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Trigonometry & Coordinate

Trigonometry And Coordinate

MODULES

Basics of Triangle
Trigonometric Formulae
Quadrants & Graph
Basics of Coordinate Geometry
Equation of a Line
Other Line Properties
Triangle, Quadrilateral & Circle
Past Questions

PRACTICE

Trigonometry & Coordinate : Level 1
Trigonometry & Coordinate : Level 2
Trigonometry & Coordinate : Level 3
ALL MODULES

CAT 2025 Lesson : Trigonometry & Coordinate - Trigonometric Formulae

bookmarked

2.2 Trigonometric Formulae

S.No. Formulae Expansion
1 sin2^{2}2 θ + cos2^{2}2 θ = 1
2 sin (A + B) sin A cos B + cos A sin B
3 sin (A – B) sin A cos B – cos A sin B
4 cos (A + B) cos A cos B – sin A sin B
5 cos (A – B) cos A cos B + sin A sin B
6 tan (A + B) tan A + tan B1−tan A tan B\dfrac{\mathrm{tan \ A \ + \ tan \ B}}{1 - \mathrm{tan \ A \ tan \ B}}1−tan A tan Btan A + tan B​
7 tan (A – B) tan A− tan B1+tan A tan B\dfrac{\mathrm{tan \ A - \ tan \ B}}{1 + \mathrm{tan \ A \ tan \ B}}1+tan A tan Btan A− tan B​
8 sin (A + B) + sin (A – B) 2 sin A cos B
9 sin (A + B) – sin (A – B) 2 cos A sin B
10 cos (A + B) + cos (A – B) 2 cos A cos B
11 cos (A – B) - cos (A + B) 2 sin A sin B
12 sin A + sin B 222 sin A + B2\dfrac{\mathrm{A \ + \ B}}{2}2A + B​ cos A − B2\dfrac{\mathrm{A \ - \ B}}{2}2A − B​
13 sin A – sin B 222 cos A + B2\dfrac{\mathrm{A \ + \ B}}{2}2A + B​ sin A − B2\dfrac{\mathrm{A \ - \ B}}{2}2A − B​
14 cos A + cos B 222 cos A + B2\dfrac{\mathrm{A \ + \ B}}{2}2A + B​ cos A − B2\dfrac{\mathrm{A \ - \ B}}{2}2A − B​
15 cos A – cos B −2-2−2 sin A + B2\dfrac{\mathrm{A \ + \ B}}{2}2A + B​ sin A − B2\dfrac{\mathrm{A \ - \ B}}{2}2A − B​

Additional formulae derived from the above:

S.No. Formulae Expansion
1 sin 2A 2 sin A cos A
2 cos 2A cos2^22 A - sin2^22 A
= 1 – 2 sin
2^22 A
= 2 cos
2^22 A - 1
3 tan 2A 2 tan A1−tan⁡2A\dfrac{2 \ \mathrm{tan \ A}}{1 - \tan^{2} \mathrm{A}}1−tan2A2 tan A​
4 sin 3A 3 sin⁡A−4 sin⁡3 A3 \ \sin \mathrm{A} - 4 \ \sin^{3} \ \mathrm{A}3 sinA−4 sin3 A
5 cos 3A 4 cos⁡3A−3 cos⁡A4 \ \cos^{3} \mathrm{A} - 3 \ \cos \mathrm{A}4 cos3A−3 cosA
6 tan 3A 3 tan⁡A−tan⁡3A1−3tan⁡3A\dfrac{3 \ \tan \mathrm{A} - \tan^{3} \mathrm{A}}{1 - 3 \tan^{3} \mathrm{A}}1−3tan3A3 tanA−tan3A​

Example 4

Find the following
(i) tan 75°75\degree75°       (ii) sin 15°15\degree15° + sin 75°75\degree75°

Solution

(i)
tan⁡75°=tan⁡(45°+30°)=tan⁡30°+tan⁡45°1−tan⁡30°tan⁡45°=(13)+11−(13)×1=3+13−1\tan75\degree = \tan(45\degree + 30\degree) = \dfrac{\tan30\degree + \tan45\degree}{1 - \tan30\degree \tan45\degree} = \dfrac{(\frac{1}{\sqrt{3}}) + 1}{1 - (\frac{1}{\sqrt{3}}) \times 1} = \dfrac{\sqrt{3} + 1}{\sqrt{3} - 1}tan75°=tan(45°+30°)=1−tan30°tan45°tan30°+tan45°​=1−(3​1​)×1(3​1​)+1​=3​−13​+1​

