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Cos3x Formula, which is a cosine function, is one of the six trigonometric functions, which relates the ratio of the adjacent side of a right-angled triangle to its hypotenuse. The cosine function is denoted as cos(x), where x is the angle between the adjacent side and the hypotenuse. The cosine function is periodic, meaning that it repeats itself after a certain interval.
- The period of the cosine function is 2π, which means that the function has the same value at x and x+2π.
- Trigonometric identities are equations that are true for all values of the variables involved.
- They are used to simplify expressions involving trigonometric functions.
- There are many trigonometric identities, some of which are derived from the Pythagorean theorem, while others are derived from the definitions of the trigonometric functions.
- The cos3x formula is one such identity.
Key Terms: Cosine, Pythagorean Identity, Periodic Functions, Fourier Series, Trigonometric Functions.
Cos3x Formula
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The cos3x formula is a trigonometric identity that expresses the cosine of three times an angle x in terms of the cosine of the angle x. The formula is as follows:
cos 3x = 4 cos3 x – 3 cos x
Cos3x Derivation
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This formula can be derived using various methods. One of the methods involves using the angle addition formula for cosine. According to the angle addition formula for cosine, we have:
cos(a + b) = cos(a).cos(b) - sin(a).sin(b)
cos(a - b) = cos(a).cos(b) + sin(a).sin(b)
Using these formulas, we can derive the cos3x formula as follows:
cos 3x = cos (2x+x)
=cos 2x cos x – sin 2x sin x
=(cos2 x – sin2x) cos x - 2sin x cos x sin x
=cos3 x – sin2x cos x – 2sin2 x cos x
=cos3 x – 3sin2x cos x
=cos3 x – 3(1 – cos2 x ) cos x
=cos3 x – 3 cos x + 3 cos3 x
=4 cos3 x – 3 cos x
Therefore, the cos 3x formula is 4 cos3 x – 3 cos x
Cos 3x in Terms of Sin x
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To express cos(3x) in terms of sin(x), we can use the Pythagorean identity for sine and cosine:
sin2 x + cos2 x = 1
cos2 x = 1 - sin2 x
Substituting this expression into the cos3x formula, we get:
cos 3x = 4 cos3 x – 3 cos x
=4(1-sin2x)3/2 – 3 cos x
Expanding the cube and simplifying, we get:
cos 3 x = 4 (1-3 sin2x + 3 sin4x – sin6x ) – 3 cos x
4-12 sin2x + 12 sin4x – 4 sin6x -3 cos x
To express cos(3x) in terms of sin(x) only, we can use the Pythagorean identity again to eliminate cos(x):
cos2x = 1 – sin2x
cos x = \(\sqrt 1- sin^2 x\)
We can choose the sign of the square root based on the quadrant of x, or use the identity cos(x) = cos(2π - x) to restrict the range of x to [0, π/2]. Substituting this expression into the previous formula, we get:
cos(3x) = 4 -12 sin2x +12 sin4x – 4 sin6x – 3
\(\sqrt 1- sin^2 x\)This formula expresses cos(3x) in terms of sin(x) as a polynomial function of sin(x) of degree 6. It can be simplified further using various trigonometric identities, such as the double-angle formula for sine and the Pythagorean identity for sine and cosine. For example, using the double-angle formula for sine, we can express sin22x in terms of sin(x) as:
sin22x =(1-cos 4x)/2=(1-2sin2x)2/2
Substituting this expression into the formula for cos(3x), we get:
cos 3x = 4 – 24 sin2x + 48 sin4x – 32 sin6x -3
\(\sqrt 1- sin^2 x\)cos 3x = 4-24 sin2x + 48 sin4x – 32 sin6x – 3 cos x
This formula expresses cos(3x) in terms of sin(x) and cos(2x) as a polynomial function of sin(x) of degree 6. It can be used to simplify expressions involving trigonometric functions of multiple angles or to solve trigonometric equations involving cos(3x) and sin(x).
Cos3x Graph
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The graph of cos(3x) is a periodic function that repeats every 2π/3, since the period of cos(x) is 2π and 3x covers the same interval as x over 2π/3 periods. The graph has three maxima and three minima per period, and crosses the x-axis at every 2π/6 or π/3 radians.
To sketch the graph of cos(3x), we can start by plotting the key points and using the symmetry properties of the function. The maxima occurs at x = 0, 2π/3, 4π/3 and the minima occurs at x = π/3, π, 5π/3. We can also note that cos(3x) = cos(-3x), so the graph is symmetric about the y-axis.
