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Wize High School Grade 11 Math Textbook > Trigonometric Identities & Proofs

Proofs using Trigonometric Identities

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Proofs using Trigonometric Identities


Proving an identity is different than solving an equation. Since an identity is a tautology, a statement that is always true, the main objective in proving identities is to show that both sides are equivalent for all values of xx (or θ\theta).

Tips for Proving Trigonometric Identities

  • Begin with the more complicated side
  • Where required, apply trigonometric identities
  • If necessary, change all values of tan⁡θ & cot⁡θ\tan\theta~\&~\cot\theta into sin⁡θ & cos⁡θ\sin\theta~\&~\cos\theta
  • Add/subtract any rational expression/fractions and make into a single rational expression/fraction
  • Determine if there is any factoring required
  • When required, multiply the numerator and denominator by the conjugate

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Example 1

Prove cot⁡θcsc⁡θ=cos⁡θ\frac{\cot\theta}{\csc\theta}=\cos\theta is true for all values of θ\theta.


LHSRHSQuotient & Reciprocal Identitycot⁡θcsc⁡θcos⁡θ(cos⁡θsin⁡θ)(1sin⁡θ)cos⁡θ(cos⁡θsin⁡θ)⋅(sin⁡θ)cos⁡θcos⁡θcos⁡θ\begin{array}{lc|cr} &\textbf{LHS}&&\textbf{RHS}&\\\hline\\ \colorTwo{\scriptsize{\text{Quotient \& Reciprocal Identity}}}&\dfrac{\cot\theta}{\csc\theta}&&\cos\theta\\\\\\ &\dfrac{\Bigg(\dfrac{\cos\theta}{\sin\theta}\Bigg)}{\Bigg(\dfrac{1}{\sin\theta}\Bigg)}&&\cos\theta\\\\\\ &\Bigg(\dfrac{\cos\theta}{\sin\theta}\Bigg)\cdot(\sin\theta)&&\cos\theta\\\\\\ &\cos\theta&&\cos\theta \end{array}

We manipulated the LHS using trigonometric identities to prove cot⁡θcos⁡θ=cos⁡θ\dfrac{\cot\theta}{\cos\theta}=\cos\theta for all values of θ.\theta.


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Example: Proofs using Trigonometric Identities


Prove sec⁡θ(1+sin⁡θ)=tan⁡θ(1+csc⁡θ)\sec\theta(1+\sin\theta)=\tan\theta(1+\csc\theta).


Since both sides are similar in complexity, we can work with both sides simultaneously.

LHSRHSQuotient & Reciprocal Identitysec⁡θ(1+sin⁡θ)tan⁡θ(1+csc⁡θ)Quotient & Reciprocal Identity1cos⁡θ(1+sin⁡θ)sin⁡θcos⁡θ(1+1sin⁡θ)1cos⁡θ+sin⁡θcos⁡θsin⁡θcos⁡θ+1cos⁡θ\begin{array}{ccc|ccc} &\textbf{LHS}&&&\textbf{RHS}&\\\hline\\ \colorTwo{\scriptsize{\text{Quotient \& Reciprocal Identity}}}&\sec\theta(1+\sin\theta)&&&\tan\theta(1+\csc\theta)&\colorTwo{\scriptsize{\text{Quotient \& Reciprocal Identity}}}\\\\\\ &\dfrac{1}{\cos\theta}(1+\sin\theta)&&&\dfrac{\sin\theta}{\cos\theta}\Bigg(1+\dfrac{1}{\sin\theta}\Bigg)\\\\\\ &\dfrac{1}{\cos\theta}+\dfrac{\sin\theta}{\cos\theta}&&&\dfrac{\sin\theta}{\cos\theta}+\dfrac{1}{\cos\theta} \end{array}

Since the LHS = RHS, then we have proved the identity.
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Example: Proofs using Trigonometric Identities

Prove sin⁡θ1−cos⁡θ+sin⁡θ1+cos⁡θ=2csc⁡θ\dfrac{\sin\theta}{1-\cos\theta}+\dfrac{\sin\theta}{1+\cos\theta}=2\csc\theta.


