       1. Chapter 3 Class 11 Trigonometric Functions
2. Serial order wise
3. Miscellaneous

Transcript

Misc 8 Find the value of sin 𝑥/2 , cos 𝑥/2 and tan 𝑥/2 in each of the following : tan⁡𝑥 = – 4/3 , 𝑥 in quadrant II Given that x is in quadrant II So, 90° < x < 180° Dividing with 2 all sides (90°)/2 < 𝑥/2 < (180°)/2 45° < 𝑥/2 < 90° So, 𝑥/2 lies in Ist quadrant In Ist quadrant, sin , cos & tan are positive ⇒ sin 𝑥/2 , cos 𝑥/2 and tan 𝑥/2 are positive Given tan x = −4/3 tan x = (2 tan⁡(𝑥/2))/(1 − 𝑡𝑎𝑛2(𝑥/2) ) −4/3 = (2 tan⁡(𝑥/2))/(1 − 𝑡𝑎𝑛2(𝑥/2) ) –4( 1 – tan2 (𝑥/2)) = 3× 2 tan (𝑥/2) –4 × 1 – (–4) × tan2 (𝑥/2) = 6 tan (𝑥/2) –4 + 4 tan2 (𝑥/2) = 6 tan (𝑥/2) –4 + 4 tan2 (𝑥/2) – 6 tan (𝑥/2) = 0 Replacing tan 𝑥/2 by a – 4 + 4a2 – 6a = 0 4a2 – 6a – 4 = 0 4a2 – 8a + 2a – 4 = 0 4a(a – 2) + 2 ( a – 2) = 0 (4a + 2) ( a – 2) = 0 Hence 4a + 2 = 0 4a = - 2 a = (−2)/( 4) a = −1/2 So, a = −1/2 or a = 2 Hence, tan 𝑥/2 = −1/2 or tan 𝑥/2 = 2 Since, 𝑥/2 lies in Ist quadrant tan 𝑥/2 is positive, ∴ tan 𝑥/2 = 2 Now, We know that 1 + tan2 x = sec2 x Replacing x with 𝑥/2 1 + tan2 𝑥/2 = sec2 𝑥/2 1 + (2)2 = sec2 𝑥/2 1 + 4 = sec2 x/2 5 = sec2 𝑥/2 sec2 𝑥/2 = 5 sec 𝑥/2 = ± √5 Since 𝑥/2 lie on the 1st Quadrant, sec 𝑥/2 is positive in the 1st Quadrant So sec 𝑥/2 = √5 sec 𝑥/2 = √5 ⇒ 1/cos⁡〖 𝑥/2〗 = √5 ⇒ 1/√5 = cos 𝑥/2 ⇒ cos 𝑥/2 = 1/√5 cos 𝑥/2 = 1/√5 × √5/√5 cos 𝑥/2 = √5/5 We know that sin2x + cos2x = 1 Replacing x with 𝑥/2 sin2 𝑥/2 + cos2 𝑥/2 = 1 sin2 𝑥/2 = 1 – cos2 𝑥/2 Putting cos 𝑥/2 = √5/5 sin2 𝑥/2 = 1 – (√5/5)2 sin2 𝑥/2 = 1 – 5/25 sin2 𝑥/2 = 1 – 1/5 sin2 𝑥/2 = (5 − 1)/5 sin2 𝑥/2 = 4/5 sin 𝑥/2 = ± √(4/5) = ± √4/√5 = ± 2/√5 = ± 2/√5 × √5/√5 = ± (2√5)/5 Since 𝑥/2 lie on the 1st Quadrant , sin 𝑥/2 is positive in the 1st Quadrant So sin 𝑥/2 = (2√5)/5 Hence, tan 𝑥/2 = 2 , cos 𝑥/2 = √5/5 & sin 𝑥/2 = (2√5)/5

Miscellaneous 