Download plane trig part 5

PLANE TRIG 6232e
2) Draw the portion of the graph (y = Cos(x)) from 0° to 360°.
Solve the following trigonometric equations.
Give all positive values of the angle between 0° and 360° that will satisfy each.
Give any approximate value to the nearest minute only.
3) Sin(2  ) = √3/2
ANSWER

= 30°, 60°, 120°, 150°, 210°, 240°, 300°, 330°
√3/2 can have positive or negative values
sin
-1(+/-√3/2)
= 0°+/-60°, 180°+/-60°, 360+/-60°, 540+/-60°, 720+/-60°
Thus: (2*φ) = 60°, 120°, 240°, 300°, 420°, 480°, 600°, 660°

= 30°, 60°, 120°, 150°, 210°, 240°, 300°, 330°
4) Sin(2*x) - Cos(x) = 0
ANSWER
Sin(2*x) - Cos(x) = 0
Therefore: Sin(2*x) = Cos(x)
But for opposing angles A and B in a right angled triangle Sin (A) = Cos (B)
Sin (A) = Cos (90°-A)
That is, A + B = 90° .
Hence 2*x + x = 90°
3*x = 90°
x = 90°/3 = 30°
Sin(2*x) = Cos(x) = √3/2
One other solution x = 90°
Sin(2*x) = Cos(x) = 0
x = 30°, 90°
5) Sin(2*x - 10°) = 1/2
Answer:
sin -1 (+1/2) = 30°
2*x - 10° = 30°
2*x = 40°
x = 20°
sin -1 (+1/2) = 150°
2*x - 10° = 150°
2*x = 160°
x = 80°
x = 20°, 80°
--------------------------------------------------------6) Cos(2*x) - Sin²(x/2) + 3/4 = 0
Cos(2*x) = 2*Cos²(x) - 1
Sin²(x/2) = (1 - Cos(x)) / 2
Thus Cos(2*x) - Sin²(x/2) + 3/4 = 0
(2*Cos²(x) - 1) - ((1 - Cos(x)) / 2) + 3/4 = 0
2*Cos²(x) + (1/2)*Cos(x) - 3/4 = 0
Cos²(x) + (1/4)*Cos(x) - 3/8 = 0
This is a Quadratic Equation which can be solved using the Quadratic Formula Cos(x) = [-(1/4) +/- √((1/4)²-4*1*(-3/8))] / 2
Cos(x) = +0.5 or -0.75
cos
cos
-1
-1
(+0.5) = 60°, 300°
(-0.75) = 138°35', 221°25'
x = 60°, 138°35', 221°25', 300°
7) Sin²(2*φ) = Cos²(2*φ) + 1/2
ANSWER
φ = 30°, 60°, 120°, 150°, 210°, 240°, 300°, 330°
DERIVATION
Sin²(2*φ) = Cos²(2*φ) + 1/2
Sin²(2*φ) = (1-Sin²(2*φ)) + 1/2
2*Sin²(2*φ) = 1+ 1/2 = 3/2
Sin²(2*φ) = 3/4
Sin(2*φ) = +/-√3/2
(2*φ) = Asin(+/-√3/2) = 60°, 120°, 240°, 300°, 420°, 480°, 600°, 660°
φ = 30°, 60°, 120°, 150°, 210°, 240°, 300°, 330°
---------------------------------------------------------8) Cos(4*x) = Sin(2*x)
ANSWER
x = 15°, 135°
DERIVATION
For opposing angles A and B in a right angled triangle Sin (A) = Cos (B)
Sin (A) = Cos (90°-A)
Sin(90°-A) = Cos(A)
Therefore Cos(4*x) = Sin(2*x)
Cos(4*x) = Sin(90°-4*x)
So Sin(2*x) = Sin(90°-4*x)
2*x = 90°-4*x
6*x = 90°
x = 90°/6 = 15°
[ Cos(4*x) = Sin(2*x) = 0.5 ]
