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Test: Advanced Trigonometric Formula


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10 Questions MCQ Test Mathematics For JEE | Test: Advanced Trigonometric Formula

Test: Advanced Trigonometric Formula for Airforce X Y / Indian Navy SSR 2022 is part of Mathematics For JEE preparation. The Test: Advanced Trigonometric Formula questions and answers have been prepared according to the Airforce X Y / Indian Navy SSR exam syllabus.The Test: Advanced Trigonometric Formula MCQs are made for Airforce X Y / Indian Navy SSR 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Advanced Trigonometric Formula below.
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Test: Advanced Trigonometric Formula - Question 1

What is the value of tan 25°?

Detailed Solution for Test: Advanced Trigonometric Formula - Question 1

Test: Advanced Trigonometric Formula - Question 2

Find tan 40° ?

Detailed Solution for Test: Advanced Trigonometric Formula - Question 2

Test: Advanced Trigonometric Formula - Question 3

Find the value of 1 + cos2A.

Detailed Solution for Test: Advanced Trigonometric Formula - Question 3

By formula 

Cos 2A = Cos²A-Sin²A

Cos²A-Sin²A=Cos²A-(1-Cos²A) =Cos²A -1 +Cos²A=2 Cos²A - 1

so, Cos 2A = 2 Cos²A - 1

 

 

Test: Advanced Trigonometric Formula - Question 4

What is the value of cos 4x ?

Test: Advanced Trigonometric Formula - Question 5

What is the value of cos 6X?

Test: Advanced Trigonometric Formula - Question 6

3 Sin 10° – 4 Sin3 10° = ?

Detailed Solution for Test: Advanced Trigonometric Formula - Question 6

 3sinA-4sin3A=sin3A
Given, 3 sin 10° - 4 sin 3 10°
So, sin3A i.e sin 30
   = ½

Test: Advanced Trigonometric Formula - Question 7

 =

Test: Advanced Trigonometric Formula - Question 8

What is the value of Sin 6X ?

Test: Advanced Trigonometric Formula - Question 9

In a triangle ABC, sin 2A = ?

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Test: Advanced Trigonometric Formula - Question 10

In a triangle ABC, if angle A = 72°, angle B = 48° and c = 9 cm then the Ĉ is

Detailed Solution for Test: Advanced Trigonometric Formula - Question 10

In any traingle The sum of 3 sides of a traingle is equal to 180°

Given 

A=72°

B=48°

A+B+C=180°

72°+48°+C=180°

C=180°-120°

C=60°

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