LHSRHSMultiply by the conjugatesin⁡θ1−cos⁡θ+sin⁡θ1+cos⁡θ2csc⁡θsin⁡θ1−cos⁡θ⋅(1+cos⁡θ1+cos⁡θ)+sin⁡θ1+cos⁡θ⋅(1−cos⁡θ1−cos⁡θ)2csc⁡θPythagorean Identitysin⁡θ(1+cos⁡θ)1−cos⁡2θ+sin⁡θ(1−cos⁡θ)1−cos⁡2θ2csc⁡θAdd Rational Expressionssin⁡θ(1+cos⁡θ)sin⁡2θ+sin⁡θ(1−cos⁡θ)sin⁡2θ2csc⁡θsin⁡θ(1+cos⁡θ)+sin⁡θ(1−cos⁡θ)sin⁡2θ2csc⁡θsin⁡θ+sin⁡θcos⁡θ+sin⁡θ−sin⁡θcos⁡θsin⁡2θ2csc⁡θ2sin⁡θsin⁡2θ2csc⁡θReciprocal Identity2sin⁡θ2csc⁡θ2csc⁡θ2csc⁡θ\begin{array}{ccc|ccc} &\textbf{LHS}&&&\textbf{RHS}\\\hline\\ \colorTwo{\scriptsize{\text{Multiply by the conjugate}}}&\dfrac{\sin\theta}{1-\cos\theta}+\dfrac{\sin\theta}{1+\cos\theta}&&&2\csc\theta\\\\\\ &\dfrac{\sin\theta}{1-\cos\theta}\cdot\Bigg(\dfrac{1+\cos\theta}{1+\cos\theta}\Bigg)+\dfrac{\sin\theta}{1+\cos\theta}\cdot\Bigg(\dfrac{1-\cos\theta}{1-\cos\theta}\Bigg)&&&2\csc\theta\\\\\\ \colorTwo{\scriptsize{\text{Pythagorean Identity}}}&\dfrac{\sin\theta(1+\cos\theta)}{1-\cos^2\theta}+\dfrac{\sin\theta(1-\cos\theta)}{1-\cos^2\theta}&&&2\csc\theta\\\\\\ \colorTwo{\scriptsize{\text{Add Rational Expressions}}}&\dfrac{\sin\theta(1+\cos\theta)}{\sin^2\theta}+\dfrac{\sin\theta(1-\cos\theta)}{\sin^2\theta}&&&2\csc\theta\\\\\\ &\dfrac{\sin\theta(1+\cos\theta)+\sin\theta(1-\cos\theta)}{\sin^2\theta}&&&2\csc\theta\\\\\\ &\dfrac{\sin\theta+\sin\theta\cos\theta+\sin\theta-\sin\theta\cos\theta}{\sin^2\theta}&&&2\csc\theta\\\\\\ &\dfrac{2\sin\theta}{\sin^2\theta}&&&2\csc\theta\\\\\\ \colorTwo{\scriptsize{\text{Reciprocal Identity}}}&\dfrac{2}{\sin\theta}&&&2\csc\theta\\\\\\ &2\csc\theta&&&2\csc\theta \end{array}

Practice: Proofs using Trigonometric Identities

True or False?

csc⁡θ−cot⁡θsec⁡θ=sin⁡θ\csc\theta-\dfrac{\cot\theta}{\sec\theta}=\sin\theta

Practice: Proofs using Trigonometric Identities

Which of the following is equivalent to tan⁡θ+sec⁡θ−1tan⁡θ−sec⁡θ+1\dfrac{\tan\theta+\sec\theta-1}{\tan\theta-\sec\theta+1}?
Extra Practice
Trigonometric Identities & Proofs
True or false:
1+sin⁡x1−sin⁡x−1−sin⁡x1+sin⁡x=4tan⁡xcsc⁡x\dfrac{1+\sin{x}}{1-\sin{x}}-\dfrac{1-\sin{x}}{1+\sin{x}}=4\tan{x}\csc{x}
Trigonometric Identities & Proofs
Determine an equivalent expression to 1+cos⁡xsin⁡x\dfrac{1+\cos{x}}{\sin{x}}.
Proofs using Trigonometric Identities
Practice: Proofs using Trigonometric Identities
Which of the following is equivalent to tan⁡2θ+1+tan⁡θsec⁡θ\tan^2\theta+1+\tan\theta\sec\theta?
Proofs using Trigonometric Identities
Practice: Proofs using Trigonometric Identities
Which of the following is equivalent to sin⁡2θ+4sin⁡θ+3cos⁡2θ\dfrac{\sin^2\theta+4\sin\theta+3}{\cos^2\theta}?
Proofs using Trigonometric Identities
Practice: Proofs using Trigonometric Identities
True or False?
(sin⁡θ−cos⁡θ)2+(sin⁡θ+cos⁡θ)2=2(\sin\theta-\cos\theta)^2+(\sin\theta+\cos\theta)^2=2.
Proofs using Trigonometric Identities
Practice: Proofs using Trigonometric Identities
True or False?
cos⁡2θ=csc⁡θcos⁡θtan⁡θ+cot⁡θ\cos^2\theta=\dfrac{\csc\theta\cos\theta}{\tan\theta+\cot\theta}.
Proofs using Trigonometric Identities
Practice: Proofs using Trigonometric Identities
True or False?
1tan⁡θ+tan⁡θ=sin⁡θcos⁡θ\dfrac{1}{\tan\theta}+\tan\theta=\sin\theta\cos\theta
Proofs using Trigonometric Identities
Practice: Proofs using Trigonometric Identities
True or False?
tan⁡θsin⁡θ+cos⁡θ=sec⁡θ\tan\theta\sin\theta+\cos\theta=\sec\theta
Double & Half Angle Identities
Practice: Double & Half Angle Identities
True or False? csc⁡2θ−2csc⁡2θ=sin⁡2θ\dfrac{\csc^2\theta-2}{\csc^2\theta}=\sin2\theta
Double & Half Angle Identities
Practice: Double & Half Angle Identities
True or False? sin⁡2θ1−cos⁡2θ=2csc⁡2θ−tan⁡θ\dfrac{\sin2\theta}{1-\cos2\theta}=2\csc2\theta-\tan\theta
Reciprocal & Quotient Identities
Practice: Reciprocal & Quotient Identities
Simplify cot⁡θ+12cos⁡θ\cot\theta+\dfrac{1}{2}\cos\theta.
Reciprocal & Quotient Identities
Practice: Reciprocal & Quotient Identities
True or false: tan⁡θcsc⁡2θsec⁡θ=sec⁡θ\dfrac{\tan\theta\csc^{2}\theta}{\sec\theta}=\sec\theta.
Reciprocal & Quotient Identities
Practice: Reciprocal & Quotient Identities
True or false: cot⁡2θsin⁡θcsc⁡θ=cos⁡2θ\dfrac{\cot^2\theta\sin\theta}{\csc\theta}=\cos^2\theta.