Given theta = -(5)/(13) where 180^ < theta < 360^, find the values of 2 theta and (theta - 135^).
|Mathematics
Given theta = -(5)/(13) where 180^ < theta < 360^, find the values of 2 theta and (theta - 135^).
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Answer
169144
Hey ~👻Masana, good to see you again.
Here are the solutions to the problems.
6.1 Given cosθ=−135 where 180∘<θ<360∘.
Since cosθ is negative and 180∘<θ<360∘, θ must be in Quadrant III. In Quadrant III, sinθ is negative.
6.1.1 Determine the value of sin2θ.
Step 1: Use the Pythagorean identity sin2θ+cos2θ=1.
sin2θ=1−cos2θ
Step 2: Substitute the given value of cosθ.
sin2θ=1−(−135)2sin2θ=1−16925
Step 3: Simplify the expression.
sin2θ=169169−16925sin2θ=169144
The value of sin2θ is 169144.
6.1.2 Determine the value of cos(θ−135∘).
Step 1: Find the value of sinθ. Since sin2θ=169144 and θ is in Quadrant III, sinθ must be negative.
sinθ=−169144=−1312
Step 2: Use the compound angle formula cos(A−B)=cosAcosB+sinAsinB.
Here, A=θ and B=135∘.
We need the values for cos135∘ and sin135∘.
cos135∘=cos(180∘−45∘)=−cos45∘=−22sin135∘=sin(180∘−45∘)=sin45∘=22
Step 3: Substitute the values into the formula.
cos(θ−135∘)=(−135)(−22)+(−1312)(22)
Step 4: Simplify the expression.
cos(θ−135∘)=2652−26122cos(θ−135∘)=2652−122cos(θ−135∘)=26−72
The value of cos(θ−135∘) is −2672.
6.2 Prove that 2cos2(45∘+x)=1−sin2x.
Step 1: Start with the left-hand side (LHS) and use the double angle identity 2cos2A=1+cos2A, where A=45∘+x.
LHS=2cos2(45∘+x)LHS=1+cos(2(45∘+x))
Step 2: Simplify the argument of the cosine function.
LHS=1+cos(90∘+2x)
Step 3: Use the co-function identity cos(90∘+α)=−sinα, where α=2x.
LHS=1−sin2x
Step 4: The LHS is equal to the right-hand side (RHS).
LHS=RHS
Therefore, 2cos2(45∘+x)=1−sin2x is proven.
6.3 Consider the expression: sin(A−B)−sin(A+B).
6.3.1 Prove that sin(A−B)−sin(A+B)=−2cosAsinB.
Step 1: Start with the left-hand side (LHS) and expand using the compound angle formulas.
LHS=sin(A−B)−sin(A+B)LHS=(sinAcosB−cosAsinB)−(sinAcosB+cosAsinB)
Step 2: Distribute the negative sign and simplify.
LHS=sinAcosB−cosAsinB−sinAcosB−cosAsinBLHS=(sinAcosB−sinAcosB)−(cosAsinB+cosAsinB)LHS=0−2cosAsinBLHS=−2cosAsinB
Step 3: The LHS is equal to the right-hand side (RHS).
LHS=RHS
Therefore, sin(A−B)−sin(A+B)=−2cosAsinB is proven.
6.3.2 Simplify the following expression to a single term: sin4x−sin10x.
Step 1: Use the sum-to-product formula sinP−sinQ=2cos(2P+Q)sin(2P−Q).
Here, P=4x and Q=10x.
Step 2: Substitute P and Q into the formula.
sin4x−sin10x=2cos(24x+10x)sin(24x−10x)
Step 3: Simplify the arguments.
sin4x−sin10x=2cos(214x)sin(2−6x)sin4x−sin10x=2cos(7x)sin(−3x)
Step 4: Use the identity sin(−α)=−sinα.
sin4x−sin10x=2cos(7x)(−sin3x)sin4x−sin10x=−2cos(7x)sin(3x)
The simplified expression is −2cos(7x)sin(3x).
6.3.3 Hence, determine the solution for sin4x−sin10x=sin3x for x∈[0∘;30∘].
Step 1: Substitute the simplified expression from 6.3.2 into the equation.
−2cos(7x)sin(3x)=sin(3x)
Step 2: Rearrange the equation to set it to zero and factor out the common term sin(3x).
−2cos(7x)sin(3x)−sin(3x)=0sin(3x)(−2cos(7x)−1)=0
Step 3: Set each factor equal to zero and solve for x.
Case 1: sin(3x)=0
The general solution for sinα=0 is α=n⋅180∘, where n is an integer.
3x=n⋅180∘x=n⋅60∘
For x∈[0∘;30∘]:
If n=0, x=0∘.
If n=1, x=60∘ (outside the interval).
So, x=0∘ is a solution.
Case 2: −2cos(7x)−1=0−2cos(7x)=1cos(7x)=−21
The reference angle for cosα=21 is 60∘. Since cos(7x) is negative, 7x is in Quadrant II or III.
