Finding values of expressions

Chapter 8 Class 10 Introduction to Trignometry
Concept wise

### Transcript

Example 3 Consider Δ ACB, right-angled at C, in which AB = 29 units, BC = 21 units and ∠ ABC = θ (see fig.). Determine the values of (i) cos2 θ + sin2 θ, Step 1 : Finding sides of triangle In right triangle ABC, Using Pythagoras theorem (Hypotenuse)2 = (Height)2 + (Base)2 AB2 = AC2 + BC2 AC2 = AB2 – BC2 AC2 = (29)2 – (21)2 Using a2 – b2 = (a + b) (a – b) AC2 = (29 – 21) (29 + 21) AC2 = (8) (50) AC = √(𝟖 × 𝟓𝟎) AC = √400 AC = √(20^2 ) AC = 20 Step 2 : Finding sin θ , cos θ Now, sin 𝜃 = (𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒 𝜃)/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 sin 𝜃 = 𝐴𝐶/𝐴𝐵 sin 𝜽 = 𝟐𝟎/𝟐𝟗 cos 𝜃 = (𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒 𝜃)/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 cos 𝜃 = 𝐵𝐶/𝐴𝐵 cos 𝜽 = 𝟐𝟏/𝟐𝟗 We have to find out , cos2 𝜃 + sin2 𝜃 Putting values = (𝟐𝟏/𝟐𝟗)^𝟐+(𝟐𝟎/𝟐𝟗)^𝟐 = ((21)2 + (20)2)/292 = (441 + 400)/841 = 𝟖𝟒𝟏/𝟖𝟒𝟏 = 1 So, cos2𝜽 + sin2 𝜽 = 1 Example 3 Consider Δ ACB, right-angled at C, in which AB = 29 units, BC = 21 units and ∠ ABC = θ(see fig.). Determine the values of (ii) cos2 θ – sin2 θ. cos2 𝜃 – sin2 𝜃 Putting values = (𝟐𝟏/𝟐𝟗)^𝟐−(𝟐𝟎/𝟐𝟗)^𝟐 = ((21)2 − (20)2)/292 Using a2 – b2 = (a + b) (a – b) = ((21 + 20)(21 − 20))/292 = 𝟒𝟏/𝟖𝟒𝟏

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.