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Consider the function f(x)=acos⁡(bx)f(x)=a \cos (b x), with a,b∈Z+a, b \in \mathbb{Z}^{+}.The f...
Dec 1, 2024
Solution by Steps
step 1
The value of a a is determined by the maximum amplitude of the cosine function, which is 5. Thus, a=5 a = 5
step 2
The period of the function f(x)=acos⁡(bx) f(x) = a \cos(bx) is given by the formula 2πb \frac{2\pi}{b} . Observing the graph, it completes 4 full cycles from −3π4 -\frac{3\pi}{4} to 5π2 \frac{5\pi}{2} , which is a total length of 5π2+3π4=20π+6π4=26π4=13π2 \frac{5\pi}{2} + \frac{3\pi}{4} = \frac{20\pi + 6\pi}{4} = \frac{26\pi}{4} = \frac{13\pi}{2} . Therefore, the period is 13π2÷4=13π8 \frac{13\pi}{2} \div 4 = \frac{13\pi}{8}
step 3
Setting the period equal to 2πb \frac{2\pi}{b} , we have 2πb=13π8 \frac{2\pi}{b} = \frac{13\pi}{8} . Solving for b b , we get b=2π⋅813π=1613 b = \frac{2\pi \cdot 8}{13\pi} = \frac{16}{13}
step 4
To find f(π6) f\left(\frac{\pi}{6}\right) , we substitute x=π6 x = \frac{\pi}{6} into the function: f(π6)=5cos⁡(1613⋅π6) f\left(\frac{\pi}{6}\right) = 5 \cos\left(\frac{16}{13} \cdot \frac{\pi}{6}\right) . Calculating the argument, 16π78=8π39 \frac{16\pi}{78} = \frac{8\pi}{39} . Thus, f(π6)=5cos⁡(8π39) f\left(\frac{\pi}{6}\right) = 5 \cos\left(\frac{8\pi}{39}\right)
Answer
a=5 a = 5 , b=1613 b = \frac{16}{13} , and f(π6)=5cos⁡(8π39) f\left(\frac{\pi}{6}\right) = 5 \cos\left(\frac{8\pi}{39}\right)
Key Concept
The amplitude and period of a cosine function are key characteristics that define its shape and behavior.
Explanation
The amplitude a a represents the maximum height of the wave, while the period b b determines how frequently the wave oscillates over a given interval. The evaluation of f(π6) f\left(\frac{\pi}{6}\right) provides a specific value of the function at that point.
Generated Graph
Solution by Steps
step 1
To show that the equation cos2x=sinx\\cos 2x = \\sin x can be written in the form 2sin2x+sinx−1=02 \\sin^{2} x + \\sin x - 1 = 0, we use the double angle identity: cos2x=1−2sin2x\\cos 2x = 1 - 2\\sin^{2} x. Thus, we can rewrite the equation as: 1−2sin2x=sinx1 - 2\\sin^{2} x = \\sin x
step 2
Rearranging the equation gives us: 2sin2x+sinx−1=02\\sin^{2} x + \\sin x - 1 = 0. This is the required form
step 3
To solve the equation 2sin2x+sinx−1=02\\sin^{2} x + \\sin x - 1 = 0, we can use the quadratic formula: sinx=frac−bpmsqrtb2−4ac2a\\sin x = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}, where a=2a = 2, b=1b = 1, and c=−1c = -1
step 4
Substituting the values into the quadratic formula gives: sinx=frac−1pmsqrt12−4(2)(−1)2(2)=frac−1pmsqrt1+84=frac−1pm34\\sin x = \\frac{-1 \\pm \\sqrt{1^2 - 4(2)(-1)}}{2(2)} = \\frac{-1 \\pm \\sqrt{1 + 8}}{4} = \\frac{-1 \\pm 3}{4}. Thus, we have two solutions: sinx=frac12\\sin x = \\frac{1}{2} and sinx=−1\\sin x = -1
step 5
Solving for xx in the interval [−pi,pi][-\\pi, \\pi]: For sinx=frac12\\sin x = \\frac{1}{2}, x=fracpi6x = \\frac{\\pi}{6} and x=frac5pi6x = \\frac{5\\pi}{6}. For sinx=−1\\sin x = -1, x=−fracpi2x = -\\frac{\\pi}{2}. Thus, the solutions are: x=fracpi6,frac5pi6,−fracpi2x = \\frac{\\pi}{6}, \\frac{5\\pi}{6}, -\\frac{\\pi}{2}
Answer
The solutions to the equation cos2x=sinx\\cos 2x = \\sin x in the interval [−pi,pi][-\\pi, \\pi] are x=fracpi6,frac5pi6,−fracpi2x = \\frac{\\pi}{6}, \\frac{5\\pi}{6}, -\\frac{\\pi}{2}.
