Trigonometric Functions

sin2θ = 2 sinθ cosθ

$cos2θ = cos^2 θ – sin^2 θ$
$cos2θ = 1 – 2 sin^2 θ$
$cos2θ = 2 cos^2 θ – 1$

$tan2θ = {2 tanθ}/{1 – tan^2 θ}$

$sin2θ = {2tan θ}/{1 + tan^2 θ}$

$cos2θ = {1 – tan^2 θ}/{1 + tan^2 θ}$

$sin3θ = 3 sinθ – 4 sin^3 θ$

$cos3θ = 4 cos^3 θ – 3 cosθ$

$tan3θ = {3 tanθ – tan^3 θ}/{1 – 3 tan^2 θ}$

$1 + cos2θ = 2 cos^2 θ$


$1 + sin2θ = (cosθ + sinθ)^2$

$1 – cos2θ = 2 sin^2 θ$

$1 – sin2θ = (cosθ – sinθ)^2$

$sinC + sinD = 2 sin({C+D}/{2}) cos({C-D}/{2})$

$sinC – sinD = 2 cos({C+D}/{2}) sin ({C – D}/{2})$

$cosC + cosD = 2 cos({C+D}/{2}) cos({C-D}/{2})$

$cosC – cosD = -2 sin({C+D}/{2}) sin({C-D}/{2})$

2 sinA cosB = sin(A+B) + sin(A-B)
2 cosA sinB = sin(A+B) – sin(A-B)
2 cosA cosB = cos(A+B) + cos(A-B)
2 sinA sinB = cos(A-B) – cos(A+B)

sin(A+B) = sinC         cos(A+B) = -cosC
sin(B+C) = sinA         cos(B+C) = -cosA
sin(A+C) = sinB         cos(A+C) = -cosB

$sin({A+B}/{2}) = cos(C/2)$         $cos({A+B}/{2}) = sin(C/2)$

$sin({B+C}/{2}) = cos(A/2)$         $cos({B+C}/{2}) = sin(A/2)$


$sin({A+C}/{2}) = cos(B/2)$         $cos({A+C}/{2}) = sin(B/2)$

Proof: In △ABC, A+B+C = 180°
A+B = 180° – C
cos(A+B) = cos(π – c) = – cosC

XII

sinx = 0 i.e. x = nπ
cosx = 0 i.e. x = (2n+1) $π/2$
tanx = 0 i.e. x = nπ

sinx = sinα i.e. x = nπ + $(-1)^n$ α
cosx = cosα i.e. x = nπ ± α
tanx = tanα i.e. x = nπ + α

$sin^2 x = sin^2 α$ i.e. x = nπ ± α
$cos^2 x = cos^2 α$ i.e. x = nπ ± α
$tan^2 x = tan^2 α$ i.e. x = nπ ± α

In △ABC, $a/{sinA} = b/{sinB} = c/{sinC}$ (sine rule)

In △ABC, (cosine rule)
$(i) a^2 = b^2 + c^2 – 2bc cosA$
$(ii) b^2 = a^2 + c^2 – 2ac cosB$
$(iii) c^2 = a^2 + b^2 – 2ab cosC$

In △ABC, (projection rule)
(i) a = c cosB + b cosC

(ii) b = a cosC + c cosA
(iii) c = a cosB + b cosA

In △ABC, if a+b+c = 2s then

$sin A/2 = √{{(s-b)(s-c)}/{bc}}$         $cos A/2 = √{{s(s-a)}/{bc}}$

$sin B/2 = √{{(s-a)(s-c)}/{ac}}$         $cos B/2 = √{{s(s-b)}/{ac}}$

$sin C/2 = √{{(s-a)(s-c)}/{ab}}$         $cos C/2 = √{{s(s-c)}/{ab}}$

$tan A/2 = √{{(s-b)(s-c)}/{s(s-a)}}$
$tan B/2 = √{{(s-a)(s-c)}/{s(s-b)}}$
$tan C/2 = √{{(s-a)(s-b)}/{s(s-c)}}$

For △ABC, its area is given by:
$ 1/2 ab sinC$         $1/2 bc sinA$         $1/2 ac sinB$

Heron’s formula
if a, b, c are lengths of sides BC, CA & AB, respectively. The area of △ABC is

$A = √{s(s-a)(s-b)(s-c)}$         where, 2s = a+b+C

Napier’s Analogies
In △ABC, we have:

(i) $tan ({B-C}/{2}) = ({b-c}/{b+c}) cot A/2$

(ii) $tan ({A-B}/{2}) = ({a-b}/{a+b}) cot C/2$

(iii) $tan ({A-C}/{2}) = ({a-c}/{a+c}) cot B/2$