Trigonometric Identities are equations or statements involving trigonometric functions that are always true regardless of what values are used. Oftentimes, this involves a more complex function being set equal to a simpler function.
There are various techniques that we can use to help prove these identities:
Here are some of the common trigonometric identities you may encounter:
| Quotient | Reciprocol | Pythagorean |
| \(\tan\theta = \cfrac{\sin\theta}{\cos\theta}\) | \(\csc\theta = \cfrac{1}{\sin\theta}\) | \(\sin²\theta + \cos²\theta = 1\) |
| \(\cot\theta = \cfrac{\cos\theta}{\sin\theta}\) | \(\sec\theta = \cfrac{1}{\cos\theta}\) | \(\sec²\theta - \tan²\theta = 1\) |
| \(\cot\theta = \cfrac{1}{\tan\theta}\) | \(\csc²\theta + \cot²\theta = 1\) |
Prove the following trigonometric identity:
\(\sin\theta + \cfrac{\cos\theta}{\tan\theta} = \cfrac{1}{\cos\theta \tan\theta}\)When working from the left side, we can simplify the expression by using quotient identities to convert \(\tan\theta\) to \(\sin\theta\) and \(\cos\theta\):
\(\text{LS} = \sin\theta + \cfrac{\cos\theta}{{\cfrac{\textcolor{red}{\sin\theta}}{\textcolor{red}{\cos\theta}}}}\)
Next, we can multiply the reciprocol to simplify the expression:
\(\text{LS} = \sin\theta + \left(\cfrac{\cos\theta}{1}\right)\left(\cfrac{\textcolor{red}{\cos\theta}}{\textcolor{red}{\sin\theta}}\right)\)
\(\text{LS} = \sin \theta + \cfrac{\cos^2 \theta}{\sin \theta}\)
Then, we can multiply \(\sin\theta\) by the trigonometric conjugate so that both terms have a common denominator. We can simplify the expression by setting both numerator terms under the same denominator:
\(\text{LS} = \left(\sin\theta\right)\left(\cfrac{\textcolor{red}{\sin\theta}}{\textcolor{red}{\sin\theta}}\right) + \cfrac{\cos^2\theta}{\sin\theta}\)
\(\text{LS} = \cfrac{\sin^2\theta}{\textcolor{red}{\sin\theta}} + \cfrac{\cos^2\theta}{\textcolor{red}{\sin\theta}}\)
\(\text{LS} = \cfrac{\sin²\theta + \cos^2\theta}{\sin\theta}\)
Using Pythagorean Identities, we can simplify the numerator:
\(\sin^2\theta + \cos^2\theta = 1\)
\(\text{LS} = \cfrac{\textcolor{red}{1}}{\sin\theta}\)
When working from the right side, we can separate the expression into \(2\) different terms:
\(\text{RS} = \left(\cfrac{1}{\cos\theta}\right)\left(\cfrac{1}{\tan\theta}\right)\)
Next, we can simplify this expression by finding the reciprocol of the second term:
\(\text{RS} = \left(\cfrac{1}{\cos\theta}\right)(\textcolor{red}{\cot\theta})\)
Then, we can then find the quotient of the second term to cancel out terms:
\(\text{RS} = \left(\cfrac{1}{\cos\theta}\right)\left(\cfrac{\textcolor{red}{\cos\theta}}{\textcolor{red}{\sin\theta}}\right)\)
\(\text{RS} = \left(\cfrac{1}{\cancel{\cos\theta}}\right)\left(\cfrac{\cancel{\cos\theta}}{\sin\theta}\right)\)
\(\text{RS} = \cfrac{1}{\sin\theta}\)
Therefore, we can identify that \(\textbf{LS} \boldsymbol{=} \textbf{RS}\).
\(\cos^2\theta = \cfrac{\sin^2\theta}{\tan^2\theta}\)
In this question, we don't need to work on the left side since it's already simplified enough.
When working from the right side, use Quotient Identities to convert \(\tan\theta\) to \(\sin\theta\) and \(\cos\theta\):
\(\text{RS} = \cfrac{\sin^2\theta}{\cfrac{\textcolor{red}{\sin^2\theta}}{\textcolor{red}{\cos²\theta}}}\)
Next, we can multiply the expression by its reciprocal to simplify:
\(\text{RS} = \left(\cfrac{\sin^2\theta}{1}\right)\left(\cfrac{\textcolor{red}{\cos^2\theta}}{\textcolor{red}{\sin²\theta}}\right)\)
\(\text{RS} = \left(\cfrac{ \cancel{\sin^2\theta}}{1}\right)\left(\cfrac{\cos^2\theta}{\cancel{\sin^2\theta}}\right)\)
\(\text{RS} = \cos²\theta\)
Therefore, we can identify that \(\textbf{LS} \boldsymbol{=} \textbf{RS}\).
\(\sec\theta \csc\theta - \cot\theta = \tan\theta\)
When working on the left side, we can use reciprocol and quotient identities to simplify the expression into using \(\sin\) and \(\cos\) terms:
\(\text{LS} = \cfrac{\textcolor{red}{1}}{\textcolor{red}{\cos\theta}}.\cfrac{\textcolor{green}{1}}{\textcolor{green}{\sin\theta}}-\cfrac{\textcolor{blue}{\cos\theta}}{\textcolor{blue}{\sin\theta}}\)
Next, we can use the trigonometric conjugate to give each term a common denominator. We can simplify the expression by setting both numerator terms under the same denominator:
\(\text{LS} = \cfrac{1}{\cos\theta \sin\theta} - \left(\cfrac{\cos\theta}{\sin\theta}\right)\left(\cfrac{\textcolor{red}{\cos\theta}}{\textcolor{red}{\cos\theta}}\right)\)
\(\text{LS} = \cfrac{1}{\textcolor{red}{\cos\theta \sin\theta}} - \cfrac{\cos^2\theta}{\textcolor{red}{\cos\theta \sin\theta}}\)
\(\text{LS} = \cfrac{1 - \cos^2\theta}{\cos\theta \sin\theta}\)
Then, we can use Pythagorean Identities to convert the numerator into a different term. We can then factor to cancel out terms:
\(\text{LS} = \cfrac{\textcolor{red}{\sin²\theta}}{\cos\theta \sin\theta}\)
\(\text{LS} = \cfrac{\textcolor{red}{\sin\theta} \cdot \textcolor{red}{\sin\theta}}{\cos\theta \sin\theta}\)
\(\text{LS} = \cfrac{\sin\theta \cdot \cancel{\sin\theta}}{\cos\theta \cancel{\sin\theta}}\)
\(\text{LS} = \cfrac{\sin\theta}{\cos\theta}\)
Finally, we can use Quotient Identities to convert the expression:
\(\text{LS} = \tan\theta\)
Therefore, we can determine that \(\textbf{LS} \boldsymbol{=} \textbf{RS}\).