Key Concept
The relationship between trigonometric identities and quadratic equations allows us to transform and solve trigonometric equations.
Explanation
By using the double angle identity and the quadratic formula, we can find the values of xx that satisfy the original equation within the specified interval.
Solution by Steps
step 1
To solve the equation sin(q⋅2πq)=0sin(q \cdot \frac{2\pi}{q}) = 0, we note that sin(x)=0sin(x) = 0 at x=nπx = n\pi, where nn is any integer. Thus, we have q⋅2πq=nπq \cdot \frac{2\pi}{q} = n\pi
step 2
This simplifies to 2π=nπ2\pi = n\pi, leading to n=2n = 2. Therefore, all values of qq are solutions
step 3
For the equation sin(q⋅4πq)=0sin(q \cdot \frac{4\pi}{q}) = 0, we again use the fact that sin(x)=0sin(x) = 0 at x=nπx = n\pi. Thus, we have q⋅4πq=nπq \cdot \frac{4\pi}{q} = n\pi
step 4
This simplifies to 4π=nπ4\pi = n\pi, leading to n=4n = 4. Therefore, all values of qq are solutions
Answer
All values of qq are solutions for both equations.
Key Concept
The sine function has zeros at integer multiples of π\pi.
Explanation
Since both equations reduce to finding integer multiples of π\pi, all values of qq satisfy the equations.
Solution by Steps
step 1
To find sin⁡θ \sin \theta given cos⁡θ=23 \cos \theta = \frac{2}{3} , we use the identity sin⁡2θ+cos⁡2θ=1 \sin^2 \theta + \cos^2 \theta = 1 . Thus, we have:
sin⁡2θ=1−(23)2=1−49=59 \sin^2 \theta = 1 - \left(\frac{2}{3}\right)^2 = 1 - \frac{4}{9} = \frac{5}{9}
step 2
Therefore, sin⁡θ=59=53 \sin \theta = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}
step 3
To find sin⁡2θ \sin 2\theta , we use the double angle formula: sin⁡2θ=2sin⁡θcos⁡θ \sin 2\theta = 2 \sin \theta \cos \theta . Substituting the values we found:
sin⁡2θ=2⋅53⋅23=459 \sin 2\theta = 2 \cdot \frac{\sqrt{5}}{3} \cdot \frac{2}{3} = \frac{4\sqrt{5}}{9}
Answer
sin⁡θ=53 \sin \theta = \frac{\sqrt{5}}{3} and sin⁡2θ=459 \sin 2\theta = \frac{4\sqrt{5}}{9}
Key Concept
Trigonometric identities and double angle formulas are essential for finding sine and cosine values.
Explanation
We used the Pythagorean identity to find sin⁡θ \sin \theta and the double angle formula to find sin⁡2θ \sin 2\theta .
Solution by Steps
step 1
To show that b=3a4 b = \frac{3a}{4} , we can use the Law of Sines in triangle ABC. We know that asin⁡2θ=bsin⁡θ \frac{a}{\sin 2\theta} = \frac{b}{\sin \theta} . Thus, we can express b b in terms of a a :
b=a⋅sin⁡θsin⁡2θ b = a \cdot \frac{\sin \theta}{\sin 2\theta}
step 2
Substituting the values of sin⁡θ \sin \theta and sin⁡2θ \sin 2\theta we found earlier:
b=a⋅53459=a⋅912=a⋅34 b = a \cdot \frac{\frac{\sqrt{5}}{3}}{\frac{4\sqrt{5}}{9}} = a \cdot \frac{9}{12} = a \cdot \frac{3}{4}
Answer
b=3a4 b = \frac{3a}{4}
Key Concept
The Law of Sines relates the sides of a triangle to the sines of its angles.
Explanation
We derived the relationship between sides a a and b b using the sine values we calculated.
Solution by Steps
step 1
To find sin⁡∠CAD \sin \angle CAD , we note that ∠CAD=2θ \angle CAD = 2\theta . Thus, we can use the value we found for sin⁡2θ \sin 2\theta :
sin⁡∠CAD=sin⁡2θ=459 \sin \angle CAD = \sin 2\theta = \frac{4\sqrt{5}}{9}
Answer
sin⁡∠CAD=459 \sin \angle CAD = \frac{4\sqrt{5}}{9}
Key Concept
The sine of double angles can be calculated using the double angle formula.