One other solution x = 135°
[ Cos(4*x) = Sin(2*x) = -1 ]
---------------------------------------------------------9) 3*Sin(φ) - 4*Cos(φ) = 2
ANSWER
φ = 76°43'
DERIVATION
A*Sin(φ) + B*Cos(φ) = √(A²+B²)*sin(φ+C)
A=3
B = -4
where C = Atan(B/A) if (A >= 0)
C = Atan(B/A)+/-180° if (A < 0)
So C = -53.13°
√(A²+B²) = √(9+16) = √25 = 5
Therefore 3*Sin(φ) - 4*Cos(φ) = 5*Sin(φ-53.13°) = 2
5*Sin(φ-53.13°) = 2
Sin(φ-53.13°) = 2/5 = 0.4
φ-53.13° = Asin(0.4) = 23.5782°
φ = 23.5782° + 53.13° = 76.7082°
φ = 76.7082°
φ = 76°43'
---------------------------------------------------------10) Tan(x+15°) = 3*Tan(x)
ANSWER
x = 8°5', 66°55', 188°5', 246°55'
DERIVATION
Tan(x+y) = (Tan(x) + Tan(y)) / (1 - Tan(x)*Tan(y))
Thus Tan(x+15°) = (Tan(x) + Tan(15°)) / (1 - Tan(x)*Tan(15°))
Tan(x+15°) = 3*Tan(x)
(Tan(x) + Tan(15°)) / (1 - Tan(x)*Tan(15°)) = 3*Tan(x)
Tan(x) + Tan(15°) = 3*Tan(x)*(1 - Tan(x)*Tan(15°))
Tan(x) + Tan(15°) = 3*Tan(x) - 3*Tan²(x)*Tan(15°)
Tan(15°) = 2*Tan(x) - 3*Tan²(x)*Tan(15°)
3*Tan(15°)*Tan²(x) - 2*Tan(x) + Tan(15°) = 0
Tan²(x) - (2/(3*Tan(15°))*Tan(x) + (1/3) = 0
Tan²(x) - 2.488034*Tan(x) + 0.3' = 0
This is a Quadratic Equation which can be solved using the Quadratic Formula Tan(x) = [2.488034 +/- √(2.488034² - 4*1*0.3')] / 2
Tan(x) = 2.345945 or 0.142089
x = Atan(2.345945) = 66°55', 246°55'
x = Atan(0.142089) = 8°5', 188°5'
---------------------------------------------------------11) Write equivalent functions in terms of inverse functions for a) x = y + Cos(φ)
b) Cos(y) = x²
ANSWER
a) φ = Acos(x-y)
b) y = Acos(x²)
DERIVATION
a) x = y + Cos(φ)
x = y + Cos(φ)
x - y = Cos(φ)
Cos(φ) = (x-y)
φ = Acos(x-y)
b) Cos(y) = x²
Cos(y) = x²
y = Acos(x²)
---------------------------------------------------------12) Give the value of a) Asin(Cos(x)) (Limit solution to an acute angle)
b) Tan(Asin(x))
c) Tan(Atan((x+1)/(x-1)) + Atan((x-1)/(x+1)))
ANSWER
a) Asin(Cos(x)) = (90°-x)
b) Tan(Asin(x)) = x / √(1 - x²)
c) Tan(90°) = ∞
DERIVATION
a) Asin(Cos(x))
Cos(x) = Sin(90°-x)
So Asin(Cos(x)) = Asin(Sin(90°-x)) = (90°-x)
b) Tan(Asin(x))
Tan(φ) = Sin(φ) / Cos(φ) = Sin(φ) / √(1 - Sin²(φ))
So Tan(Asin(x)) = Sin(Asin(x)) / √(1 - Sin²(Asin(x)))
Tan(Asin(x)) = x / √(1 - x²)
c) Tan(Atan((x+1)/(x-1)) + Atan((x-1)/(x+1)))
Atan(z) + Atan(1/z) = 90°
Let [ z = (x+1)/(x-1) ], so Tan(Atan((x+1)/(x-1)) + Atan((x-1)/(x+1)))
= Tan(Atan(z) + Atan(1/z))
