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3 years ago

Use the unit circle to find the value of sin ³π/2 and cos ³π/2

Mathematics

2 answers:

VikaD [51]3 years ago

7 0

Answer:

SIn³ π/2 = 1³ = 1

cos³ π/2 = 0³ = 0

Step-by-step explanation:

Considering a circle of unit radius as shown in the image below.

So, OB = OA = 1 unit

Thus, by Pythagoras theorem,

OB² + OA² = AB²

So, AB² = 2

AB = √2 units

If ∠OAB = 90°

Then AB and OA approaches to OA. SO,

OB = AB = 1 unit and OA = 0

Thus,

SIn π/2 = P/H = 1/1 = 1

Cos π/2 = B/H = 0/1 = 0

So,

SIn³ π/2 = 1³ = 1

cos³ π/2 = 0³ = 0

juin [17]3 years ago

5 0

Answer:

1 and 0

Step-by-step explanation:

sin( $\frac{\pi }{2}$ ) = 1, hence

sin³( $\frac{\pi }{2}$ ) = 1³ = 1

cos( $\frac{\pi }{2}$ ) = 0, hence

cos³ ( $\frac{\pi }{2}$ ) = 0³ = 0

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Can someone help me with this question with details

Rufina [12.5K]

Answer:

I think the answer is 5 ( if wrong then do a low rating)

Step-by-step explanation:

The way I found the answer was by using PEDMAS so first I handled [(4 - (-5)) x (-7)] which is the same as 4 + -5 which equals -1 and when multiplied by -7 equals positive 7 because whenever you multiply 2 negatives you get a positive.

Next I solved 6 / (-3) which is -2 since whenever a positive is divided by a negative the quotient is negative. Now the equation is:

-2 + 7 and when solved is -2 + 7 = 5

8 0

2 years ago

If a > b > c > d, then which is larger, a+c or b+d ? Can we tell from a > b > c > d which of a+d and b+c is la

victus00 [196]

Answer:

1. a+c is larger than b+d

2. No way to tell whether a+d or b+c is larger.

Step-by-step explanation:

1. Which is larger, a+c or b+d?

Let a, b, c, and d be any numbers such that $a > b > c > d$ .

Specifically, note that $a > b$ , and subtracting b from both sides of the inequality, observe that $a-b > 0$ .

Similarly, $c > d$ , and subtracting d from both sides of the inequality, observe that $c-d > 0$ .

From this, add "a-b" (a positive number, as proven above) to both sides of the inequality.

$(a-b)+(c-d) > (a-b)+0$

Addition by zero (the additive identity) doesn't change anything, so the right side remains "a-b"...

$(a-b)+(c-d) > a-b$

... and "a-b" is positive...

$(a-b)+(c-d) > a-b > 0$

... so, by the transitive property of inequality...

$(a-b)+(c-d) > 0$

Recall that subtraction is addition by a negative number...
$a+(-b)+c+(-d) > 0$

...and that addition is associative and commutative, so things can be added in any order, so the middle two terms on the left side can be rearranged...

$a+c+(-b)+(-d) > 0$

Adding b + d to both sides of the inequality

$(a+c+(-b)+(-d))+(b+d) > 0+(b+d)$

... and simplifying

$a+c > b+d$

So, a+c is larger than b+d.

2. Which is larger, a+d or b+c?

Consider the following two examples:

Example 1

Suppose a=10; b=3; c=2; d=1.

Note that $a > b > c > d$ ( $10 > 3 > 2 > 1$ ) and, also observe that $a+d=(10)+(1)=11$ , and $b+c=(3)+(2)=5$ , so a+d is larger than b+c.

Example 2

However, suppose a=10; b=9; c=8; d=1.

Note that $a > b > c > d$ ( $10 > 9 > 8 > 1$ ) but that $a+d=(10)+(1)=11$ , and $b+c=(9)+(8)=17$ , so a+d is smaller than b+c.

So, in one example, a+d is bigger, and in the other, a+d is smaller. Therefore, there is no way to tell which of a+d or b+c is larger from only the given information.

5 0

2 years ago

Please use the image below in order to find out the answer to the following question:

vesna_86 [32]

Given:

The measure of ∠EGF is 51°

We need to determine the measure of ∠EHF

Measure of ∠EHF:

By inscribed angle theorem, we know that, "if two inscribed angles of a circle intercept the same arc then the angles are congruent".

Applying the theorem, we have that the two inscribed angles of a circle intercepting the same arc are ∠G and ∠H

Then, the two angles are congruent.

Thus, we have;

∠G ≅ ∠H

Therefore, the measure of ∠G = ∠H = 51°

Hence, the measure of ∠EHF = 51°

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4 years ago

Tom spends 30% of his monthly budget on rent. His total monthly budget is $3,000. How much does Tom spend on rent?

lilavasa [31]

Hey You!

3,000 * 30% = 900

So, Tom spends $900 on his monthly rent.

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3 years ago

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What is the area of a right triangle 9 centimeters high with a base of 9 centimeters?

SSSSS [86.1K]

Answer:C

Step-by-step explanation:

Area = 1/2 b h

Area = 1/2 * 9 * 9 = 40.5 sq cm

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3 years ago