MHF 4U Advanced Functions: UNIT ASSIGNMENT TRIGONOMETRY. Complete Q&A.

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UNIT ASSIGNMENT TRIGONOMETRY Question 1 3 sin ( x )= √ 2

a). Find all solutions for b). If 0≤ y ≤

π 2

sin ⁡( x− y)

sin ( x )=

1 3

and

sec ( y )=

5 4

, where

0≤x ≤

π 2

and

, evaluate the expression .

Answers a).

π x= +2 πn 3

,

2π + 2 πn 3

b). We know that

, where sin θ=

opp hyp

n∈Z

, and since we know

1

that sin ( x )= 3 , we can draw a triangle using that information, and we can find the unknown side using the Pythagorean theorem.

Let the unknown side be

x

.

x 2+12=32 2

x =9−1

x 2=8 x=√ 8 x=2 √ 2

Now that we know the length of the adjacent side, we can get the value of adj since cos θ= hyp , we know that cos ( x )= 2 3√2 .

Since

sec ( y )=

5 4

, and since

cos θ=

1 sec θ

cos ( x)

, we know

4

that cos ( y )= 5 . Using the same steps as before, we can find the value of sin( y ) , by finding the length of the opposite side.

Let the opposite side be x . Using the Pythagorean theorem, we get: x 2+ 42 =52 2

x =25−16

x 2=9 x=3

,

Now that we know the length of the opposite side, we can derive the value of sin ⁡( y) , which 3 is 5 .

Now let’s evaluate sin(x− y) . Using the angle difference formula, we get: sin ( x− y )=sin ( x ) cos ( y )−cos ( x ) sin ( y )

Substituting all the known values into the equation, we get: sin ( x− y )= ¿

4 2√2 − 15 5

¿

4−6 √ 2 15

( 13 × 45 )−( 23√2 × 35 )

Question 2 Solve for all values of intervals. a).

2 cos ( x )+ sin (2 x )=0

for

b).

2 sin2 ( x )=1

for

x∈R

c).

tan 2 ( x )−3=0

for

x∈R

x

0 ≤ x ≤2 π

in the given

Answers a).

Combining all the solutions, we get b).

π 3π x= + πn , + πn 4 4

where

n∈Z

.

π 3π x= , 2 2

.

c).

π 2π x= + πn , +πn 3 3

where

n∈Z

.

Question 3 Prove the following identities:

( π2 )=−sin ( x)

a).

cos x+

b).

( sin ( x )+ cos ( x ) )2=1+sin ( 2 x )

c).

2 csc ( 2 x )=sec ( x ) csc ( x )

d).

sin ( x + y ) tan ( x ) +tan ( y ) = sin ⁡( x− y) tan ( x )−tan ( y )

e).

sec (−t ) =−csc (t ) tan (−t )

Answers a).

b).

c).

d).

e).

Question 4 Describe how to use both an equivalent trigonometric identity and a diagram to demonstrate that two trigonometric ratios are equivalent. a). Use one of the following equivalent trigonometric expressions:

( 32π )=−cos ( θ )

sin θ+

( 32π )=sin ( θ)

cos θ+

( 32π )=−cot ( θ )

tan θ+ sin

( 32π −θ)=−cos (θ )

cos

( 32π −θ)=−sin (θ )

tan

( 32π −θ)=cot ( θ )

b). Using a diagram, demonstrate how the related angle formulas are true. Create an example to illustrate your findings in part a). (choose a value for θ and solve both sides to prove that they are equal.)

Answers

( 32π )=−cos ( θ)

i).

sin θ+

a). Let a be the the equation.

LHS

and

b

be the

RHS

of

a=¿

( 32π )=¿

sin θ+

sin ( θ ) cos

( 32π )+cos ( θ ) sin ( 32π )=¿

sin ( θ ) × ( 0 )+ cos ( θ ) × (−1 )=¿

−cos ( θ )=b a=b

∴ LHS=RHS

b). Let us look at the graphs of either side of the equation.

LHS

( 32π )

sin θ+

This is what the graph of the base function, sin ( θ ) , looks like:

And this is what the graph of the function 3π looks like after 2 has been added to the value:

As you can see the graph of the base function has been shifted to the left, by a 3π value of 2 units.

x

RHS −cos ( θ )

This is what the graph of the base function, cos ( θ ) , looks like:

And this is what the graph of like:

−cos ( θ )

looks

As you can see, the graph has been reflected over the x -axis.

Now let us compare the graphs of both sides of the original equation.

When drawn on the same graph, you can see that they overlap. This proves that the two sides of the equation are indeed, equal. ii).

( 32π )=sin ⁡(θ)

cos θ+

a). Let a be the the equation. a=¿

( 32π )=¿

cos θ+

cos ( θ ) cos

( 32π )−sin ( θ) sin ( 32π )=¿

cos (θ)× ( 0 )−sin ( θ ) × (−1 )=¿ −sin ( θ ) × (−1 )=¿

LHS

and

b

be the

RHS

of

sin ( θ )=b

a=b ∴ LHS=RHS

b). Let us look at the graphs of either side of the equation.

LHS This is what the graph of the base function, cos ( θ ) , looks like:

And this is what the graph of like:

( 32π ) looks

cos θ+

As you can see, the graph of the base function has been shifted to the left by a 3π value of 2 units.

RHS This is what the graph of the RHS of the equation, sin ⁡(θ) , looks like:

Let us compare the graphs of both sides of the equation.

When drawn on the same graph, we can see that they overlap. This means that both sides of the equation are equal. iii).

( 32π )=−cot ( θ )

tan θ+

a). Let a be the the equation.

LHS

and

b

be the

RHS

of

a=¿

( 32π )=¿

tan θ+

( 32π ) 3π cos (θ+ 2 ) sin θ+

Looking back at the previous two examples, we can substitute the values,

( 32π ) =¿ 3π cos (θ+ 2 ) sin θ+

−cos ( θ ) =¿ sin ( θ ) −cot ( θ )=b

a=b ∴ LHS=RHS

b). Let us look at the graphs of either side of the equation.

LHS

( 32π )

tan θ+

This is what the graph of the base function, tan ( θ ) , looks like:

And this is what the graph of like:

( 32π ) looks

tan θ+

As you can see, the graph of the base function, has been shifted to the left by a 3π value of 2 .

RHS This is what the graph of the base function, cot ( θ ) looks like:

And this is what the graph of like:

−cot ⁡( θ)

As you can see, the base function, been reflected over the x -axis.

looks

cot ⁡( θ)

, has

Now let’s compare the graphs of both sides of the equation.

When drawn on the same graph, we can see that they overlap. This proves that both sides of the equation are equal. iv).

sin

( 32π −θ)=−cos ⁡(θ)

a). Let a be the the equation. a=¿

sin

( 32π −θ )=¿

sin

( 32π ) cos ( θ)−cos( 32π ) sin (θ )=¿

(−1 ) × cos ( θ )−( 0 ) sin ( θ )=¿ −cos ( θ )=b

a=b

LHS

and

b

be the

RHS

of

∴ LHS=RHS

b). Let us look at graphs of either side of the equation.

LHS This is what the graph of the base function, sin ⁡( θ) , looks like:

And this is what the graph of like:

sin

( 32π −θ ) looks

As you can see, the graph has been reflected over the x -axis, and it has been 3π translated left by a value of 2 units.

RHS This is what the base function, like:

And this is what the graph of like:

cos ⁡( θ)

−cos ( θ )

, looks

looks

Let us compare the graphs of both sides of the original equation.

When drawn on the same graph, we can see that they overlap. This means both sides of the equation are equal. v).

cos

( 32π −θ)=−sin (θ )

a). Let a be the the equation. a=¿ cos

( 32π −θ)=¿

cos

( 32π ) cos ( θ )+ sin ( 32π ) sin ( θ ) =¿

( 0 ) cos ( θ )+ (−1 ) sin ( θ ) =¿ −sin ( θ )=b

LHS

and

b

be the

RHS

of

a=b

∴ LHS=RHS

b). Let us compare the graphs of both sides of the equation.

LHS This is what the base function, like:

cos ⁡( θ)

, looks

And this is what the graph of like:

cos

( 32π −θ) looks

As you can see, the graph has been reflected over the x -axis and has been 3π shifted left by a value of 2 units.

RHS This is what the graph of the base function, sin ( θ ) , looks like:

This is what the graph of

−sin ( θ )

looks like:

Now let’s compare the graphs of both sides of the original equation:

When drawn on the same graph, you can see that they overlap. This means that both sides of the equation are equal.

vi).

tan

( 32π −θ)=cot ( θ )

a). Let a be the the equation.

LHS

and

b

be the

RHS

of

a=¿

tan

( 32π −θ)=¿

( 32π −θ) 3π cos ( −θ) 2 sin

Looking back at the previous two examples, we can substitute the values.

( 32π −θ) =¿ 3π cos ( −θ) 2 sin

−cos ( θ ) =¿ −sin ( θ ) 1 =¿ tan ( θ ) cot ( θ )=b

a=b ∴ LHS=RHS

b). Let’s compare the graphs of both sides of the equation.

LHS

This is what the graph of the base function, tan ( θ ) , looks like:

And this is what the graph of like:

tan

( 32π −θ) looks

We can see that the graph has been reflected over the x -axis and it has been 3π shifted to the left by a value of 2 .

RHS This is what the graph of

cot ( θ )

looks like:

Now let us compare the graphs of both sides of the equation.

When drawn on the same graph, we can see that they overlap. This means that both sides of the equation are equal.