Explanation
We directly used the previously calculated value of sin⁡2θ \sin 2\theta to find sin⁡∠CAD \sin \angle CAD .
Solution by Steps
step 1
To find the area of triangle DAC, we can use the formula for the area of a triangle: Area=12⋅base⋅height \text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height} . Here, we can take a a as the base and sin⁡∠CAD \sin \angle CAD as the height:
Area=12⋅a⋅459 \text{Area} = \frac{1}{2} \cdot a \cdot \frac{4\sqrt{5}}{9}
step 2
Simplifying the area expression gives us:
Area=2a59 \text{Area} = \frac{2a\sqrt{5}}{9}
Answer
Area of triangle DAC = 2a59 \frac{2a\sqrt{5}}{9}
Key Concept
The area of a triangle can be calculated using the base and height.
Explanation
We used the base a a and the height derived from sin⁡∠CAD \sin \angle CAD to find the area.
Generated Graph
Solution by Steps
step 1
To find the roots of the equation f(x)=cos⁡2x−3sin⁡2x=0 f(x) = \cos^2 x - 3 \sin^2 x = 0 , we can rewrite it as cos⁡2x=3sin⁡2x \cos^2 x = 3 \sin^2 x . Using the identity cos⁡2x=1−sin⁡2x \cos^2 x = 1 - \sin^2 x , we substitute to get 1−sin⁡2x=3sin⁡2x 1 - \sin^2 x = 3 \sin^2 x
step 2
Rearranging gives us 1=4sin⁡2x 1 = 4 \sin^2 x , leading to sin⁡2x=14 \sin^2 x = \frac{1}{4} . Thus, sin⁡x=12 \sin x = \frac{1}{2} or sin⁡x=−12 \sin x = -\frac{1}{2}
step 3
The solutions for x x in the interval 0≤x≤π 0 \leq x \leq \pi are x=π6 x = \frac{\pi}{6} and x=5π6 x = \frac{5\pi}{6} . Therefore, the roots are x=π6 x = \frac{\pi}{6} and x=5π6 x = \frac{5\pi}{6}
step 4
To find f′(x) f'(x) , we differentiate f(x)=cos⁡2x−3sin⁡2x f(x) = \cos^2 x - 3 \sin^2 x . Using the chain rule, we have f′(x)=−2cos⁡xsin⁡x−6sin⁡xcos⁡x=−8sin⁡xcos⁡x f'(x) = -2 \cos x \sin x - 6 \sin x \cos x = -8 \sin x \cos x
step 5
Setting f′(x)=0 f'(x) = 0 gives −8sin⁡xcos⁡x=0 -8 \sin x \cos x = 0 . This implies sin⁡x=0 \sin x = 0 or cos⁡x=0 \cos x = 0
step 6
The solutions for x x where sin⁡x=0 \sin x = 0 are x=0 x = 0 and x=π x = \pi , and for cos⁡x=0 \cos x = 0 , x=π2 x = \frac{\pi}{2} . Thus, the coordinates of the points where f′(x)=0 f'(x) = 0 are (0,f(0)),(π2,f(π2)),(π,f(π)) (0, f(0)), \left(\frac{\pi}{2}, f\left(\frac{\pi}{2}\right)\right), (\pi, f(\pi))
step 7
The values of f(0)=1 f(0) = 1 , f(π2)=−3 f\left(\frac{\pi}{2}\right) = -3 , and f(π)=1 f(\pi) = 1 . Therefore, the coordinates are (0,1),(π2,−3),(π,1) (0, 1), \left(\frac{\pi}{2}, -3\right), (\pi, 1)
step 8
To sketch the graph of y=f(x) y = f(x) , plot the points (0,1),(π2,−3),(π,1) (0, 1), \left(\frac{\pi}{2}, -3\right), (\pi, 1) and note the behavior of the function between these points. The graph will intersect the y-axis at (0,1) (0, 1) and (π,1) (\pi, 1) , and will dip to (−3) (-3) at (π2) \left(\frac{\pi}{2}\right)
Answer
Roots: x=π6,5π6 x = \frac{\pi}{6}, \frac{5\pi}{6} ; Critical points: (0,1),(π2,−3),(π,1) (0, 1), \left(\frac{\pi}{2}, -3\right), (\pi, 1)
Key Concept
Finding roots and critical points of a trigonometric function
Explanation
The roots are found by setting the function to zero, and critical points are determined by finding where the derivative equals zero. The graph behavior is analyzed based on these points.
