1. Sin δ = ( cos δ + 200 )
since cosx = sin(90-x),
cos(x+200) = sin(90-(x+200)) = sin(-110-x)
so, x = -110-x
x = -55
check: sin(-55) = -sin(55) = -cos(35) = cos(145)
Sin δ = Cos δ + 200
Sin δ = Cos ( 900 - ( δ + 200 ) )
δ = 900 - δ - 20
2δ = 70
δ = 350
sorry its 90 not 900
To solve the equation sin δ = cos δ + 200, we need to use some basic trigonometric identities and algebraic manipulations. Here's an explanation of how to get the answer:
1. Start by rearranging the equation by subtracting cos δ from both sides:
sin δ - cos δ = 200
2. Use the trigonometric identity sin δ - cos δ = -√2 sin(δ + π/4). The left side of the equation becomes:
-√2 sin(δ + π/4) = 200
3. Divide both sides of the equation by -√2:
sin(δ + π/4) = -200/√2
4. Calculate the right side of the equation:
-200/√2 = -141.42 (approximately)
5. Use the inverse sine function (sin⁻¹) to solve for δ:
δ + π/4 = sin⁻¹(-141.42)
6. Calculate sin⁻¹(-141.42):
sin⁻¹(-141.42) = undefined since the arcsine function is only defined for values between -1 and 1.
Therefore, the equation does not have a real solution for δ.