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Proving Trigonometrical Identities (A level)

Proving Trigonometrical Identities (A level)

Proving trigonometrical identities requires you to use mathematical and logical steps to prove that one side of the identity can be simplified into the other side of the equation. Logically you cannot work from both sides of the identity at the same time.

Since we will be simplifying one side to prove that it is equal to the other side, it is helpful to assign names to the sides. The “left-hand side” of an identity is denoted by LHS, and the “right-hand side” is denoted as RHS.


Question 1

cos^2 \theta - sin^2 \theta \equiv 2cos^2 \theta - 1

Solution 1

cos^2 \theta - sin^2 \theta \equiv 2cos^2 \theta - 1

Take LHS

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  • cos^2 \theta - sin^2 \theta

However we know that cos^2 \theta + sin^2 \theta \equiv 1 and we can transpose it to sin^2 \theta \equiv 1 -  cos^2 \theta. By substituting this value of sin^2 \theta into our working we get:

  • cos^2 \theta -  (1 - cos^2 \theta)

  • cos^2 \theta -  1 + cos^2 \theta

  • 2cos^2 \theta -  1


Question 2

(\frac{cosec \theta + cot \theta}{tan \theta + sec \theta})tan \theta \equiv \frac{1 + cos \theta}{sin \theta + 1}

Solution 2

(\frac{cosec \theta + cot \theta}{tan \theta + sec \theta})tan \theta \equiv \frac{1 + cos \theta}{sin \theta + 1}

Take the LHS

  • (\frac{cosec \theta + cot \theta}{tan \theta + sec \theta})tan \theta

  • (\frac{ \frac{1}{sin \theta} +\frac{cos \theta}{sin \theta}}{\frac{sin \theta}{cos \theta} + \frac{1}{cos \theta}})\frac{sin \theta}{cos \theta}

  • (\frac{ \frac{1 + cos \theta}{sin \theta} }{ \frac{sin \theta + 1}{cos \theta}})\frac{sin \theta}{cos \theta}

  • ( \frac{1 + cos \theta}{sin \theta}   \div \frac{sin \theta + 1}{cos \theta})\frac{sin \theta}{cos \theta}

  • ( \frac{1 + cos \theta}{sin \theta}   \times \frac{cos \theta}{sin \theta + 1})\frac{sin \theta}{cos \theta}

  • \frac{1 + cos \theta}{sin \theta}   \times \frac{cos \theta}{sin \theta + 1} \times \frac{sin \theta}{cos \theta}

  • \frac{1 + cos \theta}{sin \theta + 1}


Question 3

\frac{1 + tan^2 \theta}{1 + cot^2 \theta } \equiv tan^2 \theta

Solution 3

\frac{1 + tan^2 \theta}{1 + cot^2 \theta } \equiv tan^2 \theta

Take the LHS

  • \frac{1 + tan^2 \theta}{1 + cot^2 \theta }

  • \frac{1 + tan^2 \theta}{1 + \frac{1}{tan^2 \theta} }

  • \frac{1 + tan^2 \theta}{\frac{tan^2 \theta +1}{tan^2 \theta} }

  • ( 1 + tan^2 \theta ) \div \frac{tan^2 \theta +1}{tan^2 \theta}

  • ( 1 + tan^2 \theta ) \times \frac{tan^2 \theta}{tan^2 \theta + 1}

  • tan^2 \theta


Question 4

(1 - cos \theta) (1 + sec \theta) \equiv sin \theta tan \theta

Solution 4

(1 - cos \theta) (1 + sec \theta) \equiv sin \theta tan \theta

Take the LHS

  • (1 - cos \theta) (1 + sec \theta)

  • 1 + sec \theta - cos \theta - cos \theta sec \theta

  • 1 + \frac{1}{cos \theta} - cos \theta - cos \theta \frac{1}{cos \theta}

  • 1 + \frac{1}{cos \theta} - cos \theta - 1

  • \frac{1}{cos \theta} - cos \theta

  • \frac{1 - cos^2 \theta}{cos \theta}

However sin^2 \theta \equiv 1 -  cos^2 \theta

  • \frac{sin^2 \theta}{cos \theta}

  • \frac{sin \theta sin \theta}{cos \theta}

  • sin \theta tan \theta


Question 5

\frac{1+cot\ y}{1+tan\ y} \equiv cot\ y

Solution 5

\frac{1+cot\ y}{1+tan\ y} \equiv cot\ y

Take the LHS

  • \frac{1+cot\ y}{1+tan\ y}

  • (1+cot\ y) \div (1+tan\ y)

  • (1+\frac{cos\ y}{sin\ y}) \div (1+\frac{sin\ y}{cos\ y})

  • (\frac{sin\ y + cos\ y}{sin\ y}) \div (\frac{cos\ y + sin\ y}{cos\ y})

  • (\frac{sin\ y + cos\ y}{sin\ y}) \times (\frac{cos\ y}{cos\ y + sin\ y})

  • \frac{cos\ y}{sin\ y}

  • cot\ y


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