1. On the interval [0, 2pi] what are the solutions to the equation sin3xcos2x =
-cos3xsin2x + 1?
pi/10 and pi/2?
2. What is the value of tan75degrees?
√(3) + 1)/(1 - √(3))?
3. Value of cos(130degrees)cos(130degrees) + sin(10degrees)sin(10degrees)?
Not sure
4. On the interval [0, 2pi] what are the solutions to sin2xcos3x = cos2xsin3x - 1/2
pi/6 and 5pi/6?
5. Value of sin(4pi/9)cos(5pi/18) - cos(4pi/9)sin(5pi/18)?
I'm not sure about my answers, is it root 3/2?
#1. Since
sin3xcos2x + cos3xsin2x = sin5x you have
sin5x = 1
So, 5x = π/2, 5π/2, ...
and x = π/10, 5π/10, 9π/10, 13π/10, 17π/10
You have to keep adding 2π/5 until x gets to 2π
#2. correct
#3. Since cos(a-b) = cosa cosb - sina sinb you have
cos120° = -cos60° = -1/2
#4. Work like #1
sin5x = 1/2
#5. sin (4/9 - 5/18)pi = sin(pi/6) = 1/2
#4.
1. To solve the equation sin3xcos2x = -cos3xsin2x + 1 on the interval [0, 2pi], we can simplify the equation using trigonometric identities.
sin3xcos2x = -cos3xsin2x + 1
Using the product-to-sum identities, we have:
[2sin(x)cos(x)][cos^2(x) - sin^2(x)] = [-cos(x)(1 - cos^2(x))][2sin(x)cos(x)] + 1
Simplifying further, we get:
2sin(x)cos(x)cos^2(x) - 2sin(x)cos(x)sin^2(x) = -2cos(x)sin(x)cos^2(x) + cos(x) + 1
Rearranging the terms, we have:
2sin(x)cos(x)(cos^2(x) + sin^2(x)) = cos(x) + 1
Using the identity cos^2(x) + sin^2(x) = 1, we get:
2sin(x)cos(x) = cos(x) + 1
Next, we can factor out cos(x) from both sides:
cos(x)(2sin(x) - 1) = 1
Now we have two possibilities:
1) cos(x) = 0, which gives x = pi/2 and x = 3pi/2.
2) 2sin(x) - 1 = 1, which gives sin(x) = 1, resulting in x = pi/2.
Thus, the solutions to the equation on the interval [0, 2pi] are x = pi/2 and x = 3pi/2.
2. To find the value of tan(75 degrees), we can use the trigonometric f
1. To solve the equation sin3x*cos2x = -cos3x*sin2x + 1 on the interval [0, 2pi], we can start by simplifying the equation.
Using the trigonometric identity sin(A-B) = sinA*cosB - cosA*sinB, we can rewrite the equation as sin(3x - 2x) = -sin(3x)*cos(2x) + cos(3x)*sin(2x) + 1.
Further simplifying, we have sin(x) = 1 + sin(3x)*cos(2x) - cos(3x)*sin(2x).
Now, we can set up an equation and solve for x.
sin(x) = 1 + sin(3x)*cos(2x) - cos(3x)*sin(2x).
Looking at the given options, pi/10 and pi/2, we can substitute them into the equation to see if they satisfy the equation. If they do, then those values are indeed the solutions.
For option pi/10:
sin(pi/10) = 1 + sin(3*pi/10)*cos(2*pi/10) - cos(3*pi/10)*sin(2*pi/10).
Calculating the left-hand side, we have sin(pi/10) ≈ 0.309.
Calculating the right-hand side, we have 1 + sin(3*pi/10)*cos(2*pi/10) - cos(3*pi/10)*sin(2*pi/10) ≈ 0.309.
Since the left-hand side and right-hand side are approximately equal, pi/10 is a solution to the equation.
For option pi/2:
sin(pi/2) = 1 + sin(3*pi/2)*cos(2*pi/2) - cos(3*pi/2)*sin(2*pi/2).
Calculating the left-hand side, we have sin(pi/2) = 1.
Calculating the right-hand side, we have 1 + sin(3*pi/2)*cos(2*pi/2) - cos(3*pi/2)*sin(2*pi/2) = 0.
Since the left-hand side and right-hand side are not equal, pi/2 is not a solution to the equation.
Therefore, the solution to the equation sin3x*cos2x = -cos3x*sin2x + 1 on the interval [0, 2pi] is pi/10.