Trigonometric Functions and Derivatives

Trigonometric Functions and Derivatives

Assessment

Interactive Video

Mathematics

11th - 12th Grade

Hard

Created by

Mia Campbell

FREE Resource

The video tutorial explores the use of trigonometric substitution in algebraic expressions, emphasizing the conversion between trigonometric and algebraic forms to simplify problem-solving. It highlights the importance of phrasing problems correctly to find effective solutions. The tutorial introduces T results, explaining their application in integration and differentiation, and provides techniques for using these results in mathematical problems.

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Why do we introduce trigonometric substitutions in algebraic expressions?

To convert them into logarithmic forms

To eliminate variables

To simplify the problem-solving process

To make the expressions more complex

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the key takeaway from the traffic analogy in problem phrasing?

Infrastructure solutions are always the best

Economic perspectives can offer new solutions

Traffic problems are unsolvable

Adding more roads is the ultimate solution

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the primary purpose of T-Results in trigonometry?

To simplify logarithmic expressions

To convert trigonometric expressions back to algebraic forms

To solve quadratic equations

To convert algebraic expressions into trigonometric ones

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Which trigonometric function is derived using the double angle result in T-Results?

Secant

Tangent

Cosine

Sine

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the relationship between sine and cosine in a right triangle?

Cosine is opposite over hypotenuse

Sine is opposite over adjacent

Sine is opposite over hypotenuse

Cosine is adjacent over opposite

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the advantage of using implicit differentiation in integration?

It eliminates the need for chain rule

It allows differentiation without making a variable the subject

It converts all functions to linear forms

It simplifies the differentiation process

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How can you express sec squared in terms of cosine?

1 over sine squared

1 over cosine squared

1 over tangent squared

1 over secant squared

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