Consider sin(x-360)sin(90-x)tan(-x)/cos(90+x)
1.A.SIMPLIFY sin(x-360)sin(90-x)tan(-x)/cos(90+x) to a single trigonometric ratio
B.hence or otherwise without using a calculator,solve for X if 0<X<360.
sin(x-360)sin(90-x)tan(-x)/cos(90+x) =0,5
2.A.prove that 8/sin^2A - 4/1+cosA=4/1-cosA
B. For which value(s) of A in the interval 0<A<360 is the identity in question 2.A
undefined?
sin(x-360)sin(90-x)tan(-x)/cos(90+x)
= sin(x)cos(x)(-sin(x)/cos(x))/(-sin(x))
= sin(x)
x = 30
8/sin^2A - 4/1+cosA
8/(1-cos^2A) - 4/(1+cosA)
(8 - 4(1-cosA))/(1-cos^2A)
(8-4+4cosA)/(1-cos^2A)
4(1+cosA)/(1-cos^2A)
4/(1-cosA)
undefined when cosA = 1
So, on 0<A<360 it is defined everywhere.
ThanQ steve
1.A. To simplify the expression sin(x-360)sin(90-x)tan(-x)/cos(90+x), let's break it down step by step:
First, use the identity sin(-x) = -sin(x) to simplify tan(-x) to -tan(x):
sin(x-360)sin(90-x)(-tan(x))/cos(90+x)
Next, use the identity tan(x) = sin(x)/cos(x) to write it as:
sin(x-360)sin(90-x)(-sin(x)/cos(x))/cos(90+x)
Simplify further by multiplying all the terms together:
-sin(x-360)sin(90-x)sin(x)/cos(x)cos(90+x)
Use the identity sin(90 - x) = cos(x) to simplify:
-sin(x-360)cos(x)sin(x)/cos(x)cos(90+x)
Now, cancel out the common terms:
-sin(x-360)sin(x)/cos(90+x)
Finally, rewrite sin(x-360) as sin(x) using the periodicity property of the sine function:
-sin(x)sin(x)/cos(90+x)
Multiply the negative signs together:
sin(x)sin(x)/cos(90+x)
Now, we have simplified the given expression to a single trigonometric ratio, which is sin(x)sin(x)/cos(90+x).
1.B. To solve the equation sin(x)sin(x)/cos(90+x) = 0.5 without a calculator, you need to use the properties of trigonometric functions.
Start by multiplying both sides of the equation by cos(90+x):
sin(x)sin(x) = 0.5cos(90+x)
Use the identity cos(90+x) = -sin(x) to simplify the right side:
sin(x)sin(x) = -0.5sin(x)
Rearrange the equation:
sin(x)sin(x) + 0.5sin(x) = 0
Factor out sin(x):
sin(x)(sin(x) + 0.5) = 0
Now, we have two equations:
1) sin(x) = 0
2) sin(x) + 0.5 = 0
For equation 1), sin(x) = 0, the solutions are x = 0, 180, 360, and any integer multiple of 180.
For equation 2), sin(x) + 0.5 = 0, we subtract 0.5 from both sides:
sin(x) = -0.5
The solution to this equation is x = 210 and x = 330.
Therefore, the solutions for 0 < x < 360 are x = 0, 180, 210, 330, and any integer multiple of 180.
2.A. To prove the identity 8/sin^2(A) - 4/(1+cos(A)) = 4/(1-cos(A)), we'll work step by step:
Multiply the second term by (1-cos(A))/(1-cos(A)) to get a common denominator:
8/sin^2(A) - (4(1-cos(A))/(1+cos(A))(1-cos(A))) = 4/(1-cos(A))
Simplify the denominator:
8/sin^2(A) - (4(1-cos(A))/(1-cos^2(A))) = 4/(1-cos(A))
Simplify further:
8/sin^2(A) - (4(1-cos(A))/(sin^2(A))) = 4/(1-cos(A))
Combine the fractions:
8/sin^2(A) - (4 - 4cos(A))/(sin^2(A)) = 4/(1-cos(A))
Combine like terms:
(8 - 4 + 4cos(A))/(sin^2(A)) = 4/(1-cos(A))
Simplify:
(4 + 4cos(A))/(sin^2(A)) = 4/(1-cos(A))
Multiply both sides by (1-cos(A))/(1-cos(A)) to get a common denominator:
(4 + 4cos(A))(1-cos(A))/(sin^2(A)(1-cos(A))) = 4/(1-cos(A))
Simplify the denominator:
(4 + 4cos(A))(1-cos(A))/(sin^2(A) - sin^2(A)cos(A)) = 4/(1-cos(A))
Simplify further:
(4 + 4cos(A))(1-cos(A))/(sin^2(A) - sin^2(A)cos(A)) = 4/(1-cos(A))
Cancel out the common terms:
4 + 4cos(A) = 4
Subtract 4 from both sides:
4cos(A) = 0
Divide by 4:
cos(A) = 0
Therefore, the identity is verified for any angle A where cos(A) = 0, which occurs at A = 90, 270, and any integer multiple of 180.
