Mathematics: Which trigonometric identity was introduced in this unit? a) sin^2 x= 1 + cos^2 x b) (sin x + cos x)^2 = 1 c) tan x = sin x/cos x d) tan x= cos x/sin x

By the chain rule, if , thenWe have , for which• • and and , so that• • Putting everything together, we get...

Which trigonometric identity was introduced in this

Ответ:

viodsenpai

08.01.2022

One way to determine if the function is a trigonometric identity or not is to use a calculator on the choices. Choice A does not fit. Choice B should be (sin x)^2 + (cos x)^2 to become equal to 1. Choice D> should be cot x, the inverse of tan x to become equal to cos x/ sin x. Hence, answer is C>

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Ответ:

kaleahlove13

16.03.2023

Mathematics

By the chain rule, if z=f(x,y)=f(x(t),y(t)) , then

$\dfrac{\mathrm dz}{\mathrm dt} = \dfrac{\partial f}{\partial x}\dfrac{\mathrm dx}{\mathrm dt} + \dfrac{\partial f}{\partial y}\dfrac{\mathrm dy}{\mathrm dt}$

We have f(x,y)=(x+y)e^y , for which

• $\dfrac{\partial f}{\partial x} = e^y$

• $\dfrac{\partial f}{\partial y} = e^y + (x+y)e^y = (x+y+1)e^y$

and x(t)=5t and y(t)=1-t^2 , so that

• $\dfrac{\mathrm dx}{\mathrm dt} = 5$

• $\dfrac{\mathrm dy}{\mathrm dt} = -2t$

Putting everything together, we get

$\dfrac{\mathrm dz}{\mathrm dt} = 5e^y - 2t(x+y+1)e^y \\\\ \dfrac{\mathrm dz}{\mathrm dt} = 5e^{1-t^2} - 2t(5t-t^2+2)e^{1-t^2} = \boxed{\left(2t^3-10t^2-4t+5\right)e^{1-t^2}}$

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Which trigonometric identity was introduced in this unit?
a) sin^2 x= 1 + cos^2 x
b) (sin x + cos x)^2 = 1
c) tan x = sin x/cos x
d) tan x= cos x/sin x

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