Answer to Question #263379 in Trigonometry for Sabelo Xulu

Question

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 ,$\pi$ )

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

Answer 1

1.1

$$y=x+8$$

1.2

$$S=x^2+y^2=x^2+(x+8)^2$$

$$=x^2+x^2+16x+64$$

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

1.3

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

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

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

$$S(-4)=32$$

2.1

$$\dfrac{1}{1-\sin \theta}=\dfrac{1}{1-\sin \theta}\cdot\dfrac{1+\sin \theta}{1+\sin \theta}$$

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

$$=\dfrac{1}{\cos^2 \theta}+\dfrac{\sin \theta}{\cos^2 \theta}=\sec^2\theta+\tan\theta\cdot\sec \theta$$

$$\dfrac{1}{1-\sin \theta}=\sec^2\theta+\tan\theta\cdot\sec \theta, \cos \theta\not=0$$

2.2

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

$$\cos \theta=0=>\theta=\dfrac{\pi}{2}+\pi n, n\in \Z$$

The identity $\dfrac{1}{1-\sin \theta}=\sec^2\theta+\tan\theta\cdot\sec \theta$ is undefined for $\theta=\dfrac{\pi}{2}+\pi n, n\in \Z.$

2.3

$$2\cos 2x\cdot\cosec2x=2\cos 2x , x\in (-\pi ,\pi )$$

$\sin 2x\not=0$

Then

$$\cos 2x=0$$

Or

$$\cosec2x=0$$

The first equation

$$\cos 2x=0$$

$$2x=\dfrac{\pi}{2}+\pi n, n\in \Z$$

$$x=\dfrac{\pi}{4}+\dfrac{\pi n}{2}, n\in \Z$$

The second equation

$$\cosec2x=0$$

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

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

Therefore we take

$$x=\dfrac{\pi}{4}+\dfrac{\pi n}{2}, n\in \Z$$

Since $x\in (-\pi, \pi)$

$$x_1=\dfrac{\pi}{4}+\dfrac{\pi (-2)}{2}=-\dfrac{3\pi}{4}$$

$$x_2=\dfrac{\pi}{4}+\dfrac{\pi (-1)}{2}=-\dfrac{\pi}{4}$$

$$x_3=\dfrac{\pi}{4}+\dfrac{\pi (0)}{2}=\dfrac{\pi}{4}$$

$$x_4=\dfrac{\pi}{4}+\dfrac{\pi (1)}{2}=\dfrac{3\pi}{4}$$

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

3.1

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

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

$$=\dfrac{\sqrt{2}}{2}(\dfrac{\sqrt{3}}{2})+\dfrac{\sqrt{2}}{2}(\dfrac{1}{2})=\dfrac{\sqrt{2}(\sqrt{3}+1)}{4}$$

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

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

3.2

(i)

3.2

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

Use the Law of Sines

$$\dfrac{\sin \angle ABC}{AC}=\dfrac{\sin \angle ACB }{AB}$$

$$AB=\dfrac{\sin 45\degree}{\sin 75\degree}\cdot10=\dfrac{\dfrac{\sqrt{2}}{2}a}{\dfrac{\sqrt{2}(\sqrt{3}+1)}{4}}\cdot 10$$

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

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