Question #263379

Suppose we have two numbers x and y whose difference is 8 and x is less than y

1.1. Find the function that models the number y in terms of the number x

1.2. Find the function that models the sum S of the squares of the two numbers in terms of x

1.3. Determine the values of the two numbers such that S is a minimum. Include the steps of your reasoning


2.1. Prove the identity 1/(1-sinθ) =sec2θ+tanθ.secθ

2.2. For which values of θ is the identity in 2.1. undefined?

2.3. Solve the equation 2cos 2x.cosec2x=2cos 2x for x\isin (-π ,π\pi )


3.1. Use the special triangles and the additional formula for sine to determine the value of sin 75°. Leave the answer in surd form if necessary

3.2. Suppose we have a triangle ABC where angle ABC is equals to 75°, angle BAC is equals to 60° and the length of AC is equals to 10 cm

(i) Briefly sketch the triangle, displaying all the given information

(ii) Use the Law of Sines, and yours answer in 3.1. to determine the length of AB. Leave the answer in surd form if necessary


1
Expert's answer
2021-11-10T07:14:46-0500

1.1


y=x+8y=x+8

1.2


S=x2+y2=x2+(x+8)2S=x^2+y^2=x^2+(x+8)^2

=x2+x2+16x+64=x^2+x^2+16x+64

S=2x2+16x+64S=2x^2+16x+64

1.3


S=2(x2+8x+32)=2(x2+8x+16+16)S=2(x^2+8x+32)=2(x^2+8x+16+16)

=2(x+4)2+32=2(x+4)^2+32

Since a=2>0,a=2>0, then the quadratic function S(x)=2(x+4)2+32S(x)=2(x+4)^2+32 has the absolute minimum at x=4.x=-4.


S(4)=32S(-4)=32

2.1


11sinθ=11sinθ1+sinθ1+sinθ\dfrac{1}{1-\sin \theta}=\dfrac{1}{1-\sin \theta}\cdot\dfrac{1+\sin \theta}{1+\sin \theta}

=1+sinθ1sin2θ=1+sinθcos2θ=\dfrac{1+\sin \theta}{1-\sin^2 \theta}=\dfrac{1+\sin \theta}{\cos^2 \theta}

=1cos2θ+sinθcos2θ=sec2θ+tanθsecθ=\dfrac{1}{\cos^2 \theta}+\dfrac{\sin \theta}{\cos^2 \theta}=\sec^2\theta+\tan\theta\cdot\sec \theta

11sinθ=sec2θ+tanθsecθ,cosθ0\dfrac{1}{1-\sin \theta}=\sec^2\theta+\tan\theta\cdot\sec \theta, \cos \theta\not=0

2.2

If cosθ=0,\cos \theta=0, then the identity 11sinθ=sec2θ+tanθsecθ\dfrac{1}{1-\sin \theta}=\sec^2\theta+\tan\theta\cdot\sec \theta is undefined.


cosθ=0=>θ=π2+πn,nZ\cos \theta=0=>\theta=\dfrac{\pi}{2}+\pi n, n\in \Z

The identity 11sinθ=sec2θ+tanθsecθ\dfrac{1}{1-\sin \theta}=\sec^2\theta+\tan\theta\cdot\sec \theta is undefined for θ=π2+πn,nZ.\theta=\dfrac{\pi}{2}+\pi n, n\in \Z.


2.3


2cos2xcosec2x=2cos2x,x(π,π)2\cos 2x\cdot\cosec2x=2\cos 2x , x\in (-\pi ,\pi )

sin2x0\sin 2x\not=0

Then

cos2x=0\cos 2x=0

Or


cosec2x=0\cosec2x=0

The first equation


cos2x=0\cos 2x=0

2x=π2+πn,nZ2x=\dfrac{\pi}{2}+\pi n, n\in \Z

x=π4+πn2,nZx=\dfrac{\pi}{4}+\dfrac{\pi n}{2}, n\in \Z



The second equation


cosec2x=0\cosec2x=0

1sin2x=0,No solution\dfrac{1}{\sin 2x}=0, No\ solution

The equation cosec2x=0\cosec2x=0 has no solution.


Therefore we take


x=π4+πn2,nZx=\dfrac{\pi}{4}+\dfrac{\pi n}{2}, n\in \Z

Since x(π,π)x\in (-\pi, \pi)


x1=π4+π(2)2=3π4x_1=\dfrac{\pi}{4}+\dfrac{\pi (-2)}{2}=-\dfrac{3\pi}{4}

x2=π4+π(1)2=π4x_2=\dfrac{\pi}{4}+\dfrac{\pi (-1)}{2}=-\dfrac{\pi}{4}

x3=π4+π(0)2=π4x_3=\dfrac{\pi}{4}+\dfrac{\pi (0)}{2}=\dfrac{\pi}{4}

x4=π4+π(1)2=3π4x_4=\dfrac{\pi}{4}+\dfrac{\pi (1)}{2}=\dfrac{3\pi}{4}


{3π4,π4,π4,3π4}\bigg\{-\dfrac{3\pi}{4}, -\dfrac{\pi}{4}, \dfrac{\pi}{4}, \dfrac{3\pi}{4}\bigg\}

3.1


sin(75°)=sin(45°+30°)\sin(75\degree)=\sin(45\degree+30\degree)

=sin(45°)cos(30°)+cos(45°)sin(30°)=\sin(45\degree)\cos(30\degree)+\cos(45\degree)\sin(30\degree)

=22(32)+22(12)=2(3+1)4=\dfrac{\sqrt{2}}{2}(\dfrac{\sqrt{3}}{2})+\dfrac{\sqrt{2}}{2}(\dfrac{1}{2})=\dfrac{\sqrt{2}(\sqrt{3}+1)}{4}

=6+24=\dfrac{\sqrt{6}+\sqrt{2}}{4}

sin(75°)=2(3+1)4=6+24\sin(75\degree)=\dfrac{\sqrt{2}(\sqrt{3}+1)}{4}=\dfrac{\sqrt{6}+\sqrt{2}}{4}

3.2

(i)



3.2


ACB=180°60°75°=45°\angle ACB=180\degree-60\degree-75\degree=45\degree

Use the Law of Sines


sinABCAC=sinACBAB\dfrac{\sin \angle ABC}{AC}=\dfrac{\sin \angle ACB }{AB}

AB=sin45°sin75°10=22a2(3+1)410AB=\dfrac{\sin 45\degree}{\sin 75\degree}\cdot10=\dfrac{\dfrac{\sqrt{2}}{2}a}{\dfrac{\sqrt{2}(\sqrt{3}+1)}{4}}\cdot 10

=2(31)3110=10(31)=\dfrac{2(\sqrt{3}-1)}{3-1}\cdot10=10(\sqrt{3}-1)

AB=10(31)AB=10(\sqrt{3}-1)


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