2.B. The identity in question 2.A is undefined when (1 - cos(A)) = 0, as division by zero is undefined.
Solve (1 - cos(A)) = 0:
1 - cos(A) = 0
cos(A) = 1
The value(s) of A in the interval 0 < A < 360 where the identity is undefined is A = 0, 360, and any integer multiple of 360.
I will provide step-by-step explanations for both questions:
1. A. To simplify the given expression, we can use trigonometric identities and properties. Let's start by simplifying each term individually:
a. sin(x - 360):
Since sin(x - 360) represents the sine function of an angle (x - 360), and the sine function has a periodicity of 360 degrees, sin(x - 360) is equal to sin(x). Therefore, sin(x - 360) can be simplified to sin(x).
b. sin(90 - x):
Using the identity sin(90 - x) = cos(x), we can simplify sin(90 - x) to cos(x).
c. tan(-x):
We know that tan(-x) = -tan(x), so tan(-x) can be simplified to -tan(x).
d. cos(90 + x):
Using the identity cos(90 + x) = -sin(x), we can simplify cos(90 + x) to -sin(x).
Now, let's substitute the simplified terms back into the original expression:
sin(x) * cos(x) * (-tan(x)) / (-sin(x))
Next, cancel out the negatives:
sin(x) * cos(x) * tan(x) / sin(x)
Now, simplify by canceling out the sin(x) term:
cos(x) * tan(x)
Therefore, the given expression sin(x - 360)sin(90 - x)tan(-x)/cos(90 + x) can be simplified to the single trigonometric ratio cos(x) * tan(x).
1. B. We need to solve the equation cos(x) * tan(x) = 0.5 over the interval 0° < x < 360°.
To solve this, we can use the fact that tan(x) = sin(x) / cos(x). So, our equation becomes:
cos(x) * (sin(x) / cos(x)) = 0.5
sin(x) = 0.5
Now, we can find the angles whose sine is 0.5 by looking at the unit circle or using a calculator. The solutions for x lie in the first quadrant and the second quadrant. So, the possible values for x in the interval 0° < x < 360° are 30° and 150°.
2. A. To prove the given identity 8/sin^2A - 4/(1 + cosA) = 4/(1 - cosA), we will start with the left-hand side (LHS) and manipulate it until it is equivalent to the right-hand side (RHS).
Starting with the LHS:
8/sin^2A - 4/(1 + cosA)
Now, let's simplify the expression:
8/sin^2A - 4/(1 + cosA) = (8 - 4sin^2A) / sin^2A
Next, we can use the identity sin^2A = 1 - cos^2A:
(8 - 4sin^2A) / sin^2A = (8 - 4(1 - cos^2A)) / sin^2A
Simplifying further:
(8 - 4 + 4cos^2A) / sin^2A = (4cos^2A + 4) / sin^2A
Now, let's simplify the RHS:
4/(1 - cosA) = 4/((1 - cosA) * (1 + cosA))
Using the difference of squares identity (a^2 - b^2 = (a + b)(a - b)):
4/((1 - cosA) * (1 + cosA)) = 4/(1 - cos^2A)
Using the Pythagorean identity sin^2A + cos^2A = 1:
4/(1 - cos^2A) = 4/sin^2A
Hence, we have shown that the LHS is equal to the RHS, which proves the given identity.
2. B. Now, let's determine the value(s) of A in the interval 0° < A < 360° for which the identity in question 2.A is undefined.
The identity is defined for all values of A as long as sin^2A ≠ 0. The only case where sin^2A = 0 is when sinA = 0. This occurs when A = 0° and A = 180°. Outside these two angles, sin^2A ≠ 0, and the identity is defined.
So, the identity in question 2.A is undefined for A = 0° and A = 180°, within the interval 0° < A < 360°.