Find the value of the trigonometric function. sin - 11 π 3

Given trigonometric function is sin - 11 π 3 We have, sin - 11 π 3 = - sin 11 π 3 = - sin 4 π - π 3 The value of sin x repeats after an interval of 2 π i.e. 2 × 2 π . = - - sin π 3 = sin π 3 = 3 2 Therefore, sin - 11 π 3 = 3 2 .

Find the value of the trigonometric function. cot ⁡ - 15 π 4

Given trigonometric function is cot ⁡ - 15 π 4 cot ⁡ - 15 π 4 = - cot ⁡ 15 π 4 = - cot ⁡ 4 π - π 4 The value of cot x repeats after an interval of 2 π i.e. 2 × 2 π . = - cot ⁡ - π 4 = - - cot ⁡ π 4 = c o t ⁡ π 4 = 1 Therefore, cot ⁡ - 15 π 4 = 1 .

Sketch the graph of following functions y = sin x 2

Given, y = sin x 2 For, x = π 6 ⇒ y = 3 - 1 2 2 ; x = π 3 ⇒ y = 1 2 ; x = π 2 ⇒ y = 1 2 ; x = π ⇒ y = 1 ; x = 3 π ⇒ y = 1 2 ; x = 2 π ⇒ y = 0 Hence, the graph becomes : Here, the frequency of the function is decreased by 0 . 5 times.

If tan θ = a b , then show that a sin θ - b cos θ a sin θ + b cos θ = a 2 - b 2 a 2 + b 2 .

Given that, tan θ = a b . We have to prove that, a sin θ - b cos θ a sin θ + b cos θ = a 2 - b 2 a 2 + b 2 LHS = a sin θ - b cos θ a sin θ + b cos θ = a sin θ cos θ - b cos θ cos θ a sin θ cos θ + b cos θ cos θ [Divide both the numerator and the denominator by cos θ ] = a tan θ - b a tan θ + b = a × a b - b a × a b + b [Substitute the value of tan θ = a b that is given] = a 2 - b 2 a a 2 + b 2 b = a 2 - b 2 a 2 + b 2 = RHS [Proved]