trigonometry

Q) prove that sin 6 θ + cos 6 θ = 1 – 3 sin 2 θ cos 2 θ Ans: Let’s start from LHS LHS = prove that sin 6 θ + cos 6 θ = (sin 2 θ) 3 + (cos 2 θ) 3 We know that a 3 + b 3 = (a + b) (a

Q) If 2 sin (A + B) = √3 and cos (A – B) = 1 , then find the measures of angles A and B. 0 <= A B, (A + B) <= 90 deg Ans: Let’s take each of the given values one by one: 2 sin(A + B) = √3 ∴ sin (A

Q) Evaluate: 2 sin2 300 sec 600 + tan2 600 Ans: We need to find the value of: 2 sin2 300 sec 600 + tan2 600 We know that sin 300 = , sec 600 = 2, tan 600 =  √3 by substituting these values in the given expression, we get: 2 sin2 300 sec 600 + tan2 600 = 2 (

  Q)  Prove that : Ans:  Let’s start from simplifying the LHS: LHS = = = Since, we need to get demnominator in simplified form, hence let’s multiply nominator and denominator by cos θ + sin θ – 1, we get: LHS = = = = = = We know that sin2 θ + cos2

Q) Evaluate: Ans: We need to find the value of: We know that cos 45 = , sec 30 = , cosec 30 = 2 by substituting these values in the given expression, we get: =  = = = …. Answer or to further simplify, we can multiply and divide the expression by (√3 –

Q) Prove that Ans: Method 1: We need to prove that Let’s start with squaring in both sides: We know that:  and ∴ ∴ Since LHS = RHS Hence Proved ! Method 2: Let’s start from LHS: Since, and ∴ LHS = = = Since, ∴ LHS = = =     = = =

Q) If x sin3 θ + y cos3 θ = sin θ cos θ and x sin θ = y cos θ, prove that x2 + y2 = 1 Ans: Given that x sin3 θ + y cos3 θ = sin θ cos θ ∴  x sin θ sin2 θ + y cos θ cos2 θ = sin θ

  Q)  Evaluate: 2√2 cos 45° sin 30° + 2√3 cos 30° Ans:  Here, let’s start values of sin & cosine for θ = 300 and 450 We know that sin 300 = , cos 300 = , cos 450 = We submit these values in the given question, we get: 2√2 cos 45° sin

  Q)  If A = 60° and B = 30°, verify that : sin (A + B) = sin A cos B + cos A sin B Ans:  Here, let’s calculate the values of LHS & RHS one by one and check if it is verified. LHS = sin (A + B) = sin (60