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prisoha [69]
3 years ago
14

Prove (cosx+cosy)^2+(sinx-siny)^2 = 2+2cos(x+y)

Mathematics
1 answer:
hram777 [196]3 years ago
7 0
<span>Here you go:
1) Applying (a + b)² = a² + b² + 2ab, 
(cosx + cosy)² = cos²x + cos²y + 2cos(x)*cos(y) and 
(sinx + siny)² = sin²x + sin²y + 2sin(x)*sin(y) 

2) ==> (cosx + cosy)² + (sinx + siny)² =  
= (cos²x + sin²x) + (cos²y + sin²y) + 2{cos(x)cos(y) + sin(x)sin(y)} 
= 2 + 2{cos(x - y)} = 2[1 + cos(x - y)]  
= 2*2cos²{(x - y)/2} [Multiple angle identity, 1 + cos(2A) = 2cos²A] 
= 4*cos²{(x - y)/2} [Proved]</span>
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Given f(x)=5x^2-2x and 3x^2+x-4. What is (f+g)(x)?
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= –15x2 + 2x + 8

\left(\small{\dfrac{f}{g}}\right)(x) = \small{\dfrac{f(x)}{g(x)}}(  

g

f

​  

)(x)=  

g(x)

f(x)

​  

 

= \small{\dfrac{3x+2}{4-5x}}=  

4−5x

3x+2

​

Step-by-step explanation:

To find the answers, all I have to do is apply the operations (plus, minus, times, and divide) that they tell me to, in the order that they tell me to.

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