UTC)

1 educator answer

trigonometry1

We have to write tan ( pi/4) - cos x as a product We know that tan x = sin x / cos x and tan (pi/4) = 1 tan ( pi/4) - cos x => 1 - cos x we can write 1 as cos 0 => cos 0 - cos x => −2 sin...

Latest answer posted January 26, 2011, 4:49 pm (UTC)

1 educator answer

trigonometry1

We have to determine the value of sin^-1(sin(5pi/3)) + tan^-1(tan(2pi/3)) sin^-1(sin(5pi/3)) or arc sin (sin (5pi/3)) = 5*pi/3 tan^-1(tan(2pi/3)) or arc tan (tan (2pi/3)) = 2*pi/3 Adding 5*pi/3 and...

Latest answer posted June 1, 2011, 11:40 pm (UTC)

1 educator answer

trigonometry1

We have to solve 5(cos x)^2 - 6 tan x*cos x = (cos x)^2- 5*( sin x)^2 - 1 5(cos x)^2 - 6 tan x*cos x = (cos x)^2- 5*( sin x)^2 - 1 => 5(cos x)^2 + 5*( sin x)^2 - 6 tan x*cos x = (cos x)^2 - 1 we...

Latest answer posted January 21, 2011, 3:09 am (UTC)

1 educator answer

trigonometry1

We have to solve for y in the interval(0 , 2*pi) if : 11*(sin y)^2 = 13 - (sin y)^2 11*(sin y)^2 = 13 - (sin y)^2 => 12(sin y)^2 = 13 => (sin y)^2 = 13/12 This is not possible as sin y...

Latest answer posted May 11, 2011, 5:51 am (UTC)

1 educator answer

trigonometry1

Given that: sinx + cosx = 1 We need to find tan 2x We know that: tan2x = 2tanx / (1-tan^2 x) Then, we will determine tanx Let us divide the equation by cosx ==> sinx + cosx = 1 ==>...

Latest answer posted February 4, 2011, 11:51 am (UTC)

1 educator answer

trigonometry1

This is a weird question, it is not possible to determine "the value of sum of sqrt2/2+sinx=?" as sin x is a variable. I guess you want the value of x for which sqrt2/2+sinx= 0 (sqrt 2)/2 + sin x =...

Latest answer posted April 30, 2011, 4:32 pm (UTC)

1 educator answer

trigonometry1

We have the function f(x) = x*cos x + sin (-x) f(x) = x*cos x - sin x The first derivative of f(x) is f'(x) => f'(x) = cos x - x*(sin x) - cos x => f'(x) = -x*sin x When 0< x< 180 we...

Latest answer posted January 26, 2011, 1:25 am (UTC)

1 educator answer

trigonometry1

We have to solve (sin B)-3sin B+2=0 for B in the range [0, 2*pi] (sin B)-3sin B+2=0 => sin B - 3sin B + 2 = 0 => -2sin B + 2 = 0 => sin B = -2 / -2 => sin B = 1 => B = arc sin 1...

Latest answer posted February 18, 2011, 7:49 pm (UTC)

1 educator answer

trigonometry1

Given that cos(A) = -3/5 We need to find sin(A) suchthat A is in the second quadrant. First let us calculate the value of sin(A). We know that sin^2 A + cos^2 A = 1 ==> sin^2 A = 1- cos^2 A...

Latest answer posted February 4, 2011, 11:55 am (UTC)

1 educator answer

trigonometry1

To simplify the expression, it is easier to express everything as sines and cosines. Note the following identities: cot(x) = cos(x)/sin(x) and tan(x) = sin(x)/cos(x) cos^2(x) + sin^2(x) = 1 Hence,...

Latest answer posted July 2, 2013, 9:22 pm (UTC)

1 educator answer

trigonometry1

We have to find the values of x for which sin x = -1/2 in the range [0, 2*pi) sin x = -1/2 => x = arc sin (-1/2) x = 210 degrees x = 330 degrees The values of x in the range [0, 2*pi) for which...

Latest answer posted April 30, 2011, 12:22 am (UTC)

1 educator answer

trigonometry1

We have to solve for x given sin2x-(cos 2x)/2 -1/2=cosx-2cosx*tanx sin 2x - (cos 2x)/2 -1/2 = cos x - 2cos x * tan x => 2 sin x cos x - (1 - 2 (sin x)^2)/2 - 1/2 = cos x - 2cos x * sin x/ cos...

