FORMULAE:
1) sin 2x= 2sin x cos x
2) Sin 3x = 3 sin x - 4 sin³x
3) Cos 2x = cos²x - sin²x
= 2 cos²x - 1
= 1 - 2 sin²x
= (1- tan²x)/(1+ tan²x)
4) 1 + cos 2x = 2 cos²x OR
Cos x= ± √{(1+ cos 2x)/2}
5) 1 - cos 2x= 2 sin²x OR
Cos x= ± √{(1- cos 2x)/2}
6) Cos 3x = 4 cos²x - 3 cos x
7) Tan x= (1- cos 2x)/sin 2x
= Sin 2x/(1+ cos 2x)
=± √{(1- cos 2x)/(1+ cos 2x)
8) Tan 2x = 2 tanx/(1- tan²x)
9) tan 3x=(3tan x -tan³x)/(1-3 tan²x)
10) sin18°=(√5 -1)/4=cos72°= sin π/10
11) cos 36°= (√5+2)/4= sin54°= cos π/5
12) sin15°= (√3-1)/2√2 = cos75°= sin π/2
13) cos15°= (√3+1)/2√2 = sin75°= cos π/2
14) tan π/12 = 2 -√3 =(√3-1)/(√3+1) = cot 5π/12
15) tan 5π/12 = 2 + √3 =(√3-1)/(√3+1) = cot π/12
16) tan 225°= √2 - 1 = cot 67.5°= cot 3π/8 = tan π/8
17) tan 67.5°= √2+ 1 = cot 22.5°
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EXERCISE -1
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1) Find Sin 2θ, cos2θ, tan2θ if
a) cosθ=⅗. 24/25, -7/25, -24/7
b) sinθ=⅘. 24/25, -7/25, -24/7
c) tanθ=½. 4/5, 3/5, 2
d) if cosA=⅗ find Sin2A
e) if sinA=⅗ find tan3A
f) If cos2a =24/25 find tan3A
2) EXPRESS:
a) cos4A in terms of sinA. 1 - 8 sin²A + 8 sin⁴A
b) tan4A in terms of tanA. (4 tanA - 4 tan⁴A)/(1+tan⁴A-6 tan²A)
c) sin4A in terms of tanA.
3) Prove:
a) Sin2α/(1-cos2α)= cotα
b) sin2θ/(1+cos2θ) = tanθ
c) cotβ + tanβ = 2csc2β
d) 1+ tanα tan2α = sec2α
e) tan²α = (sec2α -1)/(sec2α +1)
f) sec²θ(1+sec2θ) = 2sec2θ
g) (cosα+sinα)/(cosα-sinα) = tan2α + sec2α
h) tanβ(1+sec2β)=tan2β
i) tan2α/tanα = 1+sec2α
j) (cotθ+tanθ)/(cotθ-tanθ)=sec2θ
k) 2/(cotα-tanα) = tan2α
l) {sinα - √(1+sin2α)}/{cosα-√(1+sin2α)= cotα
m) tan(π/4 + α)=(1+sin2α)/cos2α
n) cos²(α+π/4)+cos²(α-π/4)=1
o) {(1-tanθ)/(1+tanθ)}² = (1-sin2θ)/(1+sin2θ)
p) tan(45+α)- tan(45-α) =2tan2α
q) tan(45+α)+ tan(45-α) =2sec2α
r) sec(45+α)sec(45-α) = 2sec2α
s) 4cosβcos(2π/3 +β) cos(4π/3+β) = cos(3β)
t) tan(π/4 +α/2)= secα +tanα = √{(1+sinα)/(1-sinα)}
u) (cotθ-tanθ)/(1-2sin2θ) = secθ cosec 2θ
v) cot15 - tan15 = 2tan60
w) tan15 + cot15 =4
x) (2cosα+1)(2cosα -1)=2cos2α +1
y) cos⁴α - sin⁴α = cos2α
z) cos⁶β -sin⁶β = cos2β(1- ¼ sin² 2β)
