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3 years ago

What is the value of sin-1(1)? negative startfraction pi over 2 endfraction 0 startfraction pi over 2 endfraction π

1 answer:

3 0

The value of sin-1(1) is negative startfraction pi over 2 endfraction, startfraction -pi over 2 endfraction.

<h3>What is the value of sin (π/2)?</h3>

The value of sin (π/2) is equal to the number 1. The value of the sin-1(1) has to be find out.

Suppose the value of this function is <em>x</em>. Thus,

$x=\sin^{-1}(1)$

Solve it further,

$\sin(x)=(1)$ ......1

The value of sin (π/2) and -sin (-π/2) is equal to 1 such that,

$\sin\dfrac{\pi}{2}=-\sin\left(-\dfrac{\pi}{2}\right)=1$

Put this value in the equation 1,

$\sin(x)=\sin\left(\dfrac{\pi}{2}\right)=-\sin\left(-\dfrac{\pi}{2}\right)$

Thus, the range will be,

$\left(\dfrac{\pi}{2},-\dfrac{\pi}{2}\right)$

Thus, the value of sin-1(1) is negative startfraction pi over 2 endfraction, startfraction -pi over 2 endfraction.

Learn more about the sine values here;

brainly.com/question/10711389

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3 years ago

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2 years ago

PLEASE HELP ME WITH THIS QUESTION ITS URGENT!!!

Alex Ar [27]

Start by distributing the exponent to each of the terms in $(2u)^{3}$ . This will become $2^{3} u^{3}$ , and $2^{3} =8$ . Now, the expression is:

$\frac{2u^{3} v^{2} }{8u^{3} * u^{2} }$ .

We can now simplify the bottom to read: $8u^{5}$ because when multiplying variables raised to an exponent, we add the exponents. The expression now looks like:

$\frac{2u^{3} v^{2} }{8u^{5} }$

The 2/8 simplifies to 1/4:

$\frac{u^{3} v^{2} }{4u^{5} }$

Now, we have two u terms on the top and bottom. When dividing variables raised to an exponent, we subtract the exponents. However, since 3-5=-2, the term will be on the bottom to avoid the negative exponent. The final answer is:

$\frac{v^{2} }{4u^{2} }$