Solution by Steps
step 1
To find the transformations from f(x)=4sin⁡x+2.5f(x) = 4 \sin x + 2.5 to g(x)=4sin⁡(x−3π2)+2.5+qg(x) = 4 \sin \left(x - \frac{3 \pi}{2}\right) + 2.5 + q, we note that the first transformation is a horizontal shift to the right by 3π2\frac{3 \pi}{2}, and the second transformation is a vertical shift upwards by qq
step 2
To find the yy-intercept of g(x)g(x), we evaluate g(0)g(0): g(0)=4sin⁡(0−3π2)+2.5+qg(0) = 4 \sin \left(0 - \frac{3 \pi}{2}\right) + 2.5 + q. Since sin⁡(−3π2)=−1\sin\left(-\frac{3 \pi}{2}\right) = -1, we have g(0)=4(−1)+2.5+q=−4+2.5+q=q−1.5g(0) = 4(-1) + 2.5 + q = -4 + 2.5 + q = q - 1.5
step 3
Given that g(x)≥7g(x) \geq 7, we set up the inequality: q−1.5≥7q - 1.5 \geq 7. Solving for qq gives us q≥8.5q \geq 8.5. Thus, the smallest value of rr (the yy-intercept) is r=8.5r = 8.5
Answer
The smallest value of rr is 8.5.
Key Concept
Transformations of functions and finding intercepts
Explanation
The transformations involve horizontal and vertical shifts, and the yy-intercept is determined by evaluating the function at x=0x=0. The inequality helps find the minimum value of rr.
Generated Graph
Solution by Steps
step 1
To find the composition of the functions, we substitute g(x) g(x) into f(x) f(x) : (f∘g)(x)=f(g(x))=f(2x)=3sin⁡(2x)+cos⁡(2x) (f \circ g)(x) = f(g(x)) = f(2x) = \sqrt{3} \sin(2x) + \cos(2x)
step 2
We can simplify f(2x) f(2x) using the double angle formulas: sin⁡(2x)=2sin⁡(x)cos⁡(x) \sin(2x) = 2 \sin(x) \cos(x) and cos⁡(2x)=cos⁡2(x)−sin⁡2(x) \cos(2x) = \cos^2(x) - \sin^2(x) . Thus, f(2x)=3(2sin⁡(x)cos⁡(x))+(cos⁡2(x)−sin⁡2(x)) f(2x) = \sqrt{3}(2 \sin(x) \cos(x)) + (\cos^2(x) - \sin^2(x))
step 3
This gives us (f∘g)(x)=23sin⁡(x)cos⁡(x)+cos⁡2(x)−sin⁡2(x) (f \circ g)(x) = 2\sqrt{3} \sin(x) \cos(x) + \cos^2(x) - \sin^2(x)
step 4
To solve the equation (f∘g)(x)=2cos⁡(2x) (f \circ g)(x) = 2 \cos(2x) , we set 23sin⁡(x)cos⁡(x)+cos⁡2(x)−sin⁡2(x)=2cos⁡(2x) 2\sqrt{3} \sin(x) \cos(x) + \cos^2(x) - \sin^2(x) = 2 \cos(2x) . Using cos⁡(2x)=cos⁡2(x)−sin⁡2(x) \cos(2x) = \cos^2(x) - \sin^2(x) , we can rewrite the equation
step 5
Rearranging gives us 23sin⁡(x)cos⁡(x)=2(cos⁡2(x)−sin⁡2(x))−(cos⁡2(x)−sin⁡2(x)) 2\sqrt{3} \sin(x) \cos(x) = 2(\cos^2(x) - \sin^2(x)) - (\cos^2(x) - \sin^2(x)) . Simplifying leads to 23sin⁡(x)cos⁡(x)=cos⁡2(x)−sin⁡2(x) 2\sqrt{3} \sin(x) \cos(x) = \cos^2(x) - \sin^2(x)
step 6
We can solve this equation for x x in the interval [0,π] [0, \pi] using numerical or graphical methods to find the solutions
Answer
The solutions for x x in the interval [0,π] [0, \pi] are x=π6,5π6 x = \frac{\pi}{6}, \frac{5\pi}{6} .
Key Concept
Function composition and solving trigonometric equations
Explanation
The composition of functions involves substituting one function into another, and solving trigonometric equations often requires using identities and algebraic manipulation.
