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

[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 0^o<=x<=180^o

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

 

[MarkFL]

Hello, and welcome to MHB! (Wave)

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 0^o<=x<=180^o

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

We are given to solve:

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

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 $\sin(x)$:

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

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

 

[JPorkins]

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

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

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 $\sin(x)$:

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

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 !

 

[MarkFL]

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:

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

$$\sin(x)=1$$

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

 

[JPorkins]

You've done splendidly so far! (Yes)

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

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

$$\sin(x)=1$$

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

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

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

Thanks

 

[MarkFL]

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

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

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

$$\sin(x)=1$$

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:

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

What angle $x$ gives us:

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

 

[JPorkins]

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

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

What angle $x$ gives us:

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

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

 

[MarkFL]

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

Yes, from:

$$\sin(x)=1$$

and:

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

We then conclude:

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

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

 

[JPorkins]

Yes, from:

$$\sin(x)=1$$

and:

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

We then conclude:

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

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 !

 

[Greg]

Alternatively,

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

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

$$\cos2x=-\sin x$$

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

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

$$2x-90=x$$

$$x=90^\circ$$