Graphical representation of Cos 3x function
The graph oscillates between 1 and -1 with a period of 2π/3, and has three peaks and three valleys per period. The peaks occur at x = 0, 2π/3, and 4π/3, where cos(3x) = 1, and the valleys occur at x = π/3, π, and 5π/3, where cos(3x) = -1.
We can also observe that the graph of cos(3x) has a steeper slope than the graph of cos(x), since cos(3x) = 4 cos3 x – 3 cos x involves higher powers of cos(x). The steepness of the graph increases as we move away from the x – axis, and the amplitude decreases as we move away from the peaks and valleys.
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Things to Remember
When working with the formula for cos(3x), there are a few key things to keep in mind:
- The formula is based on the trigonometric identity cos 3x = 4cos3 x – 3 cos x. This identity can be used to express cos(3x) in terms of cos(x) or sin(x).
- The graph of cos(3x) is a periodic function with a period of 2π/3. It oscillates between 1 and -1 and has three peaks and three valleys per period.
- To solve trigonometric equations involving cos(3x), we can use the identity cos 3x =4 cos3 x – 3 cos x to transform the equation into a polynomial equation in cos(x).
- The formula for cos(3x) can be used in Fourier series to represent periodic functions as a sum of sine and cosine functions.
Sample Questions
Ques. Solve the equation cos(3x) = 0 for 0 ≤ x ≤ 2π. (2 Marks)
Ans. Since cos(3x) = 0 if and only if 3x = (2n+1)π/2 for some integer n, we have:
3x = π/2, 3π/2, 5π/2, 7π/2
So, x = π/6, π/2, 5π/6, 7π/6.
Ques. Find the exact value of cos(3π/4). (3 Marks)
Ans. We can use the identity cos(3x) = 4 cos3 x – 3 cos x to rewrite cos(3π/4) as:
cos(3π/4) =4 cos3 π/4 – 3 cos π/4
cos π/4 =
√2/2, so we have:cos(3π/4) = (4
√2/2)3 – 3 √2/2=- √2/2Therefore, cos(3π/4) = –
√2/2Ques. Find the value of cos(120°) using the formula for cos(3x). (5 Marks)
Ans. We can express 120° as a multiple of 3: 120° = 3 × 40°. Therefore, using the identity cos(3x) = 4 cos3 x – 3 cos x, we have:
cos(120°) = cos(3 × 40°) = 4 cos3 40o – 3 cos 40o
Now, using the half-angle formula for cosine, we can express cos(40°) in terms of cos(20°), and then use the formula for cos(3x) again to express cos(120°) in terms of cos(20°):
cos(20°) =
√1+cos 40/2cos(120°) =4 cos3 40o – 3 cos 40o=4(2 cos2 20-1)3 – 3 (2 cos2 20 – 1)
Substituting cos(20°) =
√1+cos 40/2, we get:cos(120°) = 4 ( 2 [ ( 1 + cos 40) / 2 ] – 1)3 – 3 [ (1 + cos 40) / 2 ] – 1
Simplifying this expression, we get:
cos(120°) = -1/2
Ques. Find the general solution of the equation cos(3x) + 4sin(x)cos(2x) = 0. (6 Marks)
Ans. Using the identity cos(3x) =4 cos3x-3 cos x, we can rewrite the equation as:
4 cos3 x - 3 cos x + 4 sin x. 2 cos2x – 1 = 0
Simplifying this expression, we get:
4 cos3 x – 6 cos x + 4 sin x cos2 x – 4 sin x=0
Dividing both sides by cos(x) (which is nonzero for any solution), we get:
4 cos2 x – 6 + 4 sin x cos x – 4 tan x =0
Using the substitution y = sin(x)/cos(x) = tan(x), we can rewrite this equation as a quadratic equation in y:
4y2 - 4y - 6 = 0
Solving this equation, we get:
y = (1 ± √7)/2
Substituting back y = sin(x)/cos(x) = tan(x), we get:
tan(x) = (1 ± √7)/2
Therefore, the general solution of the equation is:
x = nπ + arctan((1 ± √7)/2)
where n is an integer.