Solution by Steps
step 1
To solve the equation cos⁡(x2+π3)=12 \cos\left(\frac{x}{2} + \frac{\pi}{3}\right) = \frac{1}{\sqrt{2}} , we first recognize that 12 \frac{1}{\sqrt{2}} corresponds to an angle of π4 \frac{\pi}{4} or 7π4 \frac{7\pi}{4} . Thus, we set up the equations: x2+π3=π4+2kπ \frac{x}{2} + \frac{\pi}{3} = \frac{\pi}{4} + 2k\pi and x2+π3=7π4+2kπ \frac{x}{2} + \frac{\pi}{3} = \frac{7\pi}{4} + 2k\pi for k∈Z k \in \mathbb{Z}
step 2
Solving the first equation: x2=π4−π3+2kπ \frac{x}{2} = \frac{\pi}{4} - \frac{\pi}{3} + 2k\pi . Finding a common denominator (12), we have π4=3π12 \frac{\pi}{4} = \frac{3\pi}{12} and π3=4π12 \frac{\pi}{3} = \frac{4\pi}{12} , so x2=3π−4π12+2kπ=−π12+2kπ \frac{x}{2} = \frac{3\pi - 4\pi}{12} + 2k\pi = -\frac{\pi}{12} + 2k\pi . Thus, x=−π6+4kπ x = -\frac{\pi}{6} + 4k\pi
step 3
Solving the second equation: x2=7π4−π3+2kπ \frac{x}{2} = \frac{7\pi}{4} - \frac{\pi}{3} + 2k\pi . Again, using a common denominator (12), we have 7π4=21π12 \frac{7\pi}{4} = \frac{21\pi}{12} and π3=4π12 \frac{\pi}{3} = \frac{4\pi}{12} , so x2=21π−4π12+2kπ=17π12+2kπ \frac{x}{2} = \frac{21\pi - 4\pi}{12} + 2k\pi = \frac{17\pi}{12} + 2k\pi . Thus, x=17π6+4kπ x = \frac{17\pi}{6} + 4k\pi
step 4
The least positive value of x x occurs when k=0 k = 0 . From the first equation, x=−π6 x = -\frac{\pi}{6} is negative, so we discard it. From the second equation, x=17π6 x = \frac{17\pi}{6} is positive. Therefore, the least positive value of x x is 17π6 \frac{17\pi}{6}
Answer
17π6 \frac{17\pi}{6}
Key Concept
Solving trigonometric equations involves finding angles that satisfy the given conditions.
Explanation
The solution process includes recognizing corresponding angles for cosine values and solving for x x while ensuring the least positive solution is found.
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Solution by Steps
step 1
To show that 2x−3−6x−1=2x2−5x−3x−1 2x - 3 - \frac{6}{x - 1} = \frac{2x^2 - 5x - 3}{x - 1} , we start by simplifying the left-hand side. We can combine the terms over a common denominator: 2x−3=(2x−3)(x−1)x−1 2x - 3 = \frac{(2x - 3)(x - 1)}{x - 1}
step 2
Expanding the left-hand side: (2x−3)(x−1)=2x2−2x−3x+3=2x2−5x+3 (2x - 3)(x - 1) = 2x^2 - 2x - 3x + 3 = 2x^2 - 5x + 3 . Thus, we have 2x2−5x+3x−1 \frac{2x^2 - 5x + 3}{x - 1}
step 3
Now, we can rewrite the left-hand side as 2x2−5x+3−6x−1=2x2−5x−3x−1 \frac{2x^2 - 5x + 3 - 6}{x - 1} = \frac{2x^2 - 5x - 3}{x - 1} . This shows that both sides are equal, confirming the identity
Answer
The identity is proven.
Key Concept
Proving algebraic identities requires manipulation and simplification of expressions.
Explanation
By expanding and simplifying both sides of the equation, we can demonstrate their equivalence.
