If \(\cos α=\frac{2}{3}\) and \(\sinβ=\frac{1}{4}\), then what is the value of cos (α - β)?

This question was previously asked in
Bihar STET TGT (Maths) Official Paper-I (Held On: 04 Sept, 2023 Shift 1)
View all Bihar STET Papers >
  1. \(\frac{2\sqrt{15}+\sqrt5}{12}\)
  2. \(\frac{\sqrt5}{12}\)
  3. 0
  4. \(\frac{2\sqrt{15}-\sqrt5}{12}\)

Answer (Detailed Solution Below)

Option 1 : \(\frac{2\sqrt{15}+\sqrt5}{12}\)
Free
Bihar STET Paper 1 Mathematics Full Test 1
14 K Users
150 Questions 150 Marks 150 Mins

Detailed Solution

Download Solution PDF

Concept use:

cos (α - β) = cosα cosβ + sinα sinβ 

Calculations:

\(\cos α=\frac{2}{3}\) = Base/Hypotenuse 

By Pythagoras theorem  H2 = P2 + B2

⇒ 32 - 22 = P2 

⇒ P = √5

sin α = √5/3

\(\sinβ=\frac{1}{4}\) = Perpendicular/Hypotenuse 

By Pythagoras theorem  H2 = P2 + B2

⇒ 42 - 12 = B2 

⇒ B = √15

cosβ = √15/4

cos (α - β) = cosα cosβ + sinα sinβ = 2/3 × √15/4 + √5/3 × 1/4 = \(\frac{2\sqrt{15}+\sqrt5}{12}\)

Hence, The Correct Answer is \(\frac{2\sqrt{15}+\sqrt5}{12}\)

Latest Bihar STET Updates

Last updated on Jan 29, 2025

-> The Bihar STET 2025 Notification will be released soon.

->  The written exam will consist of  Paper-I and Paper-II  of 150 marks each. 

-> The candidates should go through the Bihar STET selection process to have an idea of the selection procedure in detail.

-> For revision and practice for the exam, solve Bihar STET Previous Year Papers.

Get Free Access Now
Hot Links: teen patti flush teen patti gold real cash teen patti 3a teen patti master game mpl teen patti