= Tan(90°)
=∞
---------------------------------------------------------13) Solve: Acos(x) + Acos(2*x) = 60°
ANSWER
x = 0.5
DERIVATION
(Acos(x) = Φ)
Cos(Φ) = x
(Acos(2*x) = Θ)
Cos(Θ) = 2*x
Therefore Cos(Θ) / Cos(Φ) = (2*x) / x = 2
Cos(Θ) = 2*Cos(Φ)
Also Acos(x) + Acos(2*x) = 60°
(Φ + Θ = 60°)
(Θ = 60° - Φ)
So Cos(Θ) = 2*Cos(Φ)
Cos(60° - Φ) = 2*Cos(Φ)
Now, [ Cos(A-B) = Cos(A)*Cos(B) + Sin(A)*Sin(B) ]
So Cos(60° - Φ) = Cos(60°)*Cos(Φ) + Sin(60°)*Sin(Φ)
Cos(60° - Φ) = (1/2)*Cos(Φ) + (√3/2)*Sin(Φ)
And since Cos(60° - Φ) = 2*Cos(Φ)
Then 2*Cos(Φ) = (1/2)*Cos(Φ) + (√3/2)*Sin(Φ)
(3/2)*Cos(Φ) = (√3/2)*Sin(Φ)
3*Cos(Φ) = √3*Sin(Φ)
3/√3 = Sin(Φ) / Cos(Φ)
Tan(Φ) = √3
Φ = Atan(√3) = 60°
Θ = 60° - Φ = 60° - 60° = 0°
Cos(Φ) = x = Cos(60°) = 0.5
Cos(Θ) = 2*x = Cos(0°) = 1.0
x = 0.5
---------------------------------------------------------14) Verify the identity: Tan(2*Atan(x)) = 2*Tan(Atan(x) + Atan(x^3))
ANSWER
Refer to the derivation below.
DERIVATION
Tan(2*Atan(x)) = 2*Tan(Atan(x) + Atan(x^3)) --- Eq 00
Separate the two sides of Eq 00 LHS = Tan(2*Atan(x))
RHS = 2*Tan(Atan(x) + Atan(x^3))
Let [ x = Tan(Φ) ]
So [ Atan(x) = Φ ]
LHS = Tan(2*Atan(x))
LHS = Tan(2*Φ)
RHS = 2*Tan(Atan(x) + Atan(x^3))
RHS = 2*Tan(Φ + Atan(Tan^3(Φ)))
Let [ Atan(Tan^3(Φ)) = ß ]
RHS = 2*Tan(Φ + Atan(Tan^3(Φ)))
RHS = 2*Tan(Φ + ß)
Now show that the LHS = RHS for Eq 00 We know that [ Tan(2*Φ) = 2*Tan(Φ) / (1 - Tan²(Φ)) ], so LHS = 2*Tan(Φ) / (1 - Tan²(Φ))
RHS = 2*Tan(Φ + ß)
Divide both sides by 2 LHS = Tan(Φ) / (1 - Tan²(Φ))
RHS = Tan(Φ + ß)
Now, we know that [ Tan(Φ+ß) = (Tan(Φ) + Tan(ß)) / (1 - Tan(Φ)*Tan(ß)) ], so LHS = Tan(Φ) / (1 - Tan²(Φ))
RHS = (Tan(Φ) + Tan(ß)) / (1 - Tan(Φ)*Tan(ß))
Multiply both sides by (1 - Tan²(Φ)) LHS = Tan(Φ)
RHS = (Tan(Φ) + Tan(ß))*(1 - Tan²(Φ)) / (1 - Tan(Φ)*Tan(ß))
Multiply both sides by (1 - Tan(Φ)*Tan(ß)) LHS = Tan(Φ)*(1 - Tan(Φ)*Tan(ß))
RHS = (Tan(Φ) + Tan(ß))*(1 - Tan²(Φ))
Expand terms for both sides LHS = Tan(Φ) - Tan²(Φ)*Tan(ß)
RHS = Tan(Φ) - Tan^3(Φ) + Tan(ß) - Tan²(Φ)*Tan(ß)
Subtract (Tan(Φ) - Tan²(Φ)*Tan(ß)) from both sides LHS = 0
RHS = -Tan^3(Φ) + Tan(ß)
Add (Tan^3(Φ)) to both sides LHS = Tan^3(Φ)
RHS = Tan(ß)
Substitute the definition for ß into the RHS, ie, [ Atan(Tan^3(Φ)) = ß ] LHS = Tan^3(Φ)
RHS = Tan(ß) = Tan(Atan(Tan^3(Φ))) = Tan^3(Φ)
Therefore we have shown that LHS = RHS for Eq 00.