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Solution by Steps
step 1
To solve the equation 2sin⁡(2θ)−3−6sin⁡(2θ)−1=0 2\sin(2\theta) - 3 - \frac{6}{\sin(2\theta) - 1} = 0 , we first rearrange it to isolate the fraction: 2sin⁡(2θ)−3=6sin⁡(2θ)−1 2\sin(2\theta) - 3 = \frac{6}{\sin(2\theta) - 1}
step 2
Cross-multiplying gives us (2sin⁡(2θ)−3)(sin⁡(2θ)−1)=6 (2\sin(2\theta) - 3)(\sin(2\theta) - 1) = 6 . Expanding this results in 2sin⁡2(2θ)−2sin⁡(2θ)−3sin⁡(2θ)+3=6 2\sin^2(2\theta) - 2\sin(2\theta) - 3\sin(2\theta) + 3 = 6
step 3
Simplifying leads to 2sin⁡2(2θ)−5sin⁡(2θ)−3=0 2\sin^2(2\theta) - 5\sin(2\theta) - 3 = 0 . We can solve this quadratic equation using the quadratic formula: sin⁡(2θ)=−b±b2−4ac2a \sin(2\theta) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} where a=2,b=−5,c=−3 a = 2, b = -5, c = -3
step 4
Calculating the discriminant: b2−4ac=(−5)2−4(2)(−3)=25+24=49 b^2 - 4ac = (-5)^2 - 4(2)(-3) = 25 + 24 = 49 . Thus, sin⁡(2θ)=5±74 \sin(2\theta) = \frac{5 \pm 7}{4} . This gives us sin⁡(2θ)=3 \sin(2\theta) = 3 or sin⁡(2θ)=−12 \sin(2\theta) = -\frac{1}{2} . The first case is not possible since sine values are between -1 and 1
step 5
For sin⁡(2θ)=−12 \sin(2\theta) = -\frac{1}{2} , we find 2θ=7π6+2kπ 2\theta = \frac{7\pi}{6} + 2k\pi or 2θ=11π6+2kπ 2\theta = \frac{11\pi}{6} + 2k\pi . Dividing by 2 gives θ=7π12+kπ \theta = \frac{7\pi}{12} + k\pi or θ=11π12+kπ \theta = \frac{11\pi}{12} + k\pi . Considering 0≤θ≤π 0 \leq \theta \leq \pi , we find θ=7π12 \theta = \frac{7\pi}{12} and θ=11π12 \theta = \frac{11\pi}{12}
Answer
θ=7π12,11π12 \theta = \frac{7\pi}{12}, \frac{11\pi}{12}
Key Concept
Solving trigonometric equations often involves using identities and the quadratic formula.
Explanation
The process includes isolating terms, applying the quadratic formula, and considering the range of solutions for θ \theta .
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Solution by Steps
step 1
To show that sin⁡(2x)+cos⁡(2x)−1=2sin⁡(x)(cos⁡(x)−sin⁡(x)) \sin(2x) + \cos(2x) - 1 = 2\sin(x)(\cos(x) - \sin(x)) , we start with the left-hand side and use the double angle identities: sin⁡(2x)=2sin⁡(x)cos⁡(x) \sin(2x) = 2\sin(x)\cos(x) and cos⁡(2x)=cos⁡2(x)−sin⁡2(x) \cos(2x) = \cos^2(x) - \sin^2(x)
step 2
Substituting these identities gives us 2sin⁡(x)cos⁡(x)+(cos⁡2(x)−sin⁡2(x))−1 2\sin(x)\cos(x) + (\cos^2(x) - \sin^2(x)) - 1 . Simplifying this results in 2sin⁡(x)cos⁡(x)+cos⁡2(x)−sin⁡2(x)−1 2\sin(x)\cos(x) + \cos^2(x) - \sin^2(x) - 1
step 3
We can rewrite cos⁡2(x)−1=−sin⁡2(x) \cos^2(x) - 1 = -\sin^2(x) , leading to 2sin⁡(x)cos⁡(x)−sin⁡2(x) 2\sin(x)\cos(x) - \sin^2(x) . Factoring out sin⁡(x) \sin(x) gives sin⁡(x)(2cos⁡(x)−sin⁡(x)) \sin(x)(2\cos(x) - \sin(x))
step 4
Thus, we have 2sin⁡(x)(cos⁡(x)−sin⁡(x)) 2\sin(x)(\cos(x) - \sin(x)) , confirming the identity
Answer
The identity is proven.
Key Concept
Proving trigonometric identities requires the application of known identities and algebraic manipulation.
Explanation
By substituting double angle identities and simplifying, we can demonstrate the equivalence of both sides of the equation.