Ques. Find the value of cos(3π/8). (6 Marks)
Ans. We can use the formula for cos(3x) in terms of cos(x) to express cos(3π/8) as follows:
cos(3π/8) = 4 cos3 (π/8) - 3 cos(π/8)
Using the double angle formula for cos(2x), we can express cos(π/4) in terms of cos(π/8) as follows:
cos(π/4) = cos2 (π/8) - sin2 (π/8) = 2 cos2 (π/8) - 1
Solving for cos(π/8), we get:
cos(π/8) = ±
√cos π/4 +1/2 = ± √2+1/4Taking the positive root, we get:
cos(π/8) = √ 2+1/4
Substituting this into the formula for cos(3π/8), we get:
cos(3π/8) =4 cos3 π/8 – 3 cos π/8 = 4( √2+1/4)3 – 3√(√2+1)/4
Simplifying this expression, we get:
cos(3π/8) = (1/4)(2√2 - √(2 + √2))
Ques. Find the solutions of the equation cos(3x) = 1/2 for 0 ≤ x ≤ 2π. (5 Marks)
Ans. Using the identity cos(3x) = 4 cos3 x - 3cos x,
we can rewrite the equation as 4 cos3x - 3cosx = 1/2.
Letting y = cos(x), we can rewrite this equation as 4y3 - 3y = 1/2.
Multiplying both sides by 2, we get 8y3 - 6y - 1 = 0.
This is a cubic equation that can be solved using standard methods, such as the Cardano formula or numerical methods like Newton's method.
The solutions are y = cos(x) = -1/2, cos(x) = 1/2(cos(2π/9) ± i sin(2π/9)), and cos(x) = 1/2(cos(4π/9) ± i sin(4π/9)).
Since we are only interested in real solutions, we have cos(x) = ½ cos (2π/9) and cos(x) = ½ cos (4π/9).
Thus, the solutions are x = 2π/9, 4π/9, 8π/9, and 10π/9.
Ques. Evaluate ∫cos(3x)dx. (6 marks)
Ans. Using the identity cos(3x) = 4 cos3x- 3cos(x), we can rewrite the integral as ∫(4cos3x - 3cos(x))dx. Integrating term by term, we get:
∫(4 cos3x - 3cos(x))dx = 4∫cos3(x)dx - 3∫cos(x) dx
To evaluate these integrals, we can use the substitution u = sin(x), du = cos(x)dx:
∫cos(x)dx = ∫du = sin(x) + C
∫cos3(x)dx = ∫cos2(x)cos(x)dx = ∫(1 - sin2(x))cos(x)dx = ∫cos(x)dx - ∫sin2x cos(x)dx
Using the identity sin2(x) = 1 - cos2(x), we have:
∫sin2(x)cos(x)dx = ∫(1 - cos2(x))cos(x)dx = ∫cos(x)dx - ∫cos3(x)dx
Substituting these integrals back into the original equation, we get:
∫cos(3x)dx = 4(∫cos(x)dx - ∫cos3(x)dx) - 3(sin(x) + C) = 4sin(x) - 4/3cos3(x) - 3sin(x) + C = -4/3 cos3x + sin(x) + C.
Ques. Find the general solution of the equation 2cos(3x) + 3 = 0. (5 Marks)
Ans. Using the formula for cos(3x), we can write 2cos(3x) = 4cos3(x) - 3cos(x). Substituting this expression into the equation, we get:
4cos3(x) - 3cos(x) + 3 = 0
Dividing both sides by 4, we get:
cos3(x) - (3/4)cos(x) - (3/4) = 0
Letting y = cos(x), we can rewrite this equation as y3- (3/4)y - (3/4) = 0.
This is a cubic equation that can be solved using standard methods, such as the Cardano formula or numerical methods like Newton's method.
The solutions are y = cos(x) = cos(2π/9), cos(2π/9 + 2π/3), and cos(2π/9 + 4π/3).
Therefore, the general solution of the equation is x = (2kπ ± 2π/9)/3, (2kπ ± 2π/9 + 2π/3)/3, and (2kπ ± 2π/9 + 4π/3)/3, where k is an integer.
Ques. Find the exact value of cos(2π/9). (4 Marks)
Ans. Using the formula cos(3x) = 4cos3(x) - 3cos(x),
We can write cos(2π/3) = cos(3π/9) = 4cos3(π/9) - 3cos(π/9).
Letting y = cos(π/9), we can rewrite this equation as 4y3 - 3y = cos(2π/9).
Solving this cubic equation using standard methods, such as the Cardano formula or numerical methods like Newton's method,
we get cos(2π/9) = (1/2)(1 + √5/5).
Ques. Find the period of the function f(x) = cos(3x). (2 Marks)
Ans. The period of f(x) is given by T = 2π/|3| = 2π/3. Therefore, the function f(x) has a period of 2π/3.
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