Latest answer posted January 22, 2011, 12:31 am (UTC)

1 educator answer

trigonometry1

We have to find tan(x + x), given that sin x + cos x = 1. We know that (sin x)^2 + (cos x)^2 = 1 sin x + cos x = 1 => (sin x + cos x)^2 = 1 => (sin x)^2 + (cos x)^2 + 2 sin x cos x = 1 =>...

Latest answer posted January 22, 2011, 1:22 am (UTC)

1 educator answer

trigonometry1

The exact value of sin(60+210) is required. sin (60 + 210) = sin 270 => sin (360 - 90) => sin (-90) For the sine function sin (-x) = -sin x => -sin 90 => -1 sin(60 + 210) is exactly...

Latest answer posted April 30, 2011, 8:56 pm (UTC)

1 educator answer

trigonometry1

We know that sin 3x = 3 sin x - 4 (sin x)^3 Now sin B is given as 1/4. sin 3B = 3 sin B - 4(sin B)^3 => 3* (1/4) - 4(1/4)^3 => 3/4 - 1/16 => (12 - 1)/16 => 11/16 Therefore sin 3B = 11/16

Latest answer posted January 21, 2011, 12:11 am (UTC)

1 educator answer

trigonometry1

cos 165 = cos(180 - 15) = -cos 15 We use the value of cos 30 to find cos 15. The relation cos 2x = 2*(cos x)^2 - 1 can be used here cos 30 = 2*(cos 15)^2 - 1 = (sqrt 3)/2 (cos 15)^2 = [(sqrt 3)/2 +...

Latest answer posted April 8, 2011, 2:45 pm (UTC)

1 educator answer

trigonometry1

We have to solve (1 + cos 2x) / 2 = sin (-2x) (1 + cos 2x) / 2 = sin (-2x) => -(1 + cos 2x) / 2 = sin (2x) => -( 1 + (cos x)^2 - (sin x)^2)/ 2 = 2 sin x cos x => -1 - (cos x)^2 + (sin x)^2...

Latest answer posted January 22, 2011, 3:39 am (UTC)

1 educator answer

trigonometry1

We have to find the solutions of the equation: (2*cos x - sqrt 3) (11*sin x - 9 ) = 0. If 2*cos x - sqrt 3 =0 => cos x = sqrt 3/2 => x arc cos (sqrt 3 / 2) => x = pi/6 + 2*n*pi If 11*sin...

Latest answer posted March 10, 2011, 7:31 pm (UTC)

1 educator answer

trigonometry1

We have to prove that cot x*sin x = cos x /((cos x)^2 + (sin x)^2) Now we know that (cos x)^2 + (sin x)^2 = 1 Also, cot x = cos x / sin x So cot x*sin x = (cos x / sin x)* sin x = cos x cos x...

Latest answer posted January 21, 2011, 1:54 am (UTC)

1 educator answer

trigonometry1

We have to determine (sin x)^2 given that sin x = 3*(cos x) sin x = 3*(cos x) => (sin x)^2 = 9*(cos x)^2 => (cos x)^2 = (sin x)^2/9 substituting in (sin x)^2 + (cos x)^2 = 1 => (sin x)^2 +...

Latest answer posted May 6, 2011, 6:32 pm (UTC)

1 educator answer

trigonometry1

You mean determine t from the equation (sin t)^2 = 4/25 (sin t)^2 = 4/25 => sin t = 2/5 and sin t = -2/5 sin t = 2/5 => t = arc sin (2/5) => t = 23.57 degrees sin t = -2/5 => t = arc...

Latest answer posted April 13, 2011, 9:41 pm (UTC)

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trigonometry1

We have to solve (tan x)^2 = 8 - 8*sec x. (tan x)^2 = 8 - 8*sec x => (sin x)^2 / (cos x)^2 = 8 - 8/(cos x) => (1 - (cos x)^2) / (cos x)^2 = (8*(cos x)^2 - 8* cos x)/ (cos x)^2 => (1 - (cos...

Latest answer posted March 8, 2011, 12:26 am (UTC)

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trigonometry1

We have cot x= 1/5. We need the value of (cos x)^2 (sin x)^2 + (cos x)^2 = 1 divide all terms by (sin x)^2 => 1 + (cot x)^2 = 1/(sin x)^2 substitute cot x = 1/5 => 1 + (1/5)^2 = 1/(sin x)^2...

Latest answer posted April 13, 2011, 9:48 pm (UTC)

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