Addition and Double Angle Formulae

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Addition and Double Angle Formulae

We’re now about to take a look at some formulae which describe angle addition.

If you don’t know your key trig values already, now would be the time to learn!

Make sure you are happy with the following topics before continuing.

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Finding Expressions for Addition Formulae

Here’s three new formulae in sin⁡textcolor{blue}{sin}, cos⁡textcolor{limegreen}{cos} and tan⁡textcolor{red}{tan}:

sin⁡(A±B)=sin⁡Acos⁡B±sin⁡Bcos⁡Atextcolor{blue}{sin} (textcolor{purple}{A} ± textcolor{orange}{B}) = textcolor{blue}{sin} textcolor{purple}{A} textcolor{limegreen}{cos} textcolor{orange}{B} ± textcolor{blue}{sin} textcolor{orange}{B} textcolor{limegreen}{cos} textcolor{purple}{A}

cos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡Btextcolor{limegreen}{cos} (textcolor{purple}{A} ± textcolor{orange}{B}) = textcolor{limegreen}{cos} textcolor{purple}{A} textcolor{limegreen}{cos} textcolor{orange}{B} mp textcolor{blue}{sin} textcolor{purple}{A} textcolor{blue}{sin} textcolor{orange}{B}

tan⁡(A±B)=tan⁡A±tan⁡B1∓tan⁡Atan⁡Btextcolor{red}{tan} (textcolor{purple}{A} ± textcolor{orange}{B}) = dfrac{textcolor{red}{tan} textcolor{purple}{A} ± textcolor{red}{tan} textcolor{orange}{B}}{1 mp textcolor{red}{tan} textcolor{purple}{A} textcolor{red}{tan} textcolor{orange}{B}}

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Note:

You might have noticed the “minus-plus” symbols above (∓mp). This is no mistake, and it is not the same as “plus-minus, ±pm“. The important thing to remember with this notation is that whichever symbol is chosen (top or bottom), must be used on the other side of the equation.

So, for example,

cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡Btextcolor{limegreen}{cos} (textcolor{purple}{A} + textcolor{orange}{B}) = textcolor{limegreen}{cos} textcolor{purple}{A} textcolor{limegreen}{cos} textcolor{orange}{B} – textcolor{blue}{sin} textcolor{purple}{A} textcolor{blue}{sin} textcolor{orange}{B}

and

cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡Btextcolor{limegreen}{cos} (textcolor{purple}{A} – textcolor{orange}{B}) = textcolor{limegreen}{cos} textcolor{purple}{A} textcolor{limegreen}{cos} textcolor{orange}{B} + textcolor{blue}{sin} textcolor{purple}{A} textcolor{blue}{sin} textcolor{orange}{B}

Double Angle Formulae

We can extend our addition formulae to two equal angles, also.

So, we have

sin⁡(2A)=2sin⁡Acos⁡Atextcolor{blue}{sin (2A)} = 2textcolor{blue}{sin A} textcolor{limegreen}{cos A}

cos⁡(2A)=cos⁡2A−sin⁡2A=2cos⁡2A−1=1−2sin⁡2Abegin{aligned}textcolor{limegreen}{cos (2A)} &= textcolor{limegreen}{cos ^2 A} – textcolor{blue}{sin ^2 A}[1.2em]&=2textcolor{limegreen}{cos^2 A}-1[1.2em]&=1-2textcolor{blue}{sin^2 A}end{aligned}

tan⁡(2A)=2tan⁡A1−tan⁡2Atextcolor{red}{tan (2A)} = dfrac{2textcolor{red}{tan A}}{1 – textcolor{red}{tan ^2 A}}

No worries if you forget these, you can just derive them from the addition formulae by setting B=AB = A.

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Example: Finding Exact Values

Find the exact value of sin⁡75°textcolor{blue}{sin} 75°, in the form 1ab(c+d)dfrac{1}{asqrt{b}}(c + sqrt{d}).

[3 marks]

sin⁡75°=sin⁡(30°+45°)textcolor{blue}{sin} 75° = textcolor{blue}{sin} (30° + 45°)

=sin⁡30°cos⁡45°+sin⁡45°cos⁡30°= textcolor{blue}{sin} 30° textcolor{limegreen}{cos} 45° + textcolor{blue}{sin} 45° textcolor{limegreen}{cos} 30°

=(12×12)+(12×32)= left( dfrac{1}{2} times dfrac{1}{sqrt{2}} right) + left( dfrac{1}{sqrt{2}} times dfrac{sqrt{3}}{2} right)

=122+322=122(1+3)= dfrac{1}{2sqrt{2}} + dfrac{sqrt{3}}{2sqrt{2}} = dfrac{1}{2sqrt{2}}(1 + sqrt{3})

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Addition and Double Angle Formulae Example Questions

Question 1: Find the exact value of cos⁡165°cos 165°.

[3 marks]

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cos⁡165°=cos⁡(210°−45°)cos 165° = cos (210° – 45°)

 

=cos⁡210°cos⁡45°+sin⁡210°sin⁡45°= cos 210° cos 45° + sin 210° sin 45°

 

=(−32×12)+(−12×12)= left( dfrac{-sqrt{3}}{2} times dfrac{1}{sqrt{2}}right) + left( dfrac{-1}{2} times dfrac{1}{sqrt{2}}right)

 

=−3−122= dfrac{-sqrt{3} – 1}{2sqrt{2}}

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Question 2: Given that tan⁡75°=2+3tan 75° = 2 + sqrt{3}, find the exact value of tan⁡150°tan 150°.

[2 marks]

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tan⁡150°=2(2+3)1−(2+3)2tan 150° = dfrac{2(2 + sqrt{3})}{1 – (2 + sqrt{3})^2}

 

=4+23−6−43=2+3−3−23= dfrac{4 + 2sqrt{3}}{-6 – 4sqrt{3}} = dfrac{2 + sqrt{3}}{-3 – 2sqrt{3}}

 

=(2+3)(−3+23)(−3−23)(−3+23)= dfrac{(2 + sqrt{3})(-3 + 2sqrt{3})}{(-3 – 2sqrt{3})(-3 + 2sqrt{3})}

 

=−6−33+43+69−12= dfrac{-6 – 3sqrt{3} + 4sqrt{3} + 6}{9 – 12}

 

=3−3=−13= dfrac{sqrt{3}}{-3} = dfrac{-1}{sqrt{3}}

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Question 3: Using angle addition formulae, prove that sin⁡(x+π2)=cos⁡xsin left( x + dfrac{pi}{2}right) = cos x.

[2 marks]

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sin⁡(x+π2)=sin⁡xcos⁡π2+sin⁡π2cos⁡xsin left( x + dfrac{pi}{2}right) = sin x cos dfrac{pi}{2} + sin dfrac{pi}{2} cos x

 

=(sin⁡x×0)+(cos⁡x×1)= (sin x times 0) + (cos x times 1)

 

=cos⁡x= cos x

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Additional Resources

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Formula Booklet

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Specification Points Covered

E6 – Understand and use double angle formulae; use of formulae for sin⁡(A±B)sin{(Apm B)}, cos⁡(A±B)cos{(Apm B)} and tan⁡(A±B)tan{(Apm B)}; understand geometrical proofs of these formulae
Understand and use expressions for acos⁡θ+bsin⁡θacos{theta}+bsin{theta} in the equivalent forms of rcos⁡(θ±α)rcos{(theta pm alpha)} or rsin⁡(θ±α)rsin{(theta pm alpha)}

Addition and Double Angle Formulae Worksheet and Example Questions

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Double Angle Formulae

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Related Topics

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Basic Trig Identities

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