English

If θ = 15°, find the value of: cos3θ - sin6θ + 3sin(5θ + 15°) - 2 tan23θ - Mathematics

Advertisements
Advertisements

Question

If θ = 15°, find the value of: cos3θ - sin6θ + 3sin(5θ + 15°) - 2 tan2

Sum

Solution

θ = 15°

`(3)/(2)cos3θ - sin6θ + 3sin(5θ + 15°) - 2tan^2 3θ`

= `(3)/(2)cos 3 xx 15° - sin6 xx 15° + 3sin(5 xx 15° + 15°) -2tan^2 3 xx 15°`

= `(3)/(2)cos45° - sin90° + 3sin90° - 2tan^2 45°`

= `(3)/(2) xx (1)/sqrt(2) - 1 + 3 xx 1 - 2 xx (1)^2`

= `(3)/(2sqrt(2)) - 1 + 3 - 2`

= `(3)/(2sqrt(2))`

= `(3)/(2sqrt(2)) xx sqrt(2)/sqrt(2)`

= `(3sqrt(2))/(4)`.

shaalaa.com
Trigonometric Equation Problem and Solution
  Is there an error in this question or solution?
Chapter 27: Trigonometrical Ratios of Standard Angles - Exercise 27.1

APPEARS IN

Frank Mathematics [English] Class 9 ICSE
Chapter 27 Trigonometrical Ratios of Standard Angles
Exercise 27.1 | Q 14
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×