A) (sin²α-sin²β)/(sinα cosα-sinβcosβ) = tan(α+β)
B) (1-cos2α+sin2α) /(1+cos2α+sin2α) = tanα
C) cos³xcos3x+sin³xsin3x=cos³2x
D) {(cosα+sinα)/(cosα-sinα)} -
{(cosα-sinα)/(cosα+sinα)}=2tan2α
E) (sinα+sin2α)/(1+cosα+cos2α) = tanα
F*) (sinα/cosα) (1-cos2α)/(1-cos4α) = tanα
G) (cos³α+sin³α)/(cosα +sinα) =1 - ½ sin2α
H) 1/sin10 - √3/cos10 = 4
I) cotα/(cotα -cot3α) = tanα/(tan3α - tanα))
J) 1/(tan3α-tanα)- 1/(cot3α -cotα) = cot2α
K) sin8β=8sinβcosβcos2βcos4β
L) cos5θ = 16cos⁵θ - 20cos³θ+ 5cosθ
M) sin5α=16sin⁵α-20sin³α+5sinα
N) cos(120-α)+cosα+cos(120+α)=0
O) cos²(α-120) +cos²α +cos²(α+12) = 3/2
P) tan4α=(4tanα-4tan³α)/(1-6 tan²α+tan⁴α)
Q) (2cos2ⁿθ+1)/(2cosθ+1)=
(2cosθ-1)(2cos2θ-1)........ (2cos2ⁿ⁻¹θ-1)
R) tan2ⁿθ/tanθ = (1+sec2θ)(1+sec2²θ) ...(1+secⁿθ)
S) sin⁴α=⅜ - ½ cos2α +⅛ cos4α
T) cos⁸α+sin⁸α= 1-sin²2α+⅛ sin⁴2α
U) tan(π/4 +α)+tan(π/4 -α) =2sec2α
V) cos³α (sin3α)/3 +sin³α(cos³α)/3= (sin4α)/4
W) cos4α - cos4β= (cosα-cosβ)
(cosα+cosβ)(cosα-sinβ(cosα+sinβ)
X) tanθ+ 2tan2θ +4tan4θ +8tan8θ= cotθ
Y) (2cos8α+1)/(2cosα +1) =
(2cosα-1)(2cos2α-1)(2cos4α-1)
Z) cos²(α-β) - sin²(α+β) = cos2αcos2β
a') 4(cos³7+sin³23) =3(cos7 +sin23)
EXERCISE-2
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1) if tanx =b/a, find acos2x+bsin2x
2) If cosx +sinx =√2 cosy show that tan2y =1
3) If sin²x = siny cosy then show cos2x = 2cos(π/4 + y)
4) If cos2a=x+1/x show 2cosa =(x³+ 1/x³)
5) If tan²x 2tanx tan2y =
tan²y+2tanytan2x prove that each side =1 or tanx = ± tany
6) If tan²θ = 1+ 2 tan²α show cos2α = 1+ 2cos2θ
7) If 2 tanα = 3tanβ show tan(α-β) =sin2β/(5 -cos2β)
8) If tan(α+β-θ)/tan(α-β+θ) =
tanθ/tanβ then show that either
sin(β-θ)=0 or sin2α+sin2β+sin2θ=0
9) If α and β are acute angles and cos2α = (3cos2α-1)/(3-cos2β) then show that tanα =√2 tanβ
10) If cosα = ½(a +1/a) show that
a)cos2θ= ½(a²+ 1/a²)
b) cos3θ= ½ (a³ +1/a³)
11) If tanα = sec2α, then prove that
sin2α = (1-tan⁴α)/(1+tan⁴α)
12) If csc2α+csc2β +csc2θ = 0 show that tanα + tanβ+ tanθ + cotα + cotβ + cotθ=0
13) If tan³x=2tan²y+1 then show that cos²2x + sin²y = 0
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