Please help me with this question!! I also need the steps
1 answer:
Hi there!
We are given:
cos(7x)cos(4x) = -1 - sin(7x)sin(4x)
Begin by moving all terms with variables to one side:
cos(7x)cos(4x) + sin(7x)sin(4x) = -1
The corresponding trig identity is cos(A - B). Thus:
cos(7x - 4x) = cos(7x)cos(4x) + sin(7x)sin(4x) = -1
cos(3x) = -1
cos = -1 at π, so:
3x = π
x = π/3
We can also find another solution. Let 3π = -1:
3x = 3π
x = π
Thus, solutions on [0, 2π) are π/3 and π.
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Answer:
<h2>The area of the triangle ABC is 32 square inches.</h2>Step-by-step explanation:
We know that the given triangle is a right-isosceles triangle, which means its leg are equal.
The area of a triangle is defined as
Where and
. Replacing values, we have
Therefore, the area of the triangle ABC is 32 square inches.
25/3 ft/s is speed of the tip of his shadow moving when a man is 40 ft from the pole given that a street light is mounted at the top of a 15-ft-tall pole and the man is 6 ft tall who is walking away from the pole with a speed of 5 ft/s along a straight path. This can be obtained by considering this as a right angled triangle.
<h3>How fast is the tip of his shadow moving?</h3>Let x be the length between man and the pole, y be the distance between the tip of the shadow and the pole.
Then y - x will be the length between the man and the tip of the shadow.
Since two triangles are similar, we can write
⇒15(y-x) = 6y
15 y - 15 x = 6y
9y = 15x
y = 15/9 x
y = 5/3 x
Differentiate both sides
dy/dt = 5/3 dx/dt
dy/dt is the speed of the tip of the shadow, dx/dt is the speed of the man.
Given that dx/dt = 5 ft/s
Thus dy/dt = (5/3)×5 ft/s
dy/dt = 25/3 ft/s
Hence 25/3 ft/s is speed of the tip of his shadow moving when a man is 40 ft from the pole given that a street light is mounted at the top of a 15-ft-tall pole and the man is 6 ft tall who is walking away from the pole with a speed of 5 ft/s along a straight path.
Learn more about similar triangles here:
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