Addition formulae for sin(A±B), cos(A±B), tan(A±B). Double angle formulae and their applications.
Answer at least 3 of 3 correctly to complete this section.
The addition formulae are among the most powerful tools in A-Level trigonometry. They let you expand expressions like sin(A+B) and cos(A−B) — something that is emphatically not the same as sinA+sinB.
From these addition formulae, you derive the double angle formulae, which are used constantly in integration, solving equations, and proving identities. If you know the addition formulae cold, you can derive everything else on the spot.
Common error: sin(A+B)=sinA+sinB. You cannot distribute sin over addition any more than you can distribute a square root. Try it with numbers: sin(30°+60°)=sin90°=1, but sin30°+sin60°=21+23≈1.37.
sin(A+B)=sinAcosB+cosAsinB
sin(A−B)=sinAcosB−cosAsinB
Find the exact value of sin75°.
Write 75°=45°+30°:
sin75°=sin(45°+30°)=sin45°cos30°+cos45°sin30°
=22×23+22×21=46+2
Practice: Find the exact value of sin15° using sin(45°−30°).
Answer: sin15°=sin45°cos30°−cos45°sin30°=46−2.
cos(A+B)=cosAcosB−sinAsinB
cos(A−B)=cosAcosB+sinAsinB
Sign pattern for cosine: Notice the signs are opposite to what you might expect. cos(A+B) has a minus between the terms, while cos(A−B) has a plus.
Given that sinA=53 (A acute) and cosB=1312 (B acute), find the exact value of cos(A−B).
First find the missing values using Pythagoras:
cosA=54 (since A is acute) and sinB=135 (since B is acute).
cos(A−B)=cosAcosB+sinAsinB=6548+6515=6563
Practice: Using the same values, find cos(A+B).
Answer: cos(A+B)=cosAcosB−sinAsinB=6548−6515=6533.
tan(A+B)=1−tanAtanBtanA+tanBtan(A−B)=1+tanAtanBtanA−tanB
Note: The tangent formula has a denominator — tan(A+B)=tanA+tanB.
Answer at least 3 of 3 correctly to complete this section.
The double angle formulae are simply the addition formulae with B=A.
For sine: Set B=A in sin(A+B)=sinAcosB+cosAsinB:
sin2A=sinAcosA+cosAsinA=2sinAcosA
For cosine: Set B=A in cos(A+B)=cosAcosB−sinAsinB:
cos2A=cos2A−sin2A
This is the first form. Using sin2A+cos2A=1, we get two more:
cos2A=2cos2A−1(replacing sin2A=1−cos2A)
cos2A=1−2sin2A(replacing cos2A=1−sin2A)
For tangent: Set B=A in tan(A+B)=1−tanAtanBtanA+tanB:
tan2A=1−tan2A2tanA
| Form | Best used when… |
|---|---|
| cos2A−sin2A | Both sinA and cosA are known |
| 2cos2A−1 | You want to eliminate sin (equation involves only cos) |
| 1−2sin2A | You want to eliminate cos (equation involves only sin) |
Rearrangements you need: From cos2A=2cos2A−1, we get cos2A=21+cos2A. From cos2A=1−2sin2A, we get sin2A=21−cos2A. These are essential for integration later in the course.
Given that cosθ=53 and θ is acute, find the exact values of sin2θ and cos2θ.
First: sinθ=54 (Pythagoras, θ acute).
sin2θ=2sinθcosθ=2×54×53=2524
cos2θ=cos2θ−sin2θ=259−2516=−257
Check: sin22θ+cos22θ=625576+62549=625625=1 ✓
Answer at least 3 of 3 correctly to complete this section.
When an equation involves both sin2θ (or cos2θ) and sinθ (or cosθ), replace the double angle using the appropriate formula to get everything in terms of a single angle.
Solve cos2θ+3sinθ=2 for 0≤θ≤2π.
Use cos2θ=1−2sin2θ (because the other terms involve sinθ):
1−2sin2θ+3sinθ=2
−2sin2θ+3sinθ−1=0
Multiply by −1: 2sin2θ−3sinθ+1=0
Factorise: (2sinθ−1)(sinθ−1)=0
sinθ=21orsinθ=1
θ=6π,65π,2π
Solve sin2θ=cosθ for 0≤θ≤2π.
Replace sin2θ=2sinθcosθ:
2sinθcosθ=cosθ
2sinθcosθ−cosθ=0
cosθ(2sinθ−1)=0
Do not divide by cosθ! Factorise instead.
Either cosθ=0: θ=2π or θ=23π
Or sinθ=21: θ=6π or θ=65π
Solutions: θ=6π,2π,65π,23π.
Prove that 1+cos2Asin2A=tanA.
Replace the double angle formulae:
LHS=1+(2cos2A−1)2sinAcosA=2cos2A2sinAcosA=cosAsinA=tanA=RHS□
Why cos2A=2cos2A−1? Because the denominator has a +1, which cancels with the −1 from this form, leaving a clean 2cos2A. Choosing the right form of cos2A is the key to clean proofs.
Answer at least 3 of 3 correctly to complete this section.
(a) Show that cos(A+B)+cos(A−B)=2cosAcosB. [2 marks]
(b) Hence find the exact value of cos75°+cos15°. [3 marks]
(a) Expand each term:
cos(A+B)=cosAcosB−sinAsinB
cos(A−B)=cosAcosB+sinAsinB
Adding: cos(A+B)+cos(A−B)=2cosAcosB□
(b) Write 75°=45°+30° and 15°=45°−30°. Then A=45°, B=30°:
cos75°+cos15°=2cos45°cos30°=2×22×23=26
Find the exact value of tan75°.
tan75°=tan(45°+30°)=1−tan45°tan30°tan45°+tan30°
=1−1×311+31=1−311+31
Multiply numerator and denominator by 3:
=3−13+1
Rationalise by multiplying by 3+13+1:
=3−1(3+1)2=23+23+1=24+23=2+3
Answer at least 3 of 4 correctly to complete this section.
Lock in what you've learned with exam-style questions and spaced repetition.
Fundamental trig identities and using them to simplify expressions and solve equations.
Expressing a sin x + b cos x in the form R sin(x+a) or R cos(x+a). Solving equations and finding maxima/minima.
Constructing proofs of trigonometric identities using known results.