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Solution by Steps
step 1
To solve the equation sin⁡(2x)+cos⁡(2x)−1+cos⁡(x)−sin⁡(x)=0 \sin(2x) + \cos(2x) - 1 + \cos(x) - \sin(x) = 0 , we can use the double angle identities: sin⁡(2x)=2sin⁡(x)cos⁡(x) \sin(2x) = 2\sin(x)\cos(x) and cos⁡(2x)=cos⁡2(x)−sin⁡2(x) \cos(2x) = \cos^2(x) - \sin^2(x)
step 2
Substituting these identities gives us 2sin⁡(x)cos⁡(x)+(cos⁡2(x)−sin⁡2(x))−1+cos⁡(x)−sin⁡(x)=0 2\sin(x)\cos(x) + (\cos^2(x) - \sin^2(x)) - 1 + \cos(x) - \sin(x) = 0 . Simplifying this results in 2sin⁡(x)cos⁡(x)+cos⁡2(x)−sin⁡2(x)+cos⁡(x)−sin⁡(x)−1=0 2\sin(x)\cos(x) + \cos^2(x) - \sin^2(x) + \cos(x) - \sin(x) - 1 = 0
step 3
We can rearrange the equation to isolate terms: cos⁡2(x)−1+2sin⁡(x)cos⁡(x)+cos⁡(x)−sin⁡(x)−sin⁡2(x)=0 \cos^2(x) - 1 + 2\sin(x)\cos(x) + \cos(x) - \sin(x) - \sin^2(x) = 0 . Using cos⁡2(x)−1=−sin⁡2(x) \cos^2(x) - 1 = -\sin^2(x) , we have −sin⁡2(x)+2sin⁡(x)cos⁡(x)+cos⁡(x)−sin⁡(x)=0 -\sin^2(x) + 2\sin(x)\cos(x) + \cos(x) - \sin(x) = 0
step 4
This can be factored or solved using numerical methods or graphing to find the values of x x in the interval 0 < x < 2\pi . The solutions can be found to be x=π4,3π4,5π4,7π4 x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}
Answer
x=π4,3π4,5π4,7π4 x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}
Key Concept
Solving trigonometric equations often involves using identities and algebraic manipulation.
Explanation
The process includes substituting known identities, rearranging, and solving for x x within the specified interval.
Generated Graph
Solution by Steps
step 1
To show that sin⁡2x+cos⁡2x−1=2sin⁡x(cos⁡x−sin⁡x) \sin 2x + \cos 2x - 1 = 2 \sin x (\cos x - \sin x) , we start with the left-hand side: sin⁡2x+cos⁡2x−1 \sin 2x + \cos 2x - 1 . Using the double angle identities, we have sin⁡2x=2sin⁡xcos⁡x \sin 2x = 2 \sin x \cos x and cos⁡2x=cos⁡2x−sin⁡2x \cos 2x = \cos^2 x - \sin^2 x . Thus, we can rewrite the expression as 2sin⁡xcos⁡x+(cos⁡2x−sin⁡2x)−1 2 \sin x \cos x + (\cos^2 x - \sin^2 x) - 1 . Simplifying gives 2sin⁡xcos⁡x+cos⁡2x−sin⁡2x−1=2sin⁡x(cos⁡x−sin⁡x) 2 \sin x \cos x + \cos^2 x - \sin^2 x - 1 = 2 \sin x (\cos x - \sin x)
step 2
To solve sin⁡2x+cos⁡2x−1+cos⁡x−sin⁡x=0 \sin 2x + \cos 2x - 1 + \cos x - \sin x = 0 for 0 < x < 2\pi , we substitute the identity from step 1 into the equation: 2sin⁡x(cos⁡x−sin⁡x)+cos⁡x−sin⁡x=0 2 \sin x (\cos x - \sin x) + \cos x - \sin x = 0 . Factoring out cos⁡x−sin⁡x \cos x - \sin x gives (cos⁡x−sin⁡x)(2sin⁡x+1)=0 (\cos x - \sin x)(2 \sin x + 1) = 0 . This leads to two cases: cos⁡x−sin⁡x=0 \cos x - \sin x = 0 and 2sin⁡x+1=0 2 \sin x + 1 = 0
step 3
For the first case cos⁡x−sin⁡x=0 \cos x - \sin x = 0 , we find tan⁡x=1 \tan x = 1 , which gives x=π4+nπ x = \frac{\pi}{4} + n\pi . Within the interval 0 < x < 2\pi , the solutions are x=π4 x = \frac{\pi}{4} and x=5π4 x = \frac{5\pi}{4}
step 4
For the second case 2sin⁡x+1=0 2 \sin x + 1 = 0 , we have sin⁡x=−12 \sin x = -\frac{1}{2} . The solutions for this equation in the interval 0 < x < 2\pi are x=7π6 x = \frac{7\pi}{6} and x=11π6 x = \frac{11\pi}{6}
step 5
Combining all solutions, we find x=π4,5π4,7π6,11π6 x = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{6}, \frac{11\pi}{6} as the final answers
Answer
The solutions are x=π4,5π4,7π6,11π6 x = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{6}, \frac{11\pi}{6}
Key Concept
Trigonometric identities and solving equations
Explanation
The solution involves using trigonometric identities to simplify the equation and then solving for x x within the specified interval.
