1007_Tut1_Sol

1007_Tut1_Sol - MATH 1007 A Tutorial#1 SOLUTIONS Example of a proof To show that 2 csc 2x = sec x csc x we start with one side say the left side(LS

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
MATH 1007 A Tutorial #1 Sept 22, 2006 SOLUTIONS Example of a proof: To show that 2 csc 2 x = sec x csc x , we start with one side, say the left side (LS), and show that it is equal to the other side: LS = 2 csc 2 x = 2 sin 2 x = 2 2 sin x cos x = 1 sin x cos x = ± 1 sin x ²± 1 cos x ² = csc x sec x = RS 1. [2 marks] Prove the following trigonometric identity, using the approach above. (sin x + cos x ) 2 = 1 + sin 2 x Solution: LS = (sin x + cos x ) 2 = sin 2 x + 2 sin x cos x + cos 2 x = (sin 2 x + cos 2 x ) + 2 sin x cos x = 1 + 2 sin x cos x = RS 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2. [2 marks] Find the slope and the equation of the tangent line to the curve y = x 3 at the point where x = - 1. Solution: y 0 = 3 x 2 y 0 ( - 1) = 3( - 1) 2 = 3 Since ( - 1) 3 = - 1, the point ( - 1 , - 1) is on y = x 3 . We now find the equation of the line through ( - 1 , - 1) with slope 3: y - y 1 = m ( x - x 1 ) y - ( - 1) = 3( x - ( - 1)) y + 1 = 3 x + 3 y = 3 x + 2 Therefore the equation of the tangent line to y = x 3 at the point where x = - 1 is y = 3 x + 2, and the slope of this tangent line is 3. 3. [4 marks]
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 03/22/2010 for the course MATH 1007 taught by Professor Unknown during the Fall '07 term at Carleton.

Page1 / 3

1007_Tut1_Sol - MATH 1007 A Tutorial#1 SOLUTIONS Example of a proof To show that 2 csc 2x = sec x csc x we start with one side say the left side(LS

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online