what is the exact value of
1-2sin(15degrees)?
also
2(sin2x)=2sinx solve it for 0<_x_<360
and also 2cos2x=-�ã3 solve it and find points of intersection for 0<_x_<ƒÎ
Well, 2sin15 is the same as sin30, right? and sin30 has a special value of 1/2...1-1/2=1/2...so the exact value of 1-2sin15 is 1/2.
I would answer your second question, but the characters are all messed up and I can't understand it!
2sin15 is NOT equal to sin30
let's use the identity
cos 2A = 1 - 2sin^2 A
cos 30 = 1 - 2sin^2 15
√3/2 = 1 - 2sin^2 15
sin^2 15 = 1 - √3/2
= (2-√3)/4
sin 15 = √[(2-√3)/4]
now you can form 1 - 2sin 15,
I will let you finish it.
for the second,
2(sin2x)=2sinx
sin 2x = sinx
2sinxcosx - sinx = 0
sinx(2cosx - 1) = 0
sinx = 0 ----> x = 0 or x= 180 or x=360
or
cosx = 1/2 --- x = 60 or x = 300
the third is easy,
cos 2x = √3/2
2x = 30 or 2x = 330
so x = 15 or x = 165
since the period of cos 2x is 180, add 180 to 15 and 165 to get two more answers
To find the exact value of 1-2sin(15 degrees), you can use the trigonometric identity for the sine of the difference of angles: sin(A - B) = sinA*cosB - cosA*sinB.
In this case, we have sin(15 degrees) = sin(60 degrees - 45 degrees), so using the identity mentioned above, we can write:
sin(15 degrees) = sin(60 degrees)*cos(45 degrees) - cos(60 degrees)*sin(45 degrees).
Using the exact values for sin(60 degrees), cos(45 degrees), and sin(45 degrees) (which can be found in trigonometric tables or calculators), we can solve for sin(15 degrees):
sin(15 degrees) = (√3/2)*(√2/2) - (1/2)*(√2/2)
= (√6 - √2)/4.
Therefore, 1 - 2sin(15 degrees) = 1 - 2*((√6 - √2)/4).
Simplifying further, we get:
1 - 2sin(15 degrees) = (4 - 2√6 + 2√2)/4
= (4 + 2√2 - 2√6)/4
= (2 + √2 - √6)/2.
So, the exact value of 1 - 2sin(15 degrees) is (2 + √2 - √6)/2.
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To solve the equation 2sin(2x) = 2sin(x) for 0 ≤ x ≤ 360 degrees, we can first simplify the equation:
2sin(2x) - 2sin(x) = 0.
Using the trigonometric identity for the double angle of sine: sin(2A) = 2sin(A)cos(A), we can rewrite the equation as:
2*2sin(x)cos(x) - 2sin(x) = 0.
Further simplifying and factoring out sin(x), we get:
2sin(x)(2cos(x) - 1) = 0.
Setting each factor equal to zero:
sin(x) = 0 (equation 1) or
2cos(x) - 1 = 0 (equation 2).
From equation 1, sin(x) = 0 when x is a multiple of 180 degrees:
x = 0, 180, 360.
From equation 2, we can solve for cos(x):
2cos(x) - 1 = 0
2cos(x) = 1
cos(x) = 1/2.
Using the inverse cosine (arccos) function, we find the values of x where cos(x) = 1/2:
x = arccos(1/2)
x = 60 degrees, 300 degrees.
Therefore, the solutions to the equation 2sin(2x) = 2sin(x) for 0 ≤ x ≤ 360 degrees are x = 0, 60, 180, 300, and 360.
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To solve the equation 2cos(2x) = -√3 for 0 ≤ x ≤ Π, we can start by dividing both sides of the equation by 2:
cos(2x) = -√3/2.
Using the double angle formula for cosine: cos(2A) = 2cos^2(A) - 1, we can rewrite the equation as:
2cos^2(x) - 1 = -√3/2.
Rearranging and combining like terms, we get:
2cos^2(x) + √3/2 = 1.
Subtracting 1 from both sides, we have:
2cos^2(x) + √3/2 - 1 = 0.
Finding a common denominator, we can rewrite the equation as:
(4cos^2(x) + 2√3 - 2)/2 = 0.
Multiplying both sides by 2, we get:
4cos^2(x) + 2√3 - 2 = 0.
Adding 2 to both sides, we have:
4cos^2(x) + 2√3 = 2.
Dividing both sides by 4, we get:
cos^2(x) + √3/2 = 1/2.
Subtracting √3/2 from both sides, we have:
cos^2(x) = 1/2 - √3/2.
Taking the square root of both sides, we obtain:
cos(x) = ±√(1/2 - √3/2).
Using the inverse cosine (arccos) function, we find the solutions for cos(x):
x = arccos(√(1/2 - √3/2))
x = arccos(-√(1/2 - √3/2)).
These solutions will give us the angles in the range of 0 ≤ x ≤ Π.
To find the points of intersection, substitute these values of x back into the equation:
For x = arccos(√(1/2 - √3/2)):
2cos(2*arccos(√(1/2 - √3/2))) = -√3.
For x = arccos(-√(1/2 - √3/2)):
2cos(2*arccos(-√(1/2 - √3/2))) = -√3.
Evaluate each expression to find the points of intersection.