As many people have pointed out by now, $\sin^2 x$ is simply a "nickname" for $(\sin x)^2$. Therefore, $\sin^2\ 30 = (\sin 30)^2 = (1/2)^2 = 1/4$. As it happens, though, there is another useful thing we can say about $\sin^2 x$: $$\sin^2 x = (\sin x)^2 = \frac12 (1 - \cos (2 x)).$$ We can see this using the double-angle formula for cosines ...

"sin" seems like the meat and body of the concept whereas (x) is just the input. It seems to make more sense to put modifications closer the the "sin". Or it does to me. $\sin(x)^2$ looks like the power is trying to modify the (x) and I have to look twice to verify that it is actually the [sin(x)] that is being squared.

2018年1月31日 · Prove: 1- cos(x) = 2sin^2(x/2) Digress, for a moment, to the identity cos(2u) = 1 - 2sin^2(u) Substitute u = x/2: cos(x) = 1-2sin^2(x/2) Add 2sin(x/2)-cos(x) to both sides: 2sin^2(x/2) = 1-cos(x) Return to the original equation and make the above substitution: 1- …

2016年11月23日 · See below. As xrarr0, x^2 rarr0 so we can use lim_(thetararr0) sintheta/theta = 1 with theta = x^2

2015年10月18日 · Express sin (x/2) in terms of cos x. Ans: #sin (x /2) = sqrt((1 - cos x)/2)# Explanation: By applying the trig identity: #cos 2a = 1 - 2sin^2 a#, we get:

I need to prove that $\\sin(x) > \\frac{x}{2}$ if $0<x<\\pi/2$ I've started working with the derivative, but if it's possible, I'd rather something simpler than that.

2016年9月18日 · By the chain rule. Letting y = sin^2(u) and u = x/2, we need to differentiate both functions and multiply the derivatives together. The derivative of y = sin^2u can be obtained as follows: y = (sinu)(sinu) By the product rule: y' = cosu xx sinu + cosu xx sinu y' = 2cosusinu y' = sin2u The derivative of u = x/2 can be obtained using the quotient rule: u = x/2 u' = (1 xx 2 - x xx 0)/2^2 u' = 2/4 ...

2015年5月2日 · Use the trig identity: 2.sin^2 a = (1 - cos 2a) Replace: 2.sin^2 (x/2) = (1 - cos x) f(x) = 2cos^2 x + 1 - cos x - 2 = 0. Call cos x = t --> quadratic equation. f(t) = 2t^2 - t - 1 = 0 Since a + b + c = 0, one real root is t = 1 and the other is t = c/a = -1/2 Next, solve t = cos x = 0 -> x = Pi/2 and x = 3Pi/2 Then, solve t = cos x = -1/2 -> x = 2Pi/3 and x = 4Pi/3

2014年9月8日 · Answer 2sin(x)cos(x) Explanation You would use the chain rule to solve this. To do that, you'll have to determine what the "outer" function is and what the "inner" function composed in the outer function is. In this case, sin(x) is …

2018年3月23日 · How do you use transformations of #f(x)=x^3# to graph the function #h(x)= 1/5 (x+1)^3+2#? See all questions in Introduction to Twelve Basic Functions Impact of this question