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Question: If \(\sin \theta =\dfrac{5}{13}\), \(\theta \) being in quadrant II, how do you find the exact value...

If sin⁡θ=513\sin \theta =\dfrac{5}{13}, θ\theta being in quadrant II, how do you find the exact value of each of the remaining trigonometric functions of θ\theta ?

Explanation

Solution

We explain the function sin⁡θ=513\sin \theta =\dfrac{5}{13} and the quadrant value for the angle θ\theta . We express the identity functions of other ratios with ratio of sin. It’s given that sin⁡θ=513\sin \theta =\dfrac{5}{13}. Thereafter we put the value to find the value of each of the remaining trigonometric functions. We also use the representation of a right-angle triangle with height and hypotenuse ratio being 1213\dfrac{12}{13}.

Complete step by step answer:
It’s given that sin⁡θ=513\sin \theta =\dfrac{5}{13}, θ\theta being in quadrant II. In that quadrant only sin⁡θ\sin \theta and csc⁡θ\csc \theta are positive whereas all the other ratios are negative.
We know the sum of the square law of (sin⁡x)2+(cos⁡x)2=1{{\left( \sin x \right)}^{2}}+{{\left( \cos x \right)}^{2}}=1.
We can put the value of sin⁡θ=513\sin \theta =\dfrac{5}{13} in the equation of (sin⁡x)2+(cos⁡x)2=1{{\left( \sin x \right)}^{2}}+{{\left( \cos x \right)}^{2}}=1.
Putting the value of sin⁡θ\sin \theta , we get (513)2+(cos⁡θ)2=1{{\left( \dfrac{5}{13} \right)}^{2}}+{{\left( \cos \theta \right)}^{2}}=1.
Now we perform the binary operations.
(513)2+(cos⁡θ)2=1 ⇒(cos⁡θ)2=1−25169=144169 ⇒(cos⁡θ)=±1213 {{\left( \dfrac{5}{13} \right)}^{2}}+{{\left( \cos \theta \right)}^{2}}=1 \\\ \Rightarrow {{\left( \cos \theta \right)}^{2}}=1-\dfrac{25}{169}=\dfrac{144}{169} \\\ \Rightarrow \left( \cos \theta \right)=\pm \dfrac{12}{13} \\\
The value of cos⁡θ\cos \theta in quadrant II will be negative and that’s why (cos⁡θ)=−1213\left( \cos \theta \right)=-\dfrac{12}{13}.
We can find the identity which gives tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x}.
Putting the values we get,
\tan \theta =\dfrac{\sin \theta }{\cos \theta }\\\ \Rightarrow\tan \theta =\dfrac{\dfrac{5}{13}}{-\dfrac{12}{13}}\\\ \Rightarrow\tan \theta =-\dfrac{5}{12}\\\
We can find the value of sec⁡θ,csc⁡θ,cot⁡θ\sec \theta ,\csc \theta ,\cot \theta from the relation of (sec⁡x)=1cos⁡x,(csc⁡x)=1sin⁡x,(cot⁡x)=1tan⁡x\left( \sec x \right)=\dfrac{1}{\cos x},\left( \csc x \right)=\dfrac{1}{\sin x},\left( \cot x \right)=\dfrac{1}{\tan x}.
Putting the value, we get,
⇒(sec⁡θ)=1cos⁡θ=−1312\Rightarrow \left( \sec \theta \right)=\dfrac{1}{\cos \theta }=-\dfrac{13}{12}
⇒(csc⁡θ)=1sin⁡θ=135\Rightarrow \left( \csc \theta \right)=\dfrac{1}{\sin \theta }=\dfrac{13}{5}
⇒(cot⁡θ)=1tan⁡θ=−125\Rightarrow \left( \cot \theta \right)=\dfrac{1}{\tan \theta }=-\dfrac{12}{5}

Hence, in this way we have found the values of all other trigonometric functions.

Note: In addition to the reciprocal relationships of certain trigonometric functions, two other types of relationships exist. These relationships, known as trigonometric identities, include functional relationships and Pythagorean relationships.