8.3-Trigonometric Identities

8.3-Trigonometric Identities Important Formulae

You are currently studying
Grade 10 → Math → Introduction to Trigonometry → 8.3-Trigonometric Identities

After successful completion of this topic, you should be able to:

  • Compute and apply trigonometric identities in order to simplify and solve mathematical problems.

Trigonometric identities are equations involving trigonometric functions that are true for all values of the variable for which the functions are defined. These identities are essential for simplifying trigonometric expressions and solving trigonometric equations. Here are some fundamental trigonometric identities:

1. Pythagorean Identities

Derived from the Pythagorean theorem, these identities relate the squares of the sine and cosine functions:

  • $\sin^2 \theta + \cos^2 \theta = 1$
  • $1 + \tan^2 \theta = \sec^2 \theta$
  • $1 + \cot^2 \theta = \csc^2 \theta$
2. Reciprocal Identities

These identities express each trigonometric function in terms of its reciprocal:

  • $\csc \theta = \frac{1}{\sin \theta}$
  • $\sec \theta = \frac{1}{\cos \theta}$
  • $\cot \theta = \frac{1}{\tan \theta}$
3. Quotient Identities

These identities express the tangent and cotangent functions in terms of sine and cosine:

  • $\tan \theta = \frac{\sin \theta}{\cos \theta}$
  • $\cot \theta = \frac{\cos \theta}{\sin \theta}$
4. Co-Function Identities

These identities express the relationship between trigonometric functions of complementary angles:

  • $\sin(90^\circ - \theta) = \cos \theta$
  • $\cos(90^\circ - \theta) = \sin \theta$
  • $\tan(90^\circ - \theta) = \cot \theta$
  • $\csc(90^\circ - \theta) = \sec \theta$
  • $\sec(90^\circ - \theta) = \csc \theta$
  • $\cot(90^\circ - \theta) = \tan \theta$
5. Even-Odd Identities

These identities determine the signs of trigonometric functions based on the angle's quadrant:

  • Even Functions:

    • $\cos(-\theta) = \cos \theta$

    $\sec(-\theta) = \sec \theta$

  • Odd Functions:

    • $\sin(-\theta) = -\sin \theta$

    $\tan(-\theta) = -\tan \theta$

    $\cot(-\theta) = -\cot \theta$

    $\csc(-\theta) = -\csc \theta$

6. Sum and Difference Formulas

These formulas help in finding the values of trigonometric functions for the sum or difference of two angles:

  • $\sin(a \pm b) = \sin a \cos b \pm \cos a \sin b$
  • $\cos(a \pm b) = \cos a \cos b \mp \sin a \sin b$
  • $\tan(a \pm b) = \frac{\tan a \pm \tan b}{1 \mp \tan a \tan b}$
7. Double Angle Formulas

These formulas are useful for expressing trigonometric functions of double angles:

  • $\sin(2\theta) = 2\sin \theta \cos \theta$
  • $\cos(2\theta) = \cos^2 \theta - \sin^2 \theta$
  • $\tan(2\theta) = \frac{2\tan \theta}{1 - \tan^2 \theta}$

Using these identities, we can simplify complex trigonometric expressions and solve various trigonometric equations. Understanding these identities is crucial for mastering trigonometry and applying it in real-world problems.

Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

Solution:

Express the trigonometric ratios sin A, sec A and tan A in terms of cot A

We know that:

1. $ \cot A = \frac{1}{\tan A} $

2. $ \sec A = \frac{1}{\cos A} $

To express $ \sin A $, $ \sec A $, and $ \tan A $ in terms of $ \cot A $, let us use the identity:

1. $ \tan A = \frac{1}{\cot A} $

Now, let's express each trigonometric ratio:

For $ \sin A $, we use the identity $ \sin^2 A + \cos^2 A = 1 $. First, express $ \cos A $ in terms of $ \cot A $. From $ \cot A = \frac{\cos A}{\sin A} $, we get $ \cos A = \sin A \cot A $. Substituting into the identity:

$ \sin^2 A + (\sin A \cot A)^2 = 1 $

$ \sin^2 A + \sin^2 A \cot^2 A = 1 $

$ \sin^2 A (1 + \cot^2 A) = 1 $

$ \sin^2 A = \frac{1}{1 + \cot^2 A} $

$ \sin A = \frac{1}{\sqrt{1 + \cot^2 A}} $

For $ \sec A $, we use the identity $ \sec^2 A = 1 + \tan^2 A $. Since $ \tan A = \frac{1}{\cot A} $, we have:

$ \sec^2 A = 1 + \left( \frac{1}{\cot A} \right)^2 = 1 + \frac{1}{\cot^2 A} $

$ \sec A = \sqrt{1 + \frac{1}{\cot^2 A}} $

For $ \tan A $, as noted earlier, we already have:

$ \tan A = \frac{1}{\cot A} $

Suggest improvements/Report errors

Write all the other trigonometric ratios of $\angle$A in terms of sec A.

Solution:

Trigonometric Ratios of $\angle A$ in Terms of $\sec A$

Given that $\sec A = \frac{1}{\cos A}$, we can express the other trigonometric ratios in terms of $\sec A$.

1. $\cos A = \frac{1}{\sec A}$

2. $\sin A = \sqrt{1 - \cos^2 A} = \sqrt{1 - \left(\frac{1}{\sec A}\right)^2} = \sqrt{1 - \frac{1}{\sec^2 A}} = \frac{\sqrt{\sec^2 A - 1}}{\sec A}$

3. $\tan A = \frac{\sin A}{\cos A} = \frac{\frac{\sqrt{\sec^2 A - 1}}{\sec A}}{\frac{1}{\sec A}} = \sqrt{\sec^2 A - 1}$

4. $\cot A = \frac{1}{\tan A} = \frac{1}{\sqrt{\sec^2 A - 1}}$

5. $\csc A = \frac{1}{\sin A} = \frac{\sec A}{\sqrt{\sec^2 A - 1}}$

Suggest improvements/Report errors

1. Choose the correct option. Justify your choice.

(i)  $9 \sec^2 A - 9 \tan^2 A$ =

 (A) 1
(B) 9
(C) 8
(D) 0

(ii)  $(1 + \tan \theta + \sec \theta)(1 + \cot \theta - \mathrm{cosec}\ \theta)$=
(A) 0
(B) 1
(C) 2
(D) –1

(iii)  $(\sec A + \tan A)(1 - \sin A)$ =
(A) $\sec A$
(B) $\sin A$
(C) $\mathrm{cosec}\ A$
(D) $\cos A$

(iv) $\dfrac{1 + \tan^2 A}{1 + \cot^2 A}$

(A) $\sec^2 A$
(B) –1
(C) $\cot^2 A$
(D) $\tan^2 A$

Solution:

(i) $9 \sec^2 A - 9 \tan^2 A$ =
(A) 1
(B) 9 CORRECT
(C) 8
(D) 0
(ii) $(1 + \tan \theta + \sec \theta)(1 + \cot \theta - \csc \theta)$ =
(A) 0
(B) 1
(C) 2 CORRECT
(D) –1
(iii) $(\sec A + \tan A)(1 - \sin A)$ =
(A) $\sec A$
(B) $\sin A$
(C) $\csc A$
(D) $\cos A$ CORRECT
(iv) $\dfrac{1 + \tan^2 A}{1 + \cot^2 A}$
(A) $\sec^2 A$
(B) –1
(C) $\cot^2 A$
(D) $\tan^2 A$ CORRECT

Suggest improvements/Report errors