# Trig Identity Problem: Solve cos2(x) + sin(x) = sin2(x) for 0^0<=x<=180^0

• MHB
• JPorkins
In summary, we are given to solve the equation cos2(x) + sin(x) = sin2(x) for 0o<=x<=180o. Using the trigonometric identity cos^2(x) + sin^2(x) = 1, we can substitute cos^2(x) for 1-sin^2(x). This leaves us with the equation 0=-2sin^2(x)+sin(x)+1, which can be factored using the quadratic formula. The resulting solutions are sin(x)=-1/2 and sin(x)=1. However, since the given restriction on x is 0o<=x<=180o, we must discard the negative solution and only consider the positive solution. From
JPorkins
Hello,
My teacher gave me some trig identity homework and it has completely stumped me .
Would be really grateful for some help, thanks!

The question is;
Solve the equation cos2(x) + sin(x) = sin2(x) for 0o<=x<=180o

I wasn't sure how to enter the degree symbol so i added ^0.

Hello, and welcome to MHB! (Wave)

JPorkins said:
Hello,
My teacher gave me some trig identity homework and it has completely stumped me .
Would be really grateful for some help, thanks!

The question is;
Solve the equation cos2(x) + sin(x) = sin2(x) for 0o<=x<=180o

I wasn't sure how to enter the degree symbol so i added ^0.

We are given to solve:

$$\displaystyle \cos^2(x)+\sin(x)=\sin^2(x)$$ where $$\displaystyle 0^{\circ}\le x\le180^{\circ}$$

Okay, the first thing I notice is that two of the terms are in terms of $\sin(x)$, and the other is part of a trig. identity (Pythagorean) involving $\sin(x)$:

$$\displaystyle \cos^2(x)+\sin^2(x)=1\implies\cos^2(x)=1-\sin^2(x)$$

So, what do you get when you substitute for $\cos^2(x)$ into the given equation using the identity above?

MarkFL said:
Hello, and welcome to MHB! (Wave)
We are given to solve:

$$\displaystyle \cos^2(x)+\sin(x)=\sin^2(x)$$ where $$\displaystyle 0^{\circ}\le x\le180^{\circ}$$

Okay, the first thing I notice is that two of the terms are in terms of $\sin(x)$, and the other is part of a trig. identity (Pythagorean) involving $\sin(x)$:

$$\displaystyle \cos^2(x)+\sin^2(x)=1\implies\cos^2(x)=1-\sin^2(x)$$

So, what do you get when you substitute for $\cos^2(x)$ into the given equation using the identity above?

Using the equation cos2 + sin2 = 1,
i think the cos2(x) can be substituted for ( 1-sin2(x) )
which changes the equation to 0=-2sin2(x) + sin(x) +1
from there i factorised using the quadratic formulea (A=(-2), B=1, C=1)
this left me with sin(x) = -1/2 or 1

I'm not confident on my maths here though

After this i know the roots to the quadtraic need to be entered into the equation to find the angles but I'm not sure on this part either.
Many thanks !

JPorkins said:
Using the equation cos2 + sin2 = 1,
i think the cos2(x) can be substituted for ( 1-sin2(x) )
which changes the equation to 0=-2sin2(x) + sin(x) +1
from there i factorised using the quadratic formulea (A=(-2), B=1, C=1)
this left me with sin(x) = -1/2 or 1

I'm not confident on my maths here though

After this i know the roots to the quadtraic need to be entered into the equation to find the angles but I'm not sure on this part either.
Many thanks !

You've done splendidly so far! (Yes)

So, from the quadratic in $\sin(x)$ that resulted, we obtain:

$$\displaystyle \sin(x)=-\frac{1}{2}$$

$$\displaystyle \sin(x)=1$$

Now, for the given restriction on $x$, do we need to discard any of these solutions?

MarkFL said:
You've done splendidly so far! (Yes)

So, from the quadratic in $\sin(x)$ that resulted, we obtain:

$$\displaystyle \sin(x)=-\frac{1}{2}$$

$$\displaystyle \sin(x)=1$$

Now, for the given restriction on $x$, do we need to discard any of these solutions?