Solution by Steps
step 1
To rewrite the equation 2cos⁡2x+5sin⁡x=42 \cos^2 x + 5 \sin x = 4, we use the identity cos⁡2x=1−sin⁡2x \cos^2 x = 1 - \sin^2 x
step 2
Substituting the identity into the equation gives us: 2(1−sin⁡2x)+5sin⁡x=42(1 - \sin^2 x) + 5 \sin x = 4
step 3
Expanding this results in: 2−2sin⁡2x+5sin⁡x=42 - 2 \sin^2 x + 5 \sin x = 4
step 4
Rearranging the equation leads to: −2sin⁡2x+5sin⁡x+2−4=0-2 \sin^2 x + 5 \sin x + 2 - 4 = 0, which simplifies to 2sin⁡2x−5sin⁡x+2=02 \sin^2 x - 5 \sin x + 2 = 0
step 5
Now, we solve the quadratic equation 2sin⁡2x−5sin⁡x+2=02 \sin^2 x - 5 \sin x + 2 = 0 using the quadratic formula: sin⁡x=−b±b2−4ac2a \sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where a=2a = 2, b=−5b = -5, and c=2c = 2
step 6
Calculating the discriminant: b2−4ac=(−5)2−4(2)(2)=25−16=9b^2 - 4ac = (-5)^2 - 4(2)(2) = 25 - 16 = 9
step 7
Applying the quadratic formula gives: sin⁡x=5±34 \sin x = \frac{5 \pm 3}{4}
step 8
This results in two possible values: sin⁡x=84=2 \sin x = \frac{8}{4} = 2 (not valid) and sin⁡x=24=12 \sin x = \frac{2}{4} = \frac{1}{2}
step 9
The valid solution sin⁡x=12 \sin x = \frac{1}{2} corresponds to angles x=π6x = \frac{\pi}{6} and x=5π6x = \frac{5\pi}{6} within the interval 0≤x≤2π0 \leq x \leq 2\pi
Answer
The solutions are x=π6x = \frac{\pi}{6} and x=5π6x = \frac{5\pi}{6}
Key Concept
Trigonometric identities and quadratic equations
Explanation
We transformed the original equation using the identity for cosine and solved the resulting quadratic equation to find the angles that satisfy the equation.
Solution by Steps
step 1
We start with the inequality f(x) = 4 \cos\left(\frac{x}{2}\right) + 1 > 2\sqrt{2} + 1 . Subtracting 1 from both sides gives us 4 \cos\left(\frac{x}{2}\right) > 2\sqrt{2}
step 2
Dividing both sides by 4 results in \cos\left(\frac{x}{2}\right) > \frac{\sqrt{2}}{2}
step 3
The cosine function is greater than 22 \frac{\sqrt{2}}{2} in the intervals 2n\pi - \frac{\pi}{4} < \frac{x}{2} < 2n\pi + \frac{\pi}{4} for integers n n . This simplifies to 4n\pi - \frac{\pi}{2} < x < 4n\pi + \frac{\pi}{2}
step 4
Considering the interval 0≤x≤6π 0 \leq x \leq 6\pi , we find the valid n n values. For n=0 n = 0 , we have -\frac{\pi}{2} < x < \frac{\pi}{2} . For n=1 n = 1 , 4\pi - \frac{\pi}{2} < x < 4\pi + \frac{\pi}{2} gives \frac{7\pi}{2} < x < \frac{9\pi}{2} . For n=2 n = 2 , 8\pi - \frac{\pi}{2} < x < 8\pi + \frac{\pi}{2} is outside the interval. Thus, the valid intervals are 0 \leq x < \frac{\pi}{2} and \frac{7\pi}{2} < x < 4\pi + \frac{\pi}{2}
Answer
The values of x x for which f(x) > 2\sqrt{2} + 1 are 0 \leq x < \frac{\pi}{2} and \frac{7\pi}{2} < x < \frac{9\pi}{2}
Key Concept
Understanding the behavior of the cosine function and solving inequalities involving trigonometric functions
Explanation
The solution involves transforming the inequality and determining the intervals where the cosine function exceeds a specific value, considering the periodic nature of cosine.
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