=
3+13−1×3+13+1=(3+1)2(3)2−12=4+232=2+3\dfrac{\sqrt{3} + 1}{\sqrt{3} - 1} \times \dfrac{\sqrt{3} + 1}{\sqrt{3} + 1} = \dfrac{(\sqrt{3} + 1)^{2}}{(\sqrt{3})^{2} - 1^{2}} = \dfrac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}3​−13​+1​×3​+13​+1​=(3​)2−12(3​+1)2​=24+23​​=2+3​

(ii)
sin⁡15°+sin⁡75°=sin⁡(45°−30°)+sin⁡(45°+30°)\sin 15\degree + \sin 75 \degree = \sin (45\degree - 30\degree) + \sin(45\degree + 30\degree)sin15°+sin75°=sin(45°−30°)+sin(45°+30°)

=
sin⁡45° cos⁡30°−cos⁡45° sin⁡30°+sin⁡45° cos⁡30°+cos⁡45°sin⁡30°\sin 45\degree \ \cos30\degree - \cos45\degree \ \sin 30\degree + \sin 45\degree \ \cos30\degree + \cos45\degree \sin 30\degreesin45° cos30°−cos45° sin30°+sin45° cos30°+cos45°sin30°

=
2(sin⁡45°cos⁡30°)=2(12×32)=32=322(\sin 45\degree \cos 30\degree) = 2 \left(\dfrac{1}{\sqrt{2}} \times \dfrac{\sqrt{3}}{2} \right) = \dfrac{\sqrt{3}}{\sqrt{2}} = \sqrt{\dfrac{3}{2}}2(sin45°cos30°)=2(2​1​×23​​)=2​3​​=23​​

Answer: (i)
2+32 + \sqrt{3}2+3​ (ii) 32\sqrt{\dfrac{3}{2}}23​​

Example 5

cosec⁡10°−3sec⁡10°=\cosec 10\degree - \sqrt{3}\sec 10\degree =cosec10°−3​sec10°=

(1) 1      (2) 2     (3)
3\sqrt{3}3​     (4) 4

Solution

cosec⁡10°−3sec⁡10°=1sin⁡10°−3cos⁡10°=cos⁡10°−3sin⁡10°sin⁡10°cos⁡10°\cosec 10\degree - \sqrt{3}\sec 10\degree = \dfrac{1}{\sin 10\degree} - \dfrac{\sqrt{3}}{\cos 10 \degree} = \dfrac{\cos 10\degree - \sqrt{3} \sin 10\degree}{\sin 10\degree \cos 10\degree}cosec10°−3​sec10°=sin10°1​−cos10°3​​=sin10°cos10°cos10°−3​sin10°​

=
2×(12×cos⁡10°−32×sin⁡10°)×22sin⁡10° cos⁡10°2 \times \left( \dfrac{1}{2} \times \cos10\degree - \dfrac{\sqrt{3}}{2} \times \sin10\degree \right) \times \dfrac{2}{2 \sin10\degree \ \cos10\degree}2×(21​×cos10°−23​​×sin10°)×2sin10° cos10°2​

=
2×(sin⁡30° cos⁡10°−cos⁡30° sin⁡10°)×2sin⁡20°2 \times (\sin 30\degree \ \cos10\degree - \cos 30\degree \ \sin10\degree) \times \dfrac{2}{\sin20\degree}2×(sin30° cos10°−cos30° sin10°)×sin20°2​

=
2×sin⁡(30°−10°)×2sin⁡20°=4×sin⁡20°sin⁡20°=42 \times \sin (30\degree - 10\degree) \times \dfrac{2}{\sin20\degree} = 4 \times \dfrac{\sin 20\degree}{\sin 20\degree} = 42×sin(30°−10°)×sin20°2​=4×sin20°sin20°​=4

Answer: (4) 4


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