Ex 3.3, 3 - Chapter 3 Class 11 Trigonometric Functions
Last updated at April 16, 2024 by Teachoo
Finding Value of trignometric functions, given angle
Negative Angle Identities
Value of sin, cos, tan repeats after 2π
Shifting angle by π/2, π, 3π/2 , 2π
Example 8
Ex 3.2, 9 Important
Ex 3.2, 8
Ex 3.2, 10 Important
Example 9 Important
Ex 3.2, 6
Ex 3.2, 7 Important
Example 10
Ex 3.3, 1 Important
Ex 3.3, 2 Important
Ex 3.3, 3 Important You are here
Ex 3.3, 4
Ex 3.3, 8 Important
Ex 3.3, 9 Important
Find values of sin 18, cos 18, cos 36, sin 36, sin 54, cos 54 Important
Chapter 3 Class 11 Trigonometric Functions
Concept wise
- Radian measure - Conversion
- Arc length
- Finding Value of trignometric functions, given other functions
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Finding Value of trignometric functions, given angle
- (x + y) formula
- 2x 3x formula - Proving
- 2x 3x formula - Finding value
- cos x + cos y formula
- 2 sin x sin y formula
- Finding Principal solutions
- Finding General Solutions
- Sine and Cosine Formula
Transcript
Ex 3.3, 3 Prove that cot2 π/6 + cosec 5π/6 + 3 tan2 π/6 = 6 Solving L.H.S. cot2 π/6 + cosec 5π/6 + 3 tan2 π/6 Putting π = 180° = cot2(180/6) + cosec((5 ×180)/6) + 3 tan2(180/6) = cot2 30° + cosec (150°) + 3tan2 30° Here, tan 30° = 1/√3 cot 30° = 1/tan〖30°〗 = 1/(1/√3) = √𝟑 For cosec 150° First, Finding sin 150° sin 150° = sin (180 – 30°) = sin 30° = 𝟏/𝟐 cosec 150° = 1/sin〖150°〗 = 1/(1/2) = 2 Putting values cot2 30° + cosec (150°) + 3tan2 30° = (√𝟑)2 + 2 + 3 × (𝟏/√𝟑)^𝟐 = 3 + 2 + 3 × 1/3 = 3 + 2 + 1 = 6 = R.H.S Hence proved
