2DTrigDblAngles

toc = Double Angle Formulas =

NOTE: Maths Methods students are not expected to have the double angle formulas memorised (and therefore they will not be used in Methods Exam 1). They are expected to be able to use the formulas in Exam 2.

NOTE: Specialist Maths students have the double angle formulas on their formula sheet.

Obtaining the Double Angle Formulas
{you are not required to be able to do this, but it is well withing the capabilities of a Year 12 maths student}

Given: ... ... sin(A + B) = sin(A)cos(B) + cos(A)sin(B)

If A = B, we get: ... ... sin(A + A) = sin(A)cos(A) + cos(A)sin(A) gives: ... ... sin(2A) = 2sin(A)cos(A)

Also, given: ... ... cos(A + B) = cos(A)cos(B) – sin(A)sin(B)

If A = B, we get: ... ... cos(A + A) = cos(A)cos(A) – sin(A)sin(A) gives ... ... cos(2A) = cos 2 (A) – sin 2 (A) or ... ... cos(2A) = 2cos 2 (A) – 1 or ... ... cos(2A) = 1 – 2sin 2 (A)

and

math . \qquad \tan(A+B) = \dfrac {\tan(A)+\tan(B)}{1-\tan(A)\tan(B)} math gives math . \qquad \tan(2A) = \dfrac {2\tan(A)}{1-\tan^2(A)} math

**Example 1**
math \text{For } \sin(x)=0.6, \quad x \in \left[\dfrac {\pi}{2}, \; \pi \right] math

a) Find sin(2x) correct to 2 decimal places

Using pythagoras math \\ . \qquad \sin^2(x)+\cos^2(x)=1 \\. \\ . \qquad 0.6^2+\cos^2(x)=1 \\. \\ . \qquad \cos^2(x)=0.64 \\. \\ . \qquad \cos(x)=\pm\sqrt{0.64} math

But x is in 2nd Quadrant, so math . \qquad \cos(x)=-\sqrt{0.64} math

Now use the double angle formula to find sin(2x) math \\ . \qquad \sin(2x)=2\sin(x)\cos(x) \\. \\ . \qquad \sin(2x)=2\times0.6\times-\sqrt{0.64} \\. \\ . \qquad \sin(2x)=-1.2\sqrt{0.64}=-0.96 math

b) Find sin(x/2) correct to 2 decimal places

Use cos(2A) = 1 – 2sin 2 (A) ... {with A=x/2}

math \\ . \qquad \cos(x)=1-2\sin^2 \left( \dfrac{x}{2} \right) \\. \\ . \qquad -\sqrt{0.64}=1-2\sin^2 \left( \dfrac{x}{2} \right) \\. \\ . \qquad 2\sin^2 \left( \dfrac{x}{2} \right) =1+\sqrt{0.64} math . math \\ . \qquad \sin^2 \left( \dfrac{x}{2} \right) = \dfrac {1+\sqrt{0.64}}{2} \\. \\ . \qquad \sin \left( \dfrac{x}{2} \right) = \pm \sqrt {\dfrac {1+\sqrt{0.64}}{2}} = \pm0.95 math

Since x is in 2nd Quadrant, it follows that x/2 is in 1st Quadrant, so sin is positive math . \qquad \sin \left( \dfrac{x}{2} \right) = 0.95 math

** Example 2 **
math \text{Find } \; \sin \left( \dfrac {\pi}{8} \right) math


 * Solution: **

Use cos(2A) = 1 – 2sin 2 (A) ... with A = pi/8

math . \qquad \cos \left( \dfrac {\pi}{4} \right) = 1 - 2 \sin^2 \left( \dfrac{\pi}{8} \right) math

insert exact value math . \qquad \dfrac {\sqrt{2}}{2} = 1 - 2 \sin^2 \left( \dfrac{\pi}{8} \right) math

rearrange math . \qquad 2 \sin^2 \left( \dfrac{\pi}{8} \right) = 1 - \dfrac {\sqrt{2}}{2} = \dfrac {2-\sqrt{2}}{2} math . math . \qquad \sin^2 \left( \dfrac{\pi}{8} \right) = \dfrac {2-\sqrt{2}}{4} math

take the square root of both sides math . \qquad \sin \left( \dfrac{\pi}{8} \right) = \pm \dfrac {\sqrt{2-\sqrt{2}}}{2} math

but pi/8 is in Quadrant 1 so sin(pi/8) is positive math . \qquad \sin \left( \dfrac{\pi}{8} \right) = \dfrac {\sqrt{2-\sqrt{2}}}{2} math

**See also** www.mathsonline.com.au Y12Extenstion --> Trigonometry --> Further Trigonometry --> Lesson 3, 4

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