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What is the value of y sin θ when θ 0?
sin(0)=0, therefore ysin(0)=0
What is limit as x approaches 0 of sin squared x by x?
lim(x->0) of sin(x)^2/x we use L'Hospital's Rule and derive the top and the bottomd/dx(sin(x)^2/x)=2*sin(x)*cos(x)/1lim(x->0) of 2*sin(x)*cos(x)=2*0*1=0
If x is obtuse and sin x equals b how do you show sin 2x is less than b?
you need to explain it better but with what i got i if sin x equals b obviously sin 2x is double of b hence sin 2x is more than b. {Not obvious at all, actually. And the above is false. Sin(Pi/2) = 1 and Sin(Pi)=0. But clearly 2 is not greater than 0. Contradiction.} Obtuse means Pi/2 < x < Pi So, sin(x) = b means b>0, because sin(y) > 0 if 0<y<Pi sin(2x)=2sin(x)cos(x) cos(x) < 0 because cos(Pi/2)= 0 and the derivative is negative there. Hence, sin(2x) = 2 sin(x) cos(x) = 2*b*(-K), where K is a positive constant Since b>0, -2Kb < b
What is the integral of sin t dt with the interval x 0?
Let g(x) = interval [0, x] of sin t dt, and f(t) = sin t. Since f(t) is a continuous function, the part one of the Fundamental Theorem of Calculus gives, g'(x) = sin x = f(x) (the original function). If you are interested in the interval [x, 0] of sin t dt, then just put a minus sign in front of the integral and interchange places of 0 and x. So that, g(x) = interval [x, 0] of sin t dt = -{ interval [0, x] of sin t dt}, then g'(x) = - sin x.
Approximation of sin x by x?
the value of sin(x) lies between -1 to +1. the approx value of sin(x)/x = 1 when x tends to 0 & sin(x)/x = 0 when x tends to infinity.
