# trigonometry1 Homework Help

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• 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,...

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• trigonometry1
We'll recall the half-angle identity for cosine function: cos (x/2) = sqrt[(1 + cos x)/2] [cos (x/2)]^2 = [(1 + cos x)/2] Let (a+b) = x such as: [cos(a+b)]^2 = [(1 + cos 2(a+b))/2] (1) [cos(a-b)]^2...

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• 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...

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• 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...

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• trigonometry1
We'll apply the following identities: cos (a-b) = cosa*cos b + sin b*sin a cos (a+b) = cosa*cos b - sin b*sin a We'll put a = 30 and b = x. cos (30+x) = cos 30*cos x - sin x*sin 30 cos (30+x) =...

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• trigonometry1
We notice that BC is the opposite side to angle A and AC is the opposite side to the angle B. We'll use the the law of tangents. We'll note the length of the side BC = a and the length of the side...

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• trigonometry1
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• trigonometry1
Adding the 1st equation to the 2nd one, we'll get: sin x*cos y + sin y*cos x = 0.25+0.75 sin x*cos y + sin y*cos x = 1 We'll recognize the formula: sin (x+y) = 1 x + y = pi/2 + k*pi (3) Subtracting...

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• trigonometry1
We'll move the product sin x*cos x to the left side and we'll shift (cos x)^2 to the right side: sin x*cos x = 1 - (cos x)^2 The pythagorean identity yields: 1 - (cos x)^2 = (sin x)^2 The equation...

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• 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...

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• 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 +...

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• 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 +...

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• 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...

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• trigonometry1
sin (arcsin2x) = sin(-1/2) We know that sin (arcsin2x) = 2x We'll re-write the identity: sin(-1/2) = 2x. We know that the sine function is odd, so: sin(-1/2) = - sin (1/2) 2x = - sin (1/2) We'll...

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• 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 =...

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• 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...

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• trigonometry1
2*(2 sinx cosx) = sqrt3 We recognize the double angle identity: 2*(sin 2x) = sqrt3 We'll divide by 2 to isolate sin 2x to the left: sin 2x = sqrt3/2 2x = (-1)^k*arcsin sqrt3/2 + k*pi 2x =...

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• 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...

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• trigonometry1
Since the functions are matching, we'll use the identity: cos a - cos b = 2 sin [(a+b)/2]*sin [(b-a)/2] a = 75 and b = 15 cos 75 - cos 15 = 2 sin [(75+15)/2]*sin[(15-75)/2] cos 75 - cos 15 = 2 sin...

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• trigonometry1
We'll write the angle 7pi/12 as being the summation of the angles pi/3 and pi/4. 7pi/12 = pi/3 + pi/4 We'll apply cosine function both sides: cos (7pi/12) = cos(pi/3 + pi/4) cos(pi/3 + pi/4) = cos...

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• trigonometry1
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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...

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• 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...

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• trigonometry1
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• trigonometry1
Absolutely not! We'll write cot 3x = cot (2x + x) cot 3x = (cot 2x*cot x - 1)/(cot 2x + cot x) cot 2x = [(cot x)^2 - 1]/2cot x cot 3x = [(cot x)^2 - 3]/[3(cot x)^2 - 1] we'll re-write the...

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• 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 +...

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• trigonometry1
Since the trigonometric functions from the given sums are matching, we'll transform the given sums into products. sin a + sin b = 2sin[(a+b)/2]*cos[(a-b)/2] (1) cos a + cos b =...

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• 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

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• 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...

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• trigonometry1
We have to write the cosine of an angle in terms of sine of the angle. Let the angle to be worked with be x. cos x = cos (2*x/2) => cos( x/2 + x/2) => (cos x/2)^2 - (sin x/2)^2 => 1 -...

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• trigonometry1
We'll apply the Pythagorean identity, in a right angle triangle: Hypotenuse^2 = ( opposite side)^2 + (adjacent side)^2 The values of opposite side and adjacent side are given from enunciation, so...

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• 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 -...

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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...

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• trigonometry1
First, we'll move the term 3cosx to the right side: 2 + 2sqrt((1-cos4x)/2) = 3sinx + 3cosx We'll apply the half angle identity for the term sqrt((1-cos4x)/2)= sin 2x. 2 + 2sin 2x = 3(sin x + cos...

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• 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 -...

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• trigonometry1
We'll write the formula of the tangent of difference of 2 angles. tan (a-b) = (tan a - tan b)/(1 + tan a*tan b) Now, we'll have to establish the signature of tan a and tan b. We know from...

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• trigonometry1
Since the given form of the complex number is rectangular form, we'll transform it into the polar form. z = a + bi z = 4-3i Re(z) = 4 and Im(z) = -3 The polar form: z = |z|(cos t + i sin t) |z| =...

• 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...

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• trigonometry1
First, we'll have to move the number to the right side. cos x = 1/2 Since 1/2 < 1, the equation has solutions. x = +arccos (1/2) + 2kpi x = +pi/3 + 2kpi x = -pi/3 + 2kpi The solutions of the...

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• 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 ==>...

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• trigonometry1
First, we'll substitute the function cot 45 by it's value 1. We'll transform the sum into a product. For this purpose, we'll have to express the value 1 as being the function cosine of an angle,...

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• 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...

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• 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 =...

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• 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) =...

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• trigonometry1
We have to solve 2*(cos x)^2 + 3*sin x = 3 for 0< x< pi. 2*(cos x)^2 + 3*sin x = 3 use (cos x)^2 = 1 - (sin x)^2 => 2*( 1 - (sin x)^2) + 3*sin x = 3 let sin x = y => 2(1 - y^2) + 3y =...

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• 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...

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• 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...

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• 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 -...

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• trigonometry1
We need to determine cosec (75 + x) – sec (15 – x) – tan (55 + x) + cot (35 – x) We use the relation cosec x = sec (90 – x) and cot x = tan (90 – x) cosec (75 + x) – sec (15 – x)...

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• 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...

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