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2 years ago

Prove that:cos^2(45+A)+cos (45-A)=1

Mathematics

2 answers:

krek1111 [17]2 years ago

4 0

Step-by-step explanation:

$\boxed{cos^2x=\frac{1-cos2x}{2}}\\cos^2(45+A)+cos^2(45-A)=\frac{1-cos2(45+A)}{2}+\frac{1-cos2(45-A}{2}\\=\frac{1 - cos(90 +2A) }{2} + \frac{1 - cos(90 - 2A) }{2} \\ = \frac{2- ( - sin 2A) - sin2A}{2} \\ = \frac{2 + sin2A -sin2A }{2} \\ = \frac{2}{2} \\ = 1$

liberstina [14]2 years ago

3 0

Step-by-step explanation:

Prove that

$\cos^2(45+A)+\cos^2(45-A) =1$

We know that

$\cos (\alpha \pm \beta) = \cos \alpha\cos \beta \mp \sin \alpha \sin\ beta)$

We can then write

$\cos (45+A)=\cos 45\cos A - \sin 45\sin A$

$\:\:\:\:\:\:\:\:= \frac{\sqrt{2}}{2}(\cos A - \sin A)$

Taking the square of the above expression, we get

$\cos^2(45+A) = \frac{1}{2}(\cos^2A - 2\sin A \cos A + \sin^2A)$

$= \frac{1}{2}(1 - 2\sin A\cos A)\:\:\;\:\:\:\:(1)$

Similarly, we can write

$\cos^2(45-A) =\frac{1}{2}(1 + 2\sin A\cos A)\:\:\;\:\:\:\:(2)$

Combining (1) and (2), we get

$\cos^2(45+A)+\cos^2(45-A)$

$= \frac{1}{2}(1 - 2\sin A\cos A) + \frac{1}{2}(1 + 2\sin A\cos A)$

$= 1$

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Answer:

Part 1) $2(x-3)^{2}-17=0$ (the missing steps in the explanation)

Part 3) (8, 4); The vertex represents the maximum profit

Part 4) x = 3.58, 0.42

Part 5) x = 6, x = 44; The zeros represent the number of monthly memberships where no profit is made

Part 6) 2(x − 7)2 + 118; x = $7

Part 7) The maximum height of the puck is 4 feet. −(x − 4)^2 + 6

Part 8) (x + 3)^2 − 4

Part 9) 2(x − 1)^2 = 4

Part 10) 8(x − 4)^2 + 592

Step-by-step explanation:

Part 1) we have

$2x^{2} -12x+1=0$

Convert to vertex form

step 1

Factor the leading coefficient and complete the square

$2(x^{2} -6x)+1=0$

$2(x^{2} -6x+9)+1-18=0$

step 2

$2(x^{2} -6x+9)+1-18=0$

$2(x^{2} -6x+9)-17=0$

step 3

Rewrite as perfect squares

$2(x-3)^{2}-17=0$

Part 3) we have

$f(x)=-x^{2}+16x-60$

we know that

This is the equation of a vertical parabola open downward

The vertex is a maximum

Convert to vertex form

$f(x)+60=-x^{2}+16x$

Factor the leading coefficient

$f(x)+60=-(x^{2}-16x)$

Complete the squares

$f(x)+60-64=-(x^{2}-16x+64)$

$f(x)-4=-(x^{2}-16x+64)$

Rewrite as perfect squares

$f(x)-4=-(x-8)^{2}$

$f(x)=-(x-8)^{2}+4$

The vertex is the point (8,4)

The vertex represent the maximum profit

Part 4) Solve for x

we have

$-2(x-2)^{2}+5=0$

$-2(x-2)^{2}=-5$

$(x-2)^{2}=2.5$

square root both sides

$(x-2)=(+/-)1.58$

$x=2(+/-)1.58$

$x=2(+)1.58=3.58$

$x=2(-)1.58=0.42$

Part 5) we have

$f(x)=-x^{2}+50x-264$

we know that

The zeros or x-intercepts are the value of x when the value of the function is equal to zero

In this context the zeros represent the number of monthly memberships where no profit is made

To find the zeros equate the function to zero

$-x^{2}+50x-264=0$

$-x^{2}+50x=264$

Factor -1 of the leading coefficient

$-(x^{2}-50x)=264$

Complete the squares

$-(x^{2}-50x+625)=264-625$

$-(x^{2}-50x+625)=-361$

$(x^{2}-50x+625)=361$

Rewrite as perfect squares

$(x-25)^{2}=361$

square root both sides

$(x-25)=(+/-)19$

$x=25(+/-)19$

$x=25(+)19=44$

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Part 6) we have

$-2x^{2}+28x+20$

This is a vertical parabola open downward

The vertex is a maximum

Convert the equation into vertex form

Factor the leading coefficient

$-2(x^{2}-14x)+20$

Complete the square

$-2(x^{2}-14x+49)+20+98$

$-2(x^{2}-14x+49)+118$

Rewrite as perfect square

$-2(x-7)^{2}+118$

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Part 7) we have

$f(x)=-x^{2}+8x-10$

Convert to vertex form

$f(x)+10=-x^{2}+8x$

Factor -1 the leading coefficient

$f(x)+10=-(x^{2}-8x)$

Complete the square

$f(x)+10-16=-(x^{2}-8x+16)$

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Rewrite as perfect square

$f(x)-6=-(x-4)^{2}$

$f(x)=-(x-4)^{2}+6$

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Convert to vertex form

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Rewrite as perfect squares

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Convert to vertex form

Factor 2 the leading coefficient

$2(x^{2}-2x)-2=0$

Complete the square

$2(x^{2}-2x+1)-2-2=0$

$2(x^{2}-2x+1)-4=0$

Rewrite as perfect squares

$2(x-1)^{2}-4=0$

$2(x-1)^{2}=4$

The vertex is the point (1,-4)

Part 10) we have

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This is the equation of a vertical parabola open upward

The vertex is a minimum

Convert to vertex form

Factor 8 the leading coefficient

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$8(x^{2}-8x+16)+592$

Rewrite as perfect squares

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