The $$\displaystyle \sin(x)=-\frac{1}{2}$$ is a negative so will be out of the 0<=x<=180 range ?
The $$\displaystyle \sin(x)=1$$ will stay however ?

Then the sin(x) =1 can be entered into the cos2(x) + sin(x) = sin2(x) equation afterwards ?

Thanks

JPorkins said:
The $$\displaystyle \sin(x)=-\frac{1}{2}$$ is a negative so will be out of the 0<=x<=180 range ?
The $$\displaystyle \sin(x)=1$$ will stay however ?

Yes, as $x$ varies from $0^{\circ}$ to $180^{\circ}$, $\sin(x)$ varies from zero, all the way up to one, and then back down to zero, and so we may only consider:

$$\displaystyle 0\le\sin(x)\le1$$

And the only root of our quadratic in that range is:

$$\displaystyle \sin(x)=1$$

JPorkins said:
Then the sin(x) =1 can be entered into the cos2(x) + sin(x) = sin2(x) equation afterwards ?

Thanks

No, what we want to do is consider for what value of $x$ do we have:

$$\displaystyle \sin(x)=1$$ ?

What angle $x$ gives us:

$$\displaystyle \sin(x)=1$$ ?

MarkFL said:
No, what we want to do is consider for what value of $x$ do we have:

$$\displaystyle \sin(x)=1$$ ?

What angle $x$ gives us:

$$\displaystyle \sin(x)=1$$ ?

At 90 degrees sin(x) = 1 ? Whether this is correct relevant to the question I'm not sure though !

JPorkins said:
At 90 degrees sin(x) = 1 ? Whether this is correct relevant to the question I'm not sure though !

Yes, from:

$$\displaystyle \sin(x)=1$$

and:

$$\displaystyle 0^{\circ}\le x\le180^{\circ}$$

We then conclude:

$$\displaystyle x=90^{\circ}$$

is the only solution to the given equation, subject to the constraint on the domain. :D

MarkFL said:
Yes, from:

$$\displaystyle \sin(x)=1$$

and:

$$\displaystyle 0^{\circ}\le x\le180^{\circ}$$

We then conclude:

$$\displaystyle x=90^{\circ}$$

is the only solution to the given equation, subject to the constraint on the domain. :D

Wow, you made it very simple ! I wish my teacher was as good as you aha.
Thank you very much !

Alternatively,

$$\cos^2x+\sin x=\sin^2x$$

$$\cos^2x-\sin^2x=-\sin x$$

$$\cos2x=-\sin x$$

$$\sin(90-2x)=-\sin x$$

$$\sin(2x-90)=\sin x$$

$$2x-90=x$$

$$x=90^\circ$$

## 1. What is a trig identity problem?

A trig identity problem is a mathematical equation involving trigonometric functions that needs to be simplified or solved. It typically involves using various trigonometric identities and properties to manipulate the equation and arrive at a simpler or more useful form.

## 2. Why are trig identities important?

Trig identities are important because they allow us to simplify complex trigonometric expressions and equations, making them easier to work with and solve. They also help us establish relationships between different trigonometric functions, making it easier to understand their behavior and apply them in various real-world situations.

## 3. What are some common trig identities?

Some common trig identities include the Pythagorean identities (used to express sine, cosine, and tangent in terms of each other), the double angle identities, the sum and difference identities, and the half angle identities. These are just a few examples, as there are many more identities that can be used to simplify trigonometric expressions.

## 4. How do I solve a trig identity problem?

To solve a trig identity problem, you will need to use various trigonometric identities and properties to manipulate the equation and arrive at a simpler form. This may involve using algebraic techniques, such as factoring and expanding, as well as substitution and rearrangement. It is important to be familiar with a variety of identities and to carefully work through each step to ensure accuracy.

## 5. What are some tips for solving trig identity problems?

Some tips for solving trig identity problems include:

• Start by simplifying one side of the equation using known identities.
• Look for common factors or patterns that can help simplify the equation.
• Be familiar with the different forms of trigonometric functions (e.g. sin^2x, 1-cos^2x, etc.) and know how to manipulate them using identities.
• Use substitution to simplify expressions involving multiple trigonometric functions.
• Check your work by substituting your simplified expression back into the original equation to ensure they are equivalent.

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