Double Angle Formulae

Question 1

What is the formulae for Sin2x ?

Answer 1

Sin2x = 2SinxCosx

Question 2

What are the formulas for Cos2x ?

Answer 2

a) Cos2x = Cos^2x - Sin^2x

b) Cos2x = 2Cos^2x - 1

c) Cos2x = 1 - 2Sin^2x

Question 3

What is the formulae for tan2x ?

Answer 3

tan2x = 2tanx / 1 - tan^2x

Question 4

How would you write 4Cos^2(10x) - 2

Answer 4

we know that Cos2A = 2Cos^2A - 1

so we can rewrite 4Cos^2(10x) - 2 as:

2(2Cos^2(10x) -1)
A=10

=2(Cos20x)
=2Cos20x

Question 5

How would you work out this question? :

Given that x is acute, and that cos x = 0.6, find sin2x, Cos2x, and Sin3x with out using a calculator

Answer 5

We know that Cos=0.6 = 3/5
A=3 H=5
Use this to draw a right angled triangle, and use Pythagoras to work out O
find out sinx and Cosx from this, then sub these values in to the one in the Q

Question 6

Show that Sin3x = 3Sinx - 4Sin3x

Answer 6

1)Rewrite Sin3x as Sin(2x+x)
2)Use compound angle identity to rewrite this
3)Sub in the double angle formulae where necessary
4)Simplify and factorise

Sin3x = Sin(2x + x)
Sin3x = Sin2xCosx + Cos2xSinx
Sin3x = (2SinxCosx)Cosx + (1-2Sin^2x)Sinx
=2SinxCos^2x + Sinx - 2Sin^3x
and so on

Question 7

If you have 2CosA / Cos2A, how can you introduce Sin2A ?

Answer 7

do (2CosA / Cos2A) x (Sin2A / Sin2A)

then swap it around
eg 3/4 x 2/5 = 2/4 x 3/5

(Sin2A / Cos2A) x (2CosA / Sin2A)

becomes tan2A x 2CosA / 2SinACosA

Simplifies to tan2A/SinA