Back to worksheet

Neokaal Study

Introduction to Trigonometry

Practice Worksheet

Trigonometric ratios, standard angles, complementary angles, and identities.

Name
Class
Date

Answer all questions.

Show all necessary working.

Use a separate notebook for your solutions.

  1. For an acute angle, suppose $\sin \theta = \frac{a}{b}$. Evaluate $\cos \theta$.

  2. Find the exact value of $\frac{\sec 60^\circ}{\operatorname{cosec} 30^\circ}$.

  3. Decide whether $(\tan \theta + 2)(2\tan \theta + 1) = 5\tan \theta + \sec^2 \theta$ is true or false.

  4. Given $\tan A = \sqrt{3}$ and $\cot B = 1$, find $A + B$.

  5. Use complementary-angle identities to reduce $\sin(\alpha - \beta)$, where $\cos(\alpha + \beta) = 0$.

  6. For a right triangle $\triangle ABC$ with $\angle C = 90^\circ$, what is $\cot (A + B)$?

  7. If $\sin \theta - \cos \theta = 0$, find $\sin^4 \theta + \cos^4 \theta$.

  8. Use identities to test the statement $\text{If } \cos A + \cos^2 A = 1\text{, then } \sin^2 A + \sin^4 A = 1$.

Neokaal Study

Answer Key

  1. $\cos \theta$ = $\frac{\sqrt{b^2 - a^2}}{b}$.

    1. Assume the angle is acute and 0 < a < b.
    2. Take opposite side = a and hypotenuse = b.
    3. Then the adjacent side is $\sqrt{b^2 - a^2}$.
    4. Therefore $\cos \theta$ = $\frac{\sqrt{b^2 - a^2}}{b}$.

  2. The exact value is $1$.

    1. Use the standard-angle values: $\sec 60^\circ = 2, \operatorname{cosec} 30^\circ = 2$.
    2. Substitute them into the expression: $\frac{\sec 60^\circ}{\operatorname{cosec} 30^\circ}$ = $\frac{2}{2}$.
    3. Simplifying gives $1$.

  3. The statement is false.

    1. The left side expands to $2\tan^2 \theta + 5\tan \theta + 2$.
    2. Using $\sec^2 \theta = 1 + \tan^2 \theta$, the right side is $\tan^2 \theta + 5\tan \theta + 1$.
    3. The two sides differ by $\tan^2 \theta + 1$, so they are not identically equal.
    4. The statement is false.

  4. $A + B$ = $105^\circ$.

    1. From the standard-angle table, $\tan A = \sqrt{3}$ gives $A = 60^\circ$.
    2. Similarly, $\cot B = 1$ gives $B = 45^\circ$.
    3. So $A + B$ = $60^\circ + 45^\circ$ = $105^\circ$.

  5. $\sin(\alpha - \beta)$ reduces to $\cos 2\beta$.

    1. Since $\cos(\alpha + \beta) = 0$, use $\alpha + \beta = 90^\circ$.
    2. Then $\alpha = 90^\circ - \beta$.
    3. So $\sin(\alpha - \beta)$ = $\sin(90^\circ - 2\beta)$ = $\cos 2\beta$.

  6. $\cot (A + B)$ = $0$.

    1. The angles of a triangle add to $180^\circ$.
    2. Since $\angle C = 90^\circ$, the other two angles add to $180^\circ - 90^\circ = 90^\circ$.
    3. So $A + B$ = $90^\circ$, and $\cot (A + B)$ = $\cot 90^\circ$ = $0$.

  7. $\sin^4 \theta + \cos^4 \theta$ = $\frac{1}{2}$.

    1. The condition $\sin \theta - \cos \theta = 0$ gives $\sin \theta = \cos \theta$.
    2. Since $\sin^2 \theta + \cos^2 \theta = 1$, each square is $\frac{1}{2}$.
    3. So $\sin^4 \theta + \cos^4 \theta$ = $\left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2$ = $\frac{1}{2}$.

  8. The statement is true.

    1. From $\cos A + \cos^2 A = 1$, we get $\cos A = 1 - \cos^2 A$.
    2. Using $1 - \cos^2 A = \sin^2 A$, this means $\cos A = \sin^2 A$.
    3. Therefore $\sin^2 A + \sin^4 A$ = $\cos A + \cos^2 A$ = $1$.
    4. So the statement is true.