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  5. Proving Trigonometric Identities

Proving Trigonometric Identities

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 and manipulating the expressions using various trigonometric properties and identities.

There are various techniques that we can use to help prove these identities:

  • Use Pythagorean Identities and Reciprocol Trigonometric Functions
  • Use Compound Angle and Double Angle Identities
  • Use Quotient and Reciprocol identities to convert each trigonometric term to their \(\sin\) or \(\cos\) equivalent
  • Identify and implement algebraic techniques such as factoring, expanding, common denominators, and distributive property to help simplify the expression
  • Work from both sides to reach a common expression
  • Focus on simplifying the more complex expression to match the simpler expression
  • Distribute the denominator to cancel terms if the numerator contains \(2\) or more terms
  • Use the trigonometric conjugate when necessary (multiplying a term or expression by another term where the numerator and denominator are the same)

Make sure to avoid the following mistakes when proving identities:

  • Distributing the numerator of any expression
  • Cancelling terms incorrectly – remember 'monomial cancels monomial', 'binomial cancels binomial', etc...
  • Forgetting to record brackets – especially if the question involves both multiplication and addition
  • Forgetting to write the input beside the trig function – cos, sin, tan without θ are meaningless
  • Incorrectly placing the exponent of a term or expression - sinθ^2 when you mean (sinθ)^2
  • Forgetting to explain your process step-by-step

Make sure to review Identities to better understand and solve the following examples.


Example

Prove the Identity \(\tan^2\theta + 1 = \sec^2\theta \)

When working from the left side, we can use Quotient Identities to convert \(\tan\) into its \(\sin\) and \(\cos\) equivalents:

\(\tan\theta = \cfrac{\sin\theta}{\cos\theta}\)

\(\text{LS} = \cfrac{\sin^2 \theta}{\cos^2 \theta} + 1\)

We can simplify the expression by combining the terms under a common denominator:

\(\text{LS} = \cfrac{\sin^2 \theta}{\cos^2 \theta} + (1) \left(\cfrac{\cos^2 \theta}{\cos^2 \theta}\right)\)

\(\text{LS} = \cfrac{\sin^2 \theta}{\cos^2 \theta} + \cfrac{\cos^2 \theta}{\cos^2 \theta}\)

\(\text{LS} = \cfrac{\sin^2 \theta + \cos^2 \theta}{\cos^2 \theta}\)

We can use Pythagorean Identities to convert the numerator:

\(\sin^2 \theta + \cos^2 \theta = 1\)

\(\text{LS} = \cfrac{1}{\cos^2 \theta}\)

When working from the right side, we can use Reciprocal Identities to convert the expression:

\(\sec^2 \theta = \cfrac{1}{\cos \theta}\)

\(\text{RS} = \cfrac{1}{\cos \theta}\)

Therefore, we can determine that \(\textbf{LS} \boldsymbol{=} \textbf{RS}\).

Example

Prove the Identity \(\sin(2\theta) = 2\sin\theta\cos\theta \).

In this question, we don't need to work on the right side since its simplified enough.

When working from the left side, apply the Sum of Angles formula to \(\sin(2\theta) \):

\(\text{LS} =\sin(A + B) = \sin A \cos B + \cos A \sin B \)

\(\text{LS} =\sin(\theta + \theta) = \sin\theta \cos\theta + \cos\theta \sin\theta \)

Next, we can combine like terms:

\(\text{LS} =\sin(2\theta) = 2\sin\theta\cos\theta\)

Therefore, we can determine that \(\textbf{LS} \boldsymbol{=} \textbf{RS}\).

Prove the following trigonometric identities:

\(\cfrac{\csc^2 x - 1}{\csc^2 x} = 1 - \sin^2 x \)

When working from the left side, we can simplify the expression by using quotient identities to convert \(\csc x\) to \(\sin x\):

\(\text{LS} = \cfrac{\cfrac{1}{\sin^2 x} - 1}{\cfrac{1}{\sin^2 x}}\)

To combine terms in the numerator, we can find a common denominator:

\(\text{LS} = \cfrac{\cfrac{1}{\sin^2 x} - 1\left(\cfrac{\sin^2 x}{\sin^2 x} \right)}{\cfrac{1}{\sin^2 x}}\)

\(\text{LS} = \cfrac{\cfrac{1}{\sin^2 x} - \cfrac{\sin^2 x}{\sin^2 x}}{\cfrac{1}{\sin^2 x}}\)

\(\text{LS} = \cfrac{\cfrac{1 - \sin^2 x}{\sin^2 x}}{\cfrac{1}{\sin^2 x}} \)

We can multiply the numerator by the reciprocal of the denominator to simplify the expression:

\(\text{LS}= \left(\cfrac{1 - \sin^2 x}{\cancel{\sin^2 x}} \right) \left(\cfrac{\cancel{\sin^2 x}}{1} \right)\)

\(\text{LS}= 1 - \sin^2 x \)

We can use Pythagorean Identities to convert the expression:

\(\sin^2 x + \cos^2 x = 1 \)

\(\cos^2 x = 1 - \sin^2 x \)

\(\text{LS} = \cos^2 x\)

Therefore, we can determine that \(\textbf{LS} \boldsymbol{=} \textbf{RS}\).

