Answer:
We have to prove
sin(α+β)-sin(α-β)=2 cos α sin β
We will take the left hand side to prove it equal to right hand side
So,
=sin(α+β)-sin(α-β) Eqn 1
We will use the following identities:
sin(α+β)=sin α cos β+cos α sin β
and
sin(α-β)=sin α cos β-cos α sin β
Putting the identities in eqn 1
=sin(α+β)-sin(α-β)
=[ sin α cos β+cos α sin β ]-[sin α cos β-cos α sin β ]
=sin α cos β+cos α sin β- sinα cos β+cos α sin β
sinα cosβ will be cancelled.
=cos α sin β+ cos α sin β
=2 cos α sin β
Hence,
sin(α+β)-sin(α-β)=2 cos α sin β
4. By the fundamental theorem of calculus,


is increasing on those intervals where

. So you have

which is clearly positive for

, and in the second interval you have

Together, this means

for all

.
5. When

,

reduces to

, so you have


Answer:
<h2>The number is 75</h2>
Step-by-step explanation:
Let the number be x
A third of a number is written as

A third of a number minus 8 is written as

The result is 17
We have

<u>Add 8 to both sides of the equation</u>
That's

<u>Multiply through by 3 to eliminate the fraction</u>
That's

<h3>The number is 75</h3>
Hope this helps you
Answer:
the answer is c
Step-by-step explanation:
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