Complete 1 page APA formatted article: Pre-Calculus. Pre-Calculus Opposite (5) hypotenuse 89) Adjacent (8) X = Arctan =(opposite/adjacent) = 5/8, but cotx = 1/tanxSo cot(actan) =8/5 X = arcsin (opposi

Complete 1 page APA formatted article: Pre-Calculus. Pre-Calculus Opposite (5) hypotenuse 89) Adjacent (8) X = Arctan =(opposite/adjacent) = 5/8, but cotx = 1/tanx

So cot(actan) =8/5

X = arcsin (opposite/hypotenuse) =(-√2/2), but secx = 1/sinx, hence sec(arcsin -√2/2) = -2/√2

2.

The value of the following without using calculator,

a. tan-1(-√3/3).

Let x = tan-1(-√3/3) = tan-1(-√3/3)x√3, hence tanx =-1/√3-3, tan2x=1/3√3, but 1+tan2x=sec2x, thus sec2x=1+1/3 =4/3, secx=1/cosx=2/(√3-3√3), but cos√3+ or -√3-3/2, which can be written as cosx = (√3-3)/2√3, hence x=π/6. From the quadrant, cosine is negative only on 2nd and 3rd , so the solution for will be in quadrants where tan is negative 2rd and 4th, so solutions are 11 π/6 and 5 π/6.

b. cos-1(1), from the quadrant the value of cosine ranges from 0 to 4 π. For the inverse of cosine of negative one the value = π+2 πk where k is an integer. Hence the value from the quadrant is π or 180 degrees.

3.

Prove of secx – cosx = sinxtanx, starting from the left side, secx =1/cosx, so the equation becomes, 1/cosx-cosx. Which gives, (1- cos²x)/cosx. But 1- cos²x=sin²x, making the above equation to become, sin²x/cosx. But also sinx/cosx=tanx, so sinx.sinx/cosx =tanx.sinx, hence the identity is verified from the right hand side, the same can be done from left hand side.

4.

Solutions of the following equations

a. 2sin(x)- √3 = 0, 2sin(x)= √3, sin(x)= √3/2

b. 2Cos² (2x)=1,

Cos²(2x)=1/2

Cos(2x)=1/√2 or -1/√2, divide each side by 2

Cos(2x)= √2/2 or -√2/2

5.

If tan (u) = 4/3 and sin (v)= -12/13, (u is in Q1 and v is in Q3), in Q1 both the tan and sine are positive while in the Q3 sine is negative and tangent is positive.

(a). cos(u +v). = cos(u).cos(v)-sin(u).sin(v)

b. tan(u +v)

6.

The law of sine or the cosines to solve the following

(a) a=16.4, b =8.8, c =12.2

Area =√s(s-a)(s-b)(s-c), where s=(a+b+c)/2, hence s= 37.4/2=18.7. hence the r=area is A=√18.7(18.7-16.4)(18.7-8.8)(18..7-12.2)= √18.7(2.3)(9.9)(6.5) = 52.6

(b) using cosine rule a=b+c-2bccosA

Using the sine rule, a =11.4, b=12.8, A=58

Area =1/2a.bsinc = 1/2.x 11.4x12.8. Sin 58 = 61.87 square units

7.

Cos 32 = y/x= , 0.848 = y/x. y= 0.848x

Tan 20 =x /(y+30). 0.364=x/(y+30): x =0.364(y+30)

Hence x = 0.364(0.848x+30) = 0.30867x + 10.92

0.69133x=10.92 thus, x 15.80 metres.

8.

To approximate the length of a pond, surveyor walks 95 metres, using sine, sin135=x/95

X= 67.17 metres

Work cited

Larson, Ron and Hostetler, Robert. Precalculus: A Conncise Course: Text . Stamford: Cengage Learning, 2007.