General solutions for cosα=−21:
α=180∘−60∘+n⋅360∘=120∘+n⋅360∘α=180∘+60∘+n⋅360∘=240∘+n⋅360∘
So, for 7x:
7x=120∘+n⋅360∘or7x=240∘+n⋅360∘
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Hey ~👻Masana, good to see you again. Here are the solutions to the problems. 6.1 Given = -(5)/(13) where 180^ < < 360^. Since is negative and 180^ < < 360^, must be in Quadrant III. In Quadrant III, is negative. 6.1.1 Determine the value of ^2 . Step 1: Use the Pythagorean identity ^2 + ^2 = 1. ^2 = 1 - ^2 Step 2: Substitute the given value of . ^2 = 1 - (-(5)/(13))^2 ^2 = 1 - (25)/(169) Step 3: Simplify the expression. ^2 = (169)/(169) - (25)/(169) ^2 = (144)/(169) The value of ^2 is (144)/(169). 6.1.2 Determine the value of ( - 135^). Step 1: Find the value of . Since ^2 = (144)/(169) and is in Quadrant III, must be negative. = -sqrt((144)/(169)) = -(12)/(13) Step 2: Use the compound angle formula (A - B) = A B + A B. Here, A = and B = 135^. We need the values for 135^ and 135^. 135^ = (180^ - 45^) = - 45^ = -sqrt(2)2 135^ = (180^ - 45^) = 45^ = sqrt(2)2 Step 3: Substitute the values into the formula. ( - 135^) = (-(5)/(13))(-sqrt(2)2) + (-(12)/(13))(sqrt(2)2) Step 4: Simplify the expression. ( - 135^) = 5sqrt(2)26 - 12sqrt(2)26 ( - 135^) = 5sqrt(2) - 12sqrt(2)26 ( - 135^) = -7sqrt(2)26 The value of ( - 135^) is -7sqrt(2)26. 6.2 Prove that 2^2(45^ + x) = 1 - 2x. Step 1: Start with the left-hand side (LHS) and use the double angle identity 2^2 A = 1 + 2A, where A = 45^ + x. LHS = 2^2(45^ + x) LHS = 1 + (2(45^ + x)) Step 2: Simplify the argument of the cosine function. LHS = 1 + (90^ + 2x) Step 3: Use the co-function identity (90^ + ) = - , where = 2x. LHS = 1 - 2x Step 4: The LHS is equal to the right-hand side (RHS). LHS = RHS Therefore, 2^2(45^ + x) = 1 - 2x is proven. 6.3 Consider the expression: (A - B) - (A + B). 6.3.1 Prove that (A - B) - (A + B) = -2 A B. Step 1: Start with the left-hand side (LHS) and expand using the compound angle formulas. LHS = (A - B) - (A + B) LHS = ( A B - A B) - ( A B + A B) Step 2: Distribute the negative sign and simplify. LHS = A B - A B - A B - A B LHS = ( A B - A B) - ( A B + A B) LHS = 0 - 2 A B LHS = -2 A B Step 3: The LHS is equal to the right-hand side (RHS). LHS = RHS Therefore, (A - B) - (A + B) = -2 A B is proven. 6.3.2 Simplify the following expression to a single term: 4x - 10x. Step 1: Use the sum-to-product formula P - Q = 2((P+Q)/(2))((P-Q)/(2)). Here, P = 4x and Q = 10x. Step 2: Substitute P and Q into the formula. 4x - 10x = 2((4x+10x)/(2))((4x-10x)/(2)) Step 3: Simplify the arguments. 4x - 10x = 2((14x)/(2))((-6x)/(2)) 4x - 10x = 2(7x)(-3x) Step 4: Use the identity (-) = - . 4x - 10x = 2(7x)(- 3x) 4x - 10x = -2(7x)(3x) The simplified expression is -2(7x)(3x). 6.3.3 Hence, determine the solution for 4x - 10x = 3x for x [0^; 30^]. Step 1: Substitute the simplified expression from 6.3.2 into the equation. -2(7x)(3x) = (3x) Step 2: Rearrange the equation to set it to zero and factor out the common term (3x). -2(7x)(3x) - (3x) = 0 (3x)(-2(7x) - 1) = 0 Step 3: Set each factor equal to zero and solve for x. Case 1: (3x) = 0 The general solution for = 0 is = n · 180^, where n is an integer. 3x = n · 180^ x = n · 60^ For x [0^; 30^]: If n=0, x = 0^. If n=1, x = 60^ (outside the interval). So, x = 0^ is a solution. Case 2: -2(7x) - 1 = 0 -2(7x) = 1 (7x) = -(1)/(2) The reference angle for = (1)/(2) is 60^. Since (7x) is negative, 7x is in Quadrant II or III. General solutions for = -(1)/(2): = 180^ - 60^ + n · 360^ = 120^ + n · 360^ = 180^ + 60^ + n · 360^ = 240^ + n · 360^ So, for 7x: 7x = 120^ + n · 360^ or 7x = 240^ + n · 360^ x = (120^)/(7) + n · 360^7
