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`sin theta = 0.1`
To solve to `theta` , take the arc sine of 0.1.
`theta = sin^(-1) 0.1`
And if we round it off to one decimal place, the value of `theta` becomes
`theta = 5.7^o`
since the number after .7 is less than 5.
Moreover, sine function is positive not only in Quadrant 1 but also in Quadrant 2. So the other value of theta is:
`theta = 174.3^o`
Since, there is no indicated interval for `theta` , then, we would use the general solution of sine function for all the positive values of `theta` .
Hence, `theta_1 = 5.7 + 360 k` degrees and `theta_2 = 174.3 + 360k` , where k is an integer that is `kgt=0` .
For very small vallues of θ we can say that sinθ ~= θ
So for θ=0,1 you would get sinθ=0,1
And for the respective vallue in the 2nd Quadrant you could say
θ' = pi - θ ~= 3,04
- Note that ~= means approximately equal-
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