Question

How do you use the double-angle identities to find $\cot \left( 2x \right)$ if $\cos x=-\dfrac{15}{17}$ and $\cos ecx$ is less than 0?

Answer 1

We first take the identity theorem ${{\cos }^{2}}x+{{\sin }^{2}}x=1$. From the given value of $\cos x=-\dfrac{15}{17}$, we find the value of $\sin x$. Then we use the theorem for the ratio $\cot \left( x \right)=\dfrac{\cos x}{\sin x}$. We use multiple angle formulas for $\cos 2x=2{{\cos }^{2}}x-1$ and $\sin 2x=2\sin x\cos x$. We put the values and find the solution for $\cot \left( 2x \right)$.

Complete step by step answer:
We have to find the solution for $\cot \left( 2x \right)$.
We know that the ratio can be broken into $\sin x$ and $\cos x$ where $\cot \left( x \right)=\dfrac{\cos x}{\sin x}$.
The value of $\cos x$ is given where $\cos x=-\dfrac{15}{17}$. We are going to use the identity theorem
$\Rightarrow {{\cos }^{2}}x+{{\sin }^{2}}x=1$.

$$\Rightarrow {{\left( -\dfrac{15}{17} \right)}^{2}}+{{\sin }^{2}}x=1 \\\ \Rightarrow {{\sin }^{2}}x=1-\dfrac{225}{289}=\dfrac{64}{289} \\\ \Rightarrow \sin x=\pm \dfrac{8}{17} \\\$$

Now it’s given that $\cos ecx$ is less than 0. We know that $\cos ecx=\dfrac{1}{\sin x}$.
This means the signs of $\sin x$ and $\cos ecx$ will be negative for both.
Therefore, we will omit the positive value for $\sin x$. The value will be $\sin x=-\dfrac{8}{17}$.
We have values for both $\sin x$ and $\cos x$.
Now we use the theorems of multiple angles to find the values for $\sin 2x$ and $\cos 2x$.
We know $\cos 2x=2{{\cos }^{2}}x-1$ and $\sin 2x=2\sin x\cos x$.
Substituting the values, we get
$\cos 2x=2{{\left( -\dfrac{15}{17} \right)}^{2}}-1=\dfrac{450}{289}-1=\dfrac{161}{289}$
For $\sin 2x=2\sin x\cos x$, we get
$\sin 2x=2\left( -\dfrac{15}{17} \right)\left( -\dfrac{8}{17} \right)=\dfrac{240}{289}$.
Now we find the relation,
$\cot \left( 2x \right)=\dfrac{\cos 2x}{\sin 2x}$
We put the values and get
$\cot \left( 2x \right)=\dfrac{{}^{161}/{}_{289}}{{}^{240}/{}_{289}}=\dfrac{161}{240}$

Therefore, the value for $\cot \left( 2x \right)$ is $\dfrac{161}{240}$.

Note: We need to remember that the solution for quadratic root will always give two values being positive and negative. The signs of $\sin x$ and $\cos ecx$ in any case remain the same as they are connected by inverse law. That’s why we had to omit the positive value for signs of $\sin x$.