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What else can I help you with?
What is the answer to sin of 390 degrees?
sin(360+30) =sin(30)= 1/2
How do you find value of sine when angle greater than 90?
For angles greater than 360 degrees, subtract multiples of 360 so that the relevant angle (the remainder) is between 0 and 360 degrees. Then For 90 < x ≤ 180 deg, sin(x) = sin(180-x) For 180 < x ≤ 270 deg, sin(x) = -sin(x-180) For 270 < x ≤ 360 deg, sin(x) = -sin(360-x)
What is cos 360 minus theta?
Cos(360 - X) = Trig. Identity Cos(360)Cos(x) + Sin(360)Sin(x) => 1CosX + 0Sinx => CosX + o => CosX
What is the exact values of sin 2 pi?
2*pi is one complete revolution, i.e. 360 degrees. Sin of 2*pi = sin 360º = 0
What is the reference angle sin 285?
Sin(285) is a number, not an angle. The reference angle for 285 degrees is 285-360 = -75 degrees.
How do you solve the following equation using identities sin2x-sinx equals 0?
2
What is the value of theta if 4sin theta is equal to 2?
4Sin(theta) = 2 Sin(Theta) = 2/4 = 1/2 - 0.5 Theta = Sin^(-1) [0.5] Theta = 30 degrees.
What is the answer to this problem cos 5x plus 49 sin 3x plus 57?
There are operators missing. The only question that I can see to make sense is: Solve x for: cos(5x + 49°) = sin(3x + 57°) It's been a while since I did this kind of problem, so there may be more solutions to the ones I give here: Cosθ = sin(90 - θ) → cos(5x + 49°) = sin(3x + 57°) → sin(90° - (5x + 49°)) = sin(3x + 57°) → sin(41° - 5x) = sin(3x + 57°) Thus: 41° - 5x = 3x + 57° → 8x = -16° → x = -2° But as sin and cos are cyclic with a period of 360°, -2° = 360° - 2= 358° → x = 358° + 360°n where n = 0, 1, 2, .... But sin θ = sin(180° - θ) which means that 180 - (41° - 5x) = 3x + 57° → 5x + 139° = 3x + 57° → 2x = -82° → x = -41° → x = 319° + 360°n where n = 0, 1, 2, 3,... is also a solution set. Thus the solutions are: x = 358° + 360°n x = 319° + 360°n where n = 0, 1, 2, 3, ...
Sin2x plus cosx equals 0?
sin 2x + cos x = 0 (substitute 2sin x cos x for sin 2x)2sin x cos x + cos x = 0 (divide by cos x each term to both sides)2sin x + 1 = 0 (subtract 1 to both sides)2sin x = -1 (divide by 2 to both sides)sin x = -1/2Because the period of the sine function is 360⁰, first find all solutions in [0, 360⁰].Because sin 30⁰ = 1/2 , the solutions of sin x = -1/2 in [0, 360] arex = 180⁰ + 30⁰ = 210⁰ (the sine is negative in the third quadrant)x = 360⁰ - 30⁰ = 330⁰ (the sine is negative in the fourth quadrant)Thus, the solutions of the equation are given byx = 210⁰ + 360⁰n and x = 330⁰ + 360⁰n, where n is any integer.
How do you solve sin x - 2 sin x is equal to 0?
sinx(1-sinx)=0 sinx=0 or 1 x= 0, 90, 180, 270, 360...
What are the trigonometric inverse of sine 1 over 2?
(30+360*k) and (150+360*k) degrees where k is any integer.
What is the period of sin3x plus cos3x?
The period of sin + cos is 2*pi radians (360 degrees) so the period of sin(3x) + cos(3x) is 2*pi/3 radians.
