Assumed prior knowledge - GCSE level right-angled trigonometry; Pythagoras’ theorem and the Sine
and Cosine Rules.
1 BASIC TRIGONOMETRIC IDENTITIES
1.1 Using trigonometry in a right-angled, unit hyptenuse triangle gives the following 3 results
sinθ =
opp
hyp
=
opp
1
∴ opp = sinθ
cosθ =
ad j
hyp
=
ad j
1
∴ ad j = cosθ
tanθ =
opp
ad j
∴ tanθ ≡
sinθ
cosθ
[identity 1]
1.2 Applying Pythagoras’ theorem to the same triangle as in 1.1
opp2 + ad j2 = 1 ∴ sin2 θ + cos2 θ ≡ 1 [identity 2]
2.1 Dividing identity 2 by cos2 θ
sin2 θ
cos2 θ
2
θ
+ cos
≡
cos2 θ
1
cos2 θ
∴ tan2 θ + 1 ≡ sec2 θ [identity 3]
2.2 Dividing identity 2 by sin2 θ
sin2 θ
sin2 θ
2
θ
+ cos
≡
sin2 θ
1
sin2 θ
∴ 1 + cot 2 θ ≡ cosec2 θ [identity 4]
3 THE ADDITION FORMULAE
3.1 Using the distance between two points formula on two right-angled, unit hypotenuse triangles with
different angles A and B
x2 = (cosA − cosB)2 + (sinA − sinB)2 = cos2 A − 2cosAcosB + cos2 B + sin2 A − 2sinAsinB + sin2 B
x2 = (cos2 A + sin2 A) + (cos2 B + sin2 B) − 2(cosAcosB + sinAsinB)
x2 = 2 − 2(cosAcosB + sinAsinB)
Using the cosine rule on the same two triangles as above
x2 = 12 + 12 − 2 × 1 × 1 × cos(A − B)
x2 = 2 − 2cos(A − B)
Combining both the above statements
2 − 2cos(A − B) ≡ 2 − 2(cosAcosB + sinAsinB)
∴ cos(A − B) ≡ cosAcosB + sinAsinB [identity 5a]
3.2 Replace B by −B in identity 5a
cos(A + B) ≡ cosAcos(−B) + sinAsin(−B)
since cosine is an even function cos(−B) ≡ cos(B)
since sine is an odd function sin(−B) ≡ −sin(B)
∴ cos(A + B) ≡ cosAcosB − sinAsinB [identity 5b]
3.3 Combining indentities 5a and 5b
∴ cos(A ± B) ≡ cosAcosB ∓ sinAsinB [identity 5]
3.4 Using the fact that cos( π2 − θ ) ≡ sinθ since the two functions are related translatively
sin(A + B) ≡ cos( π2 − (A + B)) ≡ cos(( π2 − A) − B)
sin(A + B) ≡ cos( π2 − A)cosB + sin( π2 − A)sinB
∴ sin(A + B) ≡ sinAcosB + cosAsinB [identity 6a]
3.5 ReplaceB by −B in identity 6a
sin(A − B) ≡ sinAcos(−B) + cosAsin(−B)
2
since cosine is an even function cos(−B) ≡ cos(B)
since sine is an odd function sin(−B) ≡ −sin(B)
∴ sin(A − B) ≡ sinAcosB − cosAsinB [identity 6b]
3.6 Combining identities 6a and 6b
∴ sin(A ± B) ≡ sinAcosB ± cosAsinB [identity 6]
3.7 For tangents
tan(A + B) ≡
sin(A+B)
cos(A+B)
≡
sinAcosB+cosAsinB
cosAcosB−sinAsinB
divide top and bottom through by cosAcosB
∴ tan(A + B) ≡
tanA+tanB
1−tanAtanB
[identity 7a]
since tan(−B) ≡ −tan(B)
∴ tan(A − B) ≡
tanA−tanB
1+tanAtanB
[identity 7b]
combining identities 7a and 7b
∴ tan(A ± B) ≡
tanA±tanB
1∓tanAtanB
[identity 7]
3.8 Related identities
cos(A + B) ≡ cosAcosB − sinAsinB
cos(A − B) ≡ cosAcosB + sinAsinB
∴ cos(A + B) + cos(A − B) ≡ 2cosAcosB [identity 8i]
∴ cos(A − B) − cos(A + B) ≡ 2sinAsinB [identity 8ii]
sin(A + B) ≡ sinAcosB + cosAsinB
sin(A − B) ≡ sinAcosB − cosAsinB
∴ sin(A + B) + sin(A − B) ≡ 2sinAcosB [identity 8iii]
∴ sin(A + B) − sin(A − B) ≡ 2cosAsinB [identity 8iv]
4 THE DOUBLE ANGLE FORMULAE
4.1 Substituting B = A into the addition formulae yields
∴ cos2A ≡ cos2 A − sin2 A [identity 9]
∴ sin2A ≡ 2sinAcosA [identity 10]
∴ tan2A ≡
2tanA
1−tan2 A
[identity 11]
4.2 Replace cos2 A by 1 − sin2 A in identity9
cos2A ≡ 1 − 2sin2 A
∴ 2sin2 A ≡ 1 − cos2A [identity 12i]
4.3 Replace sin2 A by 1 − cos2 A in identity9
cos2A ≡ 2cos2 A − 1
3
∴ 2cos2 A ≡ 1 + cos2A [identity 12ii]
5 THE FORM asinx + bcosx
5.1 Equating to identity 5 or 6
asinx + bcosx ≡ Rsin(x + α) ≡ Rsinxcosα + Rcosxsinα
asinx − bcosx ≡ Rsin(x − α) ≡ Rsinxcosα − Rcosxsinα
acosx + bsinx ≡ Rcos(x − α) ≡ Rcosxcosα + Rsinxsinα
acosx − bsinx ≡ Rcos(x + α) ≡ Rcosxcosα − Rsinxsinα
[identities 13]
a and b must be positive and α has to be acute (0 < α < π2 ) - always try to choose the form which
produces the terms in the right order and with the correct sign then α can always be acute.
Using Pythagoras’ theorem and right-angled trigonometry
√
R = (a2 + b2 )
and
Rcosα = a and Rsinα = b
∴ tanα =
b
a
or
Rsinα = b and Rcosα = a
∴ tanα =
a
b
depending on which form has been used.
4