\(\cos^4 x - \sin^4 x = \cos 2x \)

When working from the left side, we can factor the expression by using the difference of squares:

\(a^2 - b^2 = (a + b)(a - b)\)

Since we can identify \( a = \cos^2 x \) and \( b = \sin^2 x \), we can write the factored expression as:

\(\text{LS}= \cos^4 x - \sin^4 x = (\cos^2 x + \sin^2 x)(\cos^2 x - \sin^2 x)\)

We can use Pythagorean Identities to convert the first part of the factored expression. We can then simplify the expression:

\(\text{LS} = (1)(\cos^2 x - \sin^2 x)\)

\(\text{LS} = \cos^2 x - \sin^2 x\)

We can use the cosine Double Angle formula to convert the expression:

\(\cos 2x = \cos^2 x - \sin^2 x \)

\(\text{LS}= \cos 2x\)

Therefore, we can determine that \(\textbf{LS} \boldsymbol{=} \textbf{RS}\).

\(\sin 2x + \sin 2y = 2 \sin(x + y) \cos(x - y) \)

When working on the left side, we can combine both terms into one:

\(\text{LS} = \sin (2(x + y))\)

We can apply the Double-Angle formulas to convert the expression on the left side:

\(\sin(2A) = 2 \sin A \cos A\)

\(\sin(A + B) = \sin A \cos B + \sin B \cos A\)

\(\text{LS} = 2 \sin x \cos x + 2 \sin y \cos y \)

We can use Double Angle identitites to convert both terms:

\(\sin2x = 2 \sin x \cos x\)

\(\text{LS} = \sin 2x + \sin 2y\)

We can apply the sine Sum-to-Product formula to convert the current expression:

\(\sin A + \sin B = 2 \sin\left(\cfrac{A + B}{2}\right) \cos\left(\cfrac{A - B}{2}\right)\)

In this instance, we can let \( A = 2x \) and \( B = 2y \) and plug these values into the formula. We can then simplify the expression:

\(\text{LS} =\sin 2x + \sin 2y = 2 \sin\left(\cfrac{2x + 2y}{2}\right) \cos\left(\cfrac{2x - 2y}{2}\right)\)

\(\text{LS} = 2 \sin(x + y) \cos(x - y)\)

Therefore, we can determine that \(\textbf{LS} \boldsymbol{=} \textbf{RS}\).

\(\cfrac{\sin 2x \sec x}{2} = \sin x\)

When working on the left side, we can use Double-Angle Identities for sine to convert part of the numerator:

\(\text{LS} = \sin 2x = 2 \sin x \cos x\)

\(\text{LS} = \cfrac{2 \sin x \cos x \sec x}{2}\)

Next, we can use Reciprocal Identities to convert secant into its equivalent form:

\(\sec x = \cfrac{1}{\cos x} \)

\(\text{LS} = \cfrac{2 \sin x \cos x}{2} \times \cfrac{1}{\cos x}\)

Then, we can simplify the expression by cancelling out terms:

\(\text{LS} = \cfrac{2 \sin x \cancel{\cos x}}{2} \times \cfrac{1}{\cancel{\cos x}} \)

\(\text{LS} = \cfrac{\cancel{2} \sin x}{\cancel{2}} \)

\(\text{LS} = \sin x\)

Therefore, we can determine that \(\textbf{LS} \boldsymbol{=} \textbf{RS}\).

\(\cfrac{1}{\cos^2 x \sin x} - \cfrac{\sin x}{\cos^2 x} = \csc x \)

When working on the left side, we can combine the terms under a common denominator:

\(\text{LS} = \cfrac{1}{\cos^2 x \sin x} - \left(\cfrac{\sin x}{\sin x}\right) \left(\cfrac{\sin x}{\cos^2 x}\right)\)

\(\text{LS} = \cfrac{1}{\cos^2 x \sin x} - \cfrac{\sin^2 x}{\cos^2 x \sin x}\)

\(\text{LS} = \cfrac{1 - \sin^2 x}{\cos^2 x \sin x}\)

Next, we can use Pythagorean Identities to convert the numerator:

\(1 - \sin^2 x = \cos^2 x\)

\(\text{LS} = \cfrac{\cos^2 x}{\cos^2 x \sin x} \)

Then, we can simplify the expression by cancelling like terms from the numerator and denominator:

\(\text{LS} = \cfrac{\cancel{\cos^2 x}}{\cancel{\cos^2 x} \sin x} \)

\(\text{LS} = \cfrac{1}{\sin x}\)

Finally, we can use Reciprocal Identities to convert the expression:

\(\cfrac{1}{\sin x} = \csc x\)

\(\text{LS} = \csc x\)

Therefore, we can determine that \(\textbf{LS} \boldsymbol{=} \textbf{RS}\).

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