# Answer in Trigonometry for Sabelo Xulu #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-sin*θ*) =sec^{2}*θ+tanθ.secθ*

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

*2.3. Solve the equation 2cos 2x.cosec*^{2}*x=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

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}cdotdfrac{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^2theta+tanthetacdotsec theta”

“dfrac{1}{1-sin theta}=sec^2theta+tanthetacdotsec theta, cos thetanot=0”

2.2

If “cos theta=0,” then the identity “dfrac{1}{1-sin theta}=sec^2theta+tanthetacdotsec theta” is undefined.

“cos theta=0=>theta=dfrac{pi}{2}+pi n, nin Z”

The identity “dfrac{1}{1-sin theta}=sec^2theta+tanthetacdotsec theta” is undefined for “theta=dfrac{pi}{2}+pi n, nin Z.”

2.3

“2cos 2xcdotcosec2x=2cos 2x , xin (-pi ,pi )”

“sin 2xnot=0”

Then

“cos 2x=0”

Or

“cosec2x=0”

The first equation

“cos 2x=0”

“2x=dfrac{pi}{2}+pi n, nin Z”

“x=dfrac{pi}{4}+dfrac{pi n}{2}, nin 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}, nin Z”

Since “xin (-pi, pi)”

“x_1=dfrac{pi}{4}+dfrac{pi (-2)}{2}=-dfrac{3pi}{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{3pi}{4}”

“bigg{-dfrac{3pi}{4}, -dfrac{pi}{4}, dfrac{pi}{4}, dfrac{3pi}{4}bigg}”

3.1

“sin(75degree)=sin(45degree+30degree)”

“=sin(45degree)cos(30degree)+cos(45degree)sin(30degree)”

“=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(75degree)=dfrac{sqrt{2}(sqrt{3}+1)}{4}=dfrac{sqrt{6}+sqrt{2}}{4}”

3.2

(i)

3.2

“angle ACB=180degree-60degree-75degree=45degree”

Use the Law of Sines

“dfrac{sin angle ABC}{AC}=dfrac{sin angle ACB }{AB}”

“AB=dfrac{sin 45degree}{sin 75degree}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)”