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sin(360+30) =sin(30)= 1/2
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)
2*pi is one complete revolution, i.e. 360 degrees. Sin of 2*pi = sin 360º = 0
Sin(285) is a number, not an angle. The reference angle for 285 degrees is 285-360 = -75 degrees.
2
4 sin(theta) = 2 => sin(theta) = 2/4 = 0.5. Therefore theta = 30 + k*360 degrees or 150 + k*360 degrees where k is any integer.
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, ...
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.
sinx(1-sinx)=0 sinx=0 or 1 x= 0, 90, 180, 270, 360...
(30+360*k) and (150+360*k) degrees where k is any integer.
The period of sin + cos is 2*pi radians (360 degrees) so the period of sin(3x) + cos(3x) is 2*pi/3 radians.
First start by reducing 2,670 down to be greater than 0 but less than 360. We can find co-terminal angles to 2,670 by subtracting 360: 2,670-360=2,310 2,310-360=1,950 1,950-360=1,590 1,590-360=1,230 1,230-360=870 870-360=510 510-360=150 So, now the problem is Tan(150). This is equal to the Sin(150)/Cos(150). The Sin(150)=1/2 and Cos(150)=-sqrt(3)/2 So Sin(150)/Cos(150)=[1/2]/[-sqrt(3)/2]=[1/2]*[2/-sqrt(3)]=-1/sqrt(3)=-sqrt(3)/3 So Tan(2,670)=-sqrt(3)/3 ("Negative square root of three over three")