使 u = 3t u = 3 t。 然后使 du = 3dt d u = 3 d t，以便 1 3du = dt 1 3 d u = d t。 使用 u u 和 d d u u 进行重写。 点击获取更多步骤... ∫ sin(u) 1 3du ∫ sin (u) 1 3 d u. 组合 sin(u) sin (u) 和 1 3 1 3。 …

To prove the triple-angle identities, we can write \sin 3 \theta sin3θ as \sin (2 \theta + \theta) sin(2θ+θ). Then we can use the sum formula and the double-angle identities to get the desired …

2023年11月19日 · sin(2α) = 2sinαcosα. sin(α+β) = sinαcosβ+cosαsinβ. cos(α+β) = cosαcosβ - sinαsinβ

prove\:\tan^2(x)-\sin^2(x)=\tan^2(x)\sin^2(x) \frac{d}{dx}(\frac{3x+9}{2-x}) (\sin^2(\theta))' \sin(120) \lim _{x\to 0}(x\ln (x)) \int e^x\cos (x)dx \int_{0}^{\pi}\sin(x)dx

1 天前 · Solve the given initial-value problem. 2 sin(3t), x(0) = 4, x #x27;(t) = 106 cos(3t) - 27Watch the full video at:https://www.numerade.com/ask/question/solv...

I'm trying to calculate the Fourier series of $\sin^3t$ in trigonometric form. In previous excercises I have been able to use trigonometric identities to be able to calculate the coefficents, but h...

Find the period of sin(3t) sin (3 t). Tap for more steps... The period of the sin(3t) sin (3 t) function is 2π 3 2 π 3 so values will repeat every 2π 3 2 π 3 radians in both directions.

2019年7月31日 · Using de Moivre's formula, proving that sin(3t) = 3sin(t) − 4sin3(t) is straightforward. I'm curious whether this formula has a geometric proof or one that doesn't use …

2024年7月23日 · 结论：sin^3t的导数可以通过导数的基本公式和三角函数的性质得到。 以下是详细的步骤： 首先，运用幂规则，我们知道当c=0且(x^n)'=nx^(n-1)，所以对于(sint)^3，它的导 …

Let u = 3t u = 3 t. Then du = 3dt d u = 3 d t, so 1 3du = dt 1 3 d u = d t. Rewrite using u u and d d u u. Tap for more steps... ∫ sin(u) 1 3du ∫ sin (u) 1 3 d u. Combine sin(u) sin (u) and 1 3 1 3. ∫ …