# Solve the given problem that involves Trigonometry

• chwala
OK. That does appear to work. In my opinion, unless you give that sort of explanation, it's next to impossible to see that your solution is correct.I think that the more straight forward way to get the result would be to let ##A=\theta## and ##B=30^\circ##, then proceed:##\displaystyle \cos(A+B) \cos(A-B)=\left\{ \cos A \cos B - \sin A \sin B \right \} \left\{ \cos A \cos B + \sin A \sin B \right \}## ##\displaystyle = \cos^2 A \cos^2 B - \sin^2 A \sin
chwala
Gold Member
Homework Statement
See attached.
Relevant Equations
Trigonometry

For part (a),

We know that ##\cos (-θ)=\cos (θ)## and ##\sin (-θ)=-\sin (θ)##

##\cos (A-B)=\cos A\cos (-B) -\sin A\sin(-B)##

##\cos (A-B)=\cos A\cos (B) +\sin A\sin(B)##

##\cos (A-B)=\cos A\cos B+\sin A\sin B##

For part (b)

...

##f(θ)=\cos 60^0- \sin (θ+30^0)\sin (θ-30^0)##
##f(θ)=\cos 2θ+ \sin (θ+30^0)\sin (θ-30^0)##

...
##2f(θ)= \cos 60^0 + cos 2θ##

##2f(θ)=\dfrac{1}{2}+ (2\cos^2 θ -1)##

##f(θ)=\cos^2 θ - \dfrac{1}{4}##For part (c) i,

The maximum value of ##f(θ)## is given by,

##f(180^0)=\left[1-\dfrac{1}{4}\right]=\dfrac{3}{4}## at smallest positive value of ##θ=180^0##

The minimum value of ##f(θ)## is given by,

##f(90^0)=\left[0-\dfrac{1}{4}\right]=-\dfrac{1}{4}## at smallest positive value of ##θ=90^0##

Bingo!

Any insight/alternative approach is highly welcome. Cheers guys.

Last edited:
What you have for part (b) is so wrong that it's difficult to make any guess as to what you were trying to do.

Clearly it can't be that both of the following statements are true:

chwala said:
For part (b)

##f(θ)=\cos 60^0- \sin (θ+30^0)\sin (θ-30^0)##
##f(θ)=\cos 60^0+ \sin (θ+30^0)\sin (θ-30^0)##

...

##2f(θ)=2\cos 60^\circ##

Last edited:
SammyS said:
What you have for part (b) is so wrong that it's difficult to make any guess as to what you were trying to do.

Clearly it can't be that both of the following statements are true:Furthermore, on addition they give:

##2f(θ)=2\cos 60^\circ##
I just amended.

...small typo boss! That should be clear to you now.

chwala said:
I just amended.

...small typo boss! That should be clear to you now.
That does give some small hint as to what you were trying to do. The only thing clear about this is that the result does not follow from what is given.

Suggestion:

Use part (a) to get a general result for ##\cos(A+B) \cos(A-B)##. You get a difference times a sum, which factors to a difference of squares. Use that to get ##f(\theta)## plugging in ##\theta## for ##A## and ##30^\circ## for ##B## .

SammyS said:
That does give some small hint as to what you were trying to do. The only thing clear about this is that the result does not follow from what is given.

Suggestion:

Use part (a) to get a general result for ##\cos(A+B) \cos(A-B)##. You get a difference times a sum, which factors to a difference of squares. Use that to get ##f(\theta)## plugging in ##\theta## for ##A## and ##30^\circ## for ##B## .
...this should be simple sir, I just made use of part (a);
Specifically, ##\cos(A+B)## and ##\cos (A-B)##...that gave me the two simultaneous equations.

chwala said:
...this should be simple sir, I just made use of part (a);
Specifically, ##\cos(A+B)## and ##\cos (A-B)##...that gave me the two simultaneous equations.
That may be what you intended to do, but you totally messed it up.

SammyS said:
That may be what you intended to do, but you totally messed it up.
fine, i hope you're having a great day...let me just repeat what i did.Let

##(θ+30^0)=A##

and

let ##(θ-30^0)=B##

Then,

##\cos(A+B)=f(θ)-\sin A\sin B##
##\cos (A-B)=f(θ)+\sin A\sin B##

##\cos 2θ=f(θ)-\sin A\sin B##
##\cos 60^0=f(θ)+\sin A\sin B##

...

cheers boss!

Last edited:
chwala said:
fine, i hope you're having a great day...let me just repeat what i did.Let

##(θ+30^0)=A##

and

let ##(θ-30^0)=B##

Then,

##\cos(A+B)=f(θ)-\sin A\sin B##
##\cos (A-B)=f(θ)+\sin A\sin B##

##\cos 2θ=f(θ)-\sin A\sin B##
##\cos 60^0=f(θ)+\sin A\sin B##

...

cheers boss!
OK. That does appear to work. In my opinion, unless you give that sort of explanation, it's next to impossible to see that your solution is correct.

I think that the more straight forward way to get the result would be to let ##A=\theta## and ##B=30^\circ##, then proceed:

##\displaystyle \cos(A+B) \cos(A-B)=\left\{ \cos A \cos B - \sin A \sin B \right \} \left\{ \cos A \cos B + \sin A \sin B \right \}##

  ##\displaystyle = \cos^2 A \cos^2 B - \sin^2 A \sin^2 B ##

  ##\displaystyle = \cos^2 \theta \cos^2 30^\circ - \sin^2 \theta \sin^2 30^\circ ##

etc.

chwala

## What are the basic trigonometric functions and their relationships?

The basic trigonometric functions are sine (sin), cosine (cos), and tangent (tan). They are defined based on a right-angled triangle as follows: sin(θ) = opposite/hypotenuse, cos(θ) = adjacent/hypotenuse, and tan(θ) = opposite/adjacent. Additionally, the relationships between these functions include: sin²(θ) + cos²(θ) = 1, tan(θ) = sin(θ)/cos(θ), and 1 + tan²(θ) = sec²(θ).

## How do you solve a right triangle using trigonometry?

To solve a right triangle, you need to find the lengths of all sides and the measures of all angles. Given one side length and one non-right angle, you can use the trigonometric functions sine, cosine, and tangent to find the other side lengths. For example, if you know angle θ and the hypotenuse, you can find the opposite side using sin(θ) = opposite/hypotenuse and the adjacent side using cos(θ) = adjacent/hypotenuse.

## What is the Pythagorean Theorem and how is it used in trigonometry?

The Pythagorean Theorem states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. Mathematically, it is expressed as a² + b² = c², where c is the hypotenuse. This theorem is used in trigonometry to relate the sides of a right triangle, especially when solving for unknown side lengths.

## How can you use trigonometric identities to simplify expressions?

Trigonometric identities are equations involving trigonometric functions that are true for all values of the variables involved. Some common identities include the Pythagorean identities (e.g., sin²(θ) + cos²(θ) = 1), angle sum and difference identities (e.g., sin(α ± β) = sin(α)cos(β) ± cos(α)sin(β)), and double-angle identities (e.g., sin(2θ) = 2sin(θ)cos(θ)). These identities can be used to simplify complex trigonometric expressions and solve equations.

## How do you solve trigonometric equations?

To solve trigonometric equations, you first isolate the trigonometric function. Next, you use inverse trigonometric functions to solve for the angle. Finally, you consider the periodic nature of trigonometric functions to find all possible solutions within the given interval.

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