# Question 7267c

Feb 13, 2018

See below

#### Explanation:

We'll be applying one key trigonometric identity to solve this problem, which is:

${\sin}^{2} \left(\theta\right) + {\cos}^{2} \left(\theta\right) = 1$

Firstly, we want to turn the ${\sin}^{2} \left(x\right)$ into something with cosines. Rearranging the above identity gives:

${\cos}^{2} \left(\theta\right) = 1 - {\sin}^{2} \left(\theta\right)$

We plug this in:

${\sin}^{2} \left(\theta\right) + \sin \left(\theta\right) = 1$

$\implies 1 - {\cos}^{2} \left(\theta\right) + \sin \left(\theta\right) = 1$

Also, note that the ones on both sides of the equation will cancel:

$\implies \sin \left(\theta\right) - {\cos}^{2} \left(\theta\right) = 0$

Secondly, we want to turn the remaining $\sin \left(x\right)$ term into something with cosines in it. This is slightly messier, but we can use our identity for this too.

$\sin \left(\theta\right) = \sqrt{1 - {\cos}^{2} \left(\theta\right)}$

We can now plug this in:

$\implies \sqrt{1 - {\cos}^{2} \left(\theta\right)} - {\cos}^{2} \left(\theta\right) = 0$

Lastly, we move the ${\cos}^{2} \left(x\right)$ to the other side of the equation, and square everything to remove the square root:

$\implies \sqrt{1 - {\cos}^{2} \left(\theta\right)} = {\cos}^{2} \left(\theta\right)$

$\implies 1 - {\cos}^{2} \left(\theta\right) = {\cos}^{4} \left(\theta\right)$

Now, we add ${\cos}^{2} \left(\theta\right)$ to both sides:

$\implies {\cos}^{4} \left(\theta\right) + {\cos}^{2} \left(\theta\right) = 1$

And there you have it. Note that you could have done this very differently, but as long as you end up at the same answer without doing incorrect math, you should be good.

Hope that helped :)

Feb 13, 2018

See the explanation

#### Explanation:

${\sin}^{2} \left(\theta\right) + \sin \left(\theta\right) = 1$

$\sin \left(\theta\right) = 1 - {\sin}^{2} \left(\theta\right)$ ---$\textcolor{red}{\left(1\right)}$

We know , $\textcolor{g r e e n}{{\sin}^{2} \left(\theta\right) + {\cos}^{2} \left(\theta\right) = 1}$

Or $\textcolor{g r e e n}{{\cos}^{2} \left(\theta\right) = 1 - {\sin}^{2} \left(\theta\right)}$

Use this value in equation $\textcolor{red}{\left(1\right)}$

We get , $\sin \left(\theta\right) = {\cos}^{2} \left(\theta\right)$

Squaring both sides

$\textcolor{b l u e}{{\sin}^{2} \left(\theta\right) = {\cos}^{4} \left(\theta\right)}$ ---$\textcolor{red}{\left(2\right)}$

${\cos}^{2} \left(\theta\right) + {\cos}^{4} \left(\theta\right)$

Use the value of $\textcolor{red}{\left(2\right)}$

$\to {\cos}^{2} \left(\theta\right) + {\sin}^{2} \left(\theta\right)$

Now use the identity in green color.

We get , ${\cos}^{2} \left(\theta\right) + {\sin}^{2} \left(\theta\right) = 1$

Hence proved.

Feb 13, 2018

see below

#### Explanation:

we have,

${\sin}^{2} \theta$ +$\sin \theta$=1-----$\textcolor{red}{1}$

Expressing ${\sin}^{2} \theta$ as 1- ${\cos}^{2} \theta$,

We have,
$\cancel{1}$- ${\cos}^{2} \theta$ + $\sin \theta$= $\cancel{1}$
Or,
$\sin \theta$=${\cos}^{2} \theta$.

Now putting this value in the R.H.S portion of your second equation,we have,
${\cos}^{2} \theta$ +${\cos}^{4} \theta$=$\sin \theta$+${\left(\sin \theta\right)}^{2}$
Or,
${\cos}^{2} \theta$+${\cos}^{4} \theta$=1 {from $\textcolor{red}{1}$}

Hence proved a[ L.H.S=R.H.S]

Feb 13, 2018

sin^2θ+sinθ=1

plugging in the identity, sin^2θ + cos^2θ = 1

1-cos^2θ+sinθ=1

-cos^2θ+sinθ=0

color(red)(cos^2θ=sinθ

so, color(magenta)(cos^4θ=sin^2θ

we've gotta prove that, color(red)(cos^2θ)+color(magenta)(cos^4θ)=1
color(red)(sinθ)+color(magenta)(sin^2θ)=1# ; thats what we're provided with.

Hence Proved.!