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3 years ago

Use the Pythagorean identity to do the following:

Mathematics

1 answer:

nlexa [21]3 years ago

7 0

Answer: 2 - 2*sin³(θ) - √1 -sin²(θ)

Step-by-step explanation: In the expression

cos(theta)*sin2(theta) − cos(theta)

sin (2θ) = 2 sin(θ)*cos(θ) ⇒ cos(θ)*2sin(θ)cos(θ) - cos(θ)

2cos²(θ)sin(θ) - cos(θ) if we use cos²(θ) = 1-sin²(θ)

2 [ (1 - sin²(θ))*sin(θ)] - cos(θ)

2 - 2sin²(θ)sin(θ) - cos(θ) ⇒ 2-2sin³(θ)-cos(θ) ; cos(θ) = √1 -sin²(θ)

2 - 2*sin³(θ) - √1 -sin²(θ)

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The smaller triangle is a dilation of the larger triangle with a center of dilation at (2,−1) .

andrew-mc [135]

Answer:

Option B. 1/2

Step-by-step explanation:

The smaller triangle is a dilation of the larger triangle with the center of dilation at (2, -1).

we have to calculate the scale factor of the dilation.

One point of the large triangle is (-4, -1). After dilation it has become (-1, -1) of the smaller one.

If we consider the x coordinates of the point x = -4 has reduced down to -1 therefore scale factor = side of small triangle/side of the large triangle = (2 + 1)/(2 + 4) = 3/6 = 1/2

Therefore Option B. 1/2 is the answer.

3 0

3 years ago

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Which of the following expressions shows how to rewrite 4 - 5 using the addtive inverse and displays the expression correctly on

Evgesh-ka [11]

Answer: 4 - 5 = 4 + (-5) = -1

Step-by-step explanation: You have to make the 5 into a negative number using parenthesis and do the math.

5 0

3 years ago

Read 2 more answers

Please answer this ASAP

k0ka [10]

Answer:

c: x=2

Step-by-step explanation:

5x-3= 13-3x

+3 +3

5x=16-3x

+3 +3

8x=16

divide all of it by 8 and thats how you get your answer

5 0

2 years ago

Use the definition of a Taylor series to find the first three non zero terms of the Taylor series for the given function centere

Ket [755]

Answer:

$e^{4x}=e^4+4e^4(x-1)+8e^4(x-1)^2+...$

$\displaystyle e^{4x}=\sum^{\infty}_{n=0} \dfrac{4^ne^4}{n!}(x-1)^n$

Step-by-step explanation:

<u>Taylor series</u> expansions of f(x) at the point x = a

$\text{f}(x)=\text{f}(a)+\text{f}\:'(a)(x-a)+\dfrac{\text{f}\:''(a)}{2!}(x-a)^2+\dfrac{\text{f}\:'''(a)}{3!}(x-a)^3+...+\dfrac{\text{f}\:^{(r)}(a)}{r!}(x-a)^r+...$

This expansion is valid only if $\text{f}\:^{(n)}(a)$ exists and is finite for all $n \in \mathbb{N}$ , and for values of x for which the infinite series converges.

$\textsf{Let }\text{f}(x)=e^{4x} \textsf{ and }a=1$

$\text{f}(x)=\text{f}(1)+\text{f}\:'(1)(x-1)+\dfrac{\text{f}\:''(1)}{2!}(x-1)^2+...$

$\boxed{\begin{minipage}{5.5 cm}\underline{Differentiating $e^{f(x)}$}\\\\If $y=e^{f(x)}$, then $\dfrac{\text{d}y}{\text{d}x}=f\:'(x)e^{f(x)}$\\\end{minipage}}$