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agasfer [191]
3 years ago
6

The greatest value of sinA.cosA is

Mathematics
2 answers:
scZoUnD [109]3 years ago
7 0
Remark
I can't imagine you having to solve this question any other way than by graphing it, if you are in middle school mathematics.

Step One
Go to Desmos. Input y = sin(A)*Cos(A) on the input bar.

Step Two
Check to see where the highest point is. It should be somewhere around 45 degrees or pi/4 which is 0.7854 on the x axis. 

Step Three 
On my graph, you can click on the greatest point on the red line. It should come back with (pi/4,0.5) That is the highest point between 0 and 90 degrees. Depending on what your graph looks like, there is another such point a (5*pi/4,0.5)

Note
The other responder did this question using Trigonometry. If you understand the answer, I would follow it. If not then this is another way to do it. 

-BARSIC- [3]3 years ago
4 0
Use the trig identity
2*sin(A)*cos(A) = sin(2*A)
to get
sin(A)*cos(A) = (1/2)*sin(2*A)

So to find the max of sin(A)*cos(A), we can find the max of (1/2)*sin(2*A)

It turns out that sin(x) maxes out at 1 where x can be any expression you want. In this case, x = 2*A. 

So (1/2)*sin(2*A) maxes out at (1/2)*1 = 1/2 = 0.5

The greatest value of sin(A)*cos(A) is 1/2 = 0.5
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Need a little help here pls
Crank

Answer:

The statements that must be true are:

XY and JK form four right angles ⇒ B

XY ⊥ JK ⇒ C

JP = KP ⇒ E

m∠JPX = 90° ⇒ F

Step-by-step explanation:

From the given figure

∵ Line XY is the perpendicular bisector of the line segment JK

→ That means line XY is the line of symmetry of the line segment JK

∴ XY ⊥ JK ⇒ C

∵ XY ∩ JK at point P

∴ P is the midpoint of JK

∵ XY ⊥ JK

∴ ∠JPX, ∠KPX, ∠JPY, and ∠KPY are right angles

∴ XY and JK form four right angles ⇒ B

∵ The measure of the right angle is 90°

∴ m∠JPX = m∠KPX = m∠JPY = m∠KPY = 90°

∴ m∠JPX = 90° ⇒ F

∵ P is the midpoint of JK

∴ JP = KP ⇒ E

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AfilCa [17]
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DochEvi [55]

Answer:

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Step-by-step explanation:

43 +4 make sure you add them

47 +2

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49

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Answer is